Download to read Mike`s final report for his project.

INVESTIGATING PATTERNS IN THE FORMATION
OF WEAKLY COMPLETE NUMBER SEQUENCES
Michael D. Paul
Loyola College in Maryland, Mathematical Sciences ’10
Advisor
Dr. Michael Knapp
Department of Mathematical Sciences
Loyola College in Maryland
Sponsored by the Hauber Fellowship Fund
Loyola College in Maryland
Summer 2008
Abstract
This research focused on finding how to determine the terms of a
weakly complete sequence with a given set of initial terms. For
the sequences beginning with {1, n,...}, {n, n + 1,...}, or {n, 2n,...}
for most values of n, specific formulae to determine the next terms
of these sequences were found and proven. A conjecture formula
for the terms of the sequence beginning with {n, n + m} has also
been found, but has yet to be proven.
Background
A sequence of numbers is considered complete if every natural number can be written as
a sum of distinct terms of the sequence. Perhaps the most famous example of a complete
sequence is the Fibonacci sequence, {1, 1, 2, 3, 5, 8, 13, 21, 34,...}1. Any positive integer can be
obtained by adding any combination of numbers in this sequence; for example, 17 = 13 + 3 + 1
and 33 = 21 + 8 + 3 + 1. A sequence is considered to be weakly complete if every integer greater
than a given natural number (called the threshold of completeness) can be written as a sum of
distinct terms of the sequence2. In this paper, the term number will henceforth mean any natural
number.
Phase I: Investigating {1, n,...} Sequences
My research with Dr. Knapp this summer focused on finding ways to create a weakly
complete sequence beginning with a given set of natural numbers in ascending order, using the
last initial term as the threshold of completeness and having as few terms as possible. If we are
given the initial terms of the sequence, we can obtain the next term by examining every number
greater than the threshold of completeness in ascending order, and searching for the least number
that cannot be formed by adding any distinct previous terms of the sequence.
1
2
Ross Honsberger, Mathematical Gems III (Mathematical Association of America, 1985), 123.
Eric Schissel, “Characterizations of Three Types of Completeness,” Fibonacci Quarterly 27 (1989), 412-3.
To illustrate how a weakly complete sequence is formed, consider the sequence with
initial terms 1 and 5 (this was the first sequence which Dr. Knapp and I investigated). We know
that 6 is not in the sequence because 6 = 1 + 5, but 7 cannot be formed by adding any distinct
combination of 1 and 5, so 7 is the next term in the sequence, which is now S = {1, 5, 7}. 8 = 7
+ 1, so 8 is not in the sequence, but 9 cannot be formed from any combination of 1, 5, or 7, so we
add 9 to the sequence to get S = {1, 5, 7, 9}. 10 = 11 + 1, so 10 is not in the sequence, but 11
cannot be formed by adding any set of distinct elements in S, so 11 must be the next term in the
sequence, and we have S = {1, 5, 7, 9, 11}. 12 = 11 + 1 = 7 + 5, so 12 is not in the sequence, and
neither is 13 (because 13 = 7 + 5). Continuing on in this fashion, we now have the sequence
S = {1, 5, 7, 9, 11, 29, 31, 89, 91, 269, 271, 809, 811,...}
Note that this sequence consists of pairs of numbers that are two apart from each other. Starting
with 9 and 11, if one examines the number between each number in one of these pairs (10, 30,
90, 270, 810, etc.), one notices that each of these numbers is three times the one before it.
Therefore, each element of the sequence is equal to 10 * 3i ± 1 for any natural number i.
The first phase of our research was to investigate other sequences beginning with 1 and
another number n and determine if there were similar patterns in the terms of these sequences.
The sequences beginning with 1 and n for each n from 3 to 10 are shown in Figure 1. The terms
of sequence for beginning with 1 and 6 appear to be 8 followed by 11 * 3i ± 1 for i = 0, 1, 2,...
The sequence beginning with 1 and 7 has 9 as its next term, but the next terms are sets of three
numbers that are 2 apart from each other. Notice that the average of the terms in each set (i.e.,
the number around which set is centered) is 4 times the average of the terms of the previous set.
Each of these sets is equal to 13 * 4i plus or minus 2 or 0. Using mathematical notation, we can
write this sequence as

2
i 0
j 0
S  {1,7,9}   S i , where S i  {13  4 i  2  2 j} .
In each subsequent sequence that I investigated, I grouped the terms into sets of similar numbers,
found the number about which each of these set was centered and the range of the terms in each
set, and looked for patterns in these numbers. Through investigation, I found that the sequence
beginning with 1 and 8 is

2
S  {1,8,10}  14  4 i  2  2 j .
i 0 j 0
The sequence beginning with 1 and 9 is

3
S  {1,9,11}  {16  5i  3  2 j}
i 0 j 0
The sequence beginning with 1 and 10 is

3
S  {1,10,12}  {17  5i  3  2 j}
i 0 j 0
Using as a guide a proof of the conjecture for the {1, 5,...} sequence which Dr. Knapp
had already written beforehand3, I then wrote proofs of the conjectures for the {1, 6,...}, {1,
4,...}, {1, 7,...}, and {1, 9,...}sequences, given in figures in Figures 2 through 5.
Our next goal was to find a formula for the general formula for a sequence beginning
with {1, n,...} for any value of n. Finding the general formula consisted basically of looking at
each formula for a specific value of n and determining a pattern in each aspect of the formula. I
arrived at the conclusion that for the sequence beginning with {1, n,...} for any n ≥ 3, the
sequence is

b
i 0
j 0
S  {1, n, n  2}   S i , where S i  {si , j } ,
3
M. Knapp, unpublished proof.
 n  1
and si,j = axi – b + 2j, where x  
, a = n + x + 2, and b = x – 2.
 2 
One we had found the formula, our next step was to prove it. In writing this proof, I used
the previous proofs that I had written as a guide. The proof for the general case followed the
same line of reasoning as the previous proofs, except that some statements of the general proof
required more rigorous proof than their counterparts in the proofs for the more specific
examples, which one could verify by simple calculation. The actual proof for the general case is
given in Figure 6; the steps of the proof are outlined below:
1. Starting with 1 and n, prove that n + 2 and the elements of S0 are the next
terms of the sequence. Then we know that we can represent every number
from n to s0, b as a sum of distinct previous terms of the sequence.
2. Let Px  S 

S
i
, and assume that the sequence begins with Pk for some
i  x 1
k ≥ 0. We then need to show that the next terms of the sequence are the
elements of Sk+1.
3. First, we need to show that every number less than sk+1, 0 can be
represented as a sum of distinct terms of the sequence. Define the set R as
the set of all numbers represented as a sum of one or more elements
of Pk 1 . Then by definition of the sequence, we know that R contains 1
and every number from n to axk – b – 1 inclusive.
4. Let Cm be the set whose elements are the all the possible sums of terms of
Sk. We can represent every number of the same parity (even or odd) from
the minimum value of Cm to the maximum value of Cm. By adding 1 to
each element of Cm, we can represent every number from the minimum
value of Cm plus 1 to the maximum value of Cm plus 1. By combining
these two sets, we have representations for every number from the
minimum of Cm to the maximum value of Cm plus 1, and we will call this
interval Am.
5. We can also add each element cj of Cm to each of the remaining elements
of R, so we now have a representation for every number from cj + n to cj +
axk – b – 1. Once we show that the intervals of representation for
consecutive values of j intersect or are consecutive, we know that we can
represent every number from the smallest value of cj plus n to the largest
value cj plus axk – b – 1. (Note: To prove continuous representation
between two intervals of integers, we show that the difference when we
subtract the smallest element of one interval from the greatest element of
the other interval, the difference is non-negative (meaning the two
intervals intersect) or -1 (meaning the two intervals are disjoint but
consecutive).
6. Find the intervals of representation for A1 and B1, Ab and Bb, and Ab+1 and
Bb+1 (note that there are b + 1 terms in each Si). We can find
representations for any number between these intervals by adding one or
more elements of Sk to either the sum of all of the elements of Pk 1 or the
sum of all of the elements of Pk 1 except 1.
7. If n ≥ 9, then b ≥ 3, so there exists at least one integer between 1 and b – 1,
so we need to show that there is continuous representation from Am to Bm
for all values of m such that 2 ≤ m ≤ b – 1.
8. Prove that there is continuous representation from Am  Bm to Am1  Bm1
for all values of m such that 0 ≤ m < b. Then we have proven that we can
represent every integer from n to sk+1, 0.
9. We now need to show that the elements of Sk+1 cannot be represented as
sums of previous terms of the sequence. Note that because 1 is in the
sequence, once we show that an element of Sk+1 is in the sequence, we will
know that that element plus 1 is not in the sequence. We first show that
any representation of an element of Sk+1 cannot be represented using
previous elements of Sk+1.
10. Since each element of Sk+1 is larger than the sum of all the elements of
Pk 1 , we know that any representation of an element of Sk+1 must use at
least one element of Sk. We then show that unless m = b + 1 (i.e., the
number of terms in Sk), the difference between any element of Sk+1 and any
element of Cm is too large to be represented using only the elements of
Pk 1 .
11. We then show that if we subtract all of the elements of Sk+1, the difference
is an element of Sk, which by definition of the sequence cannot be
represented as a sum of elements of Pk 1 . Therefore, the elements of Sk+1
cannot be represented as sums of previous terms of S, so these elements
must be the next terms of S, and thus we have shown by induction that the
sequence follows the formula stated in the conjecture. QED.
Phase II: Developing a Program
The next phase of our research was to find formulas for weakly complete
sequences that began with any set of initial terms, such as {1, 2, n,...}, {2, n,...}, {n, n +
1}, and {n, 2n}. However, calculating these sequences by hand would be quite tedious.
Therefore, we needed a computer program that would generate sequences quickly and
efficiently. To calculate the {1,n,...} sequences (see Figure 1), Dr. Knapp had used a
program that he had created using the Maple software package, which he had to modify
manually every time he wanted to test a different sequence. Using the programming
skills that I had learned in my freshman computer science class, I created a Java version
of Dr. Knapp’s program, and modified it so that it could calculate any {1, n,...} sequence.
This was version 1 of the program, and the Java code for it is given in Figure 7. Version
2 of the program calculates the sequence beginning with any set of initial terms, and the
code for its most up-to-date incarnation, version 2.2, is given in Figure 8.
Unfortunately, version 2 had several serious flaws in the design. First, the user
had to indicate at the beginning of the program’s execution how many initial terms he/she
was entering. Secondly, the initial terms had to be entered in ascending order for the
program to run properly. Finally, every time the user wanted to change even one initial
term in the sequence, he would have to execute the program and re-enter every initial
term all over again. I concluded that the best way to resolve these issues would be to
create a version of the program that used a graphical user interface. The only problem
was that I did not yet know how to create graphics using Java. I solved this problem by
teaching myself using the Java All-in-One Desk Reference for Dummies by Doug Lowe
and Barry Burd (Wiley, 2007), and within a few weeks was able to create version 3, the
graphical version of the Sequence Program. The code for version 3.12 is given in Figure
9, and a screen shot of the program is shown in Figure 10. A description of the features
of the program follows:
•
Initial Terms List: Shows the numbers that the user has defined to be at the
beginning of the sequence.
•
Adding a Term: The user enters the term to be added to the list of initial terms in
the Next Term box and clicks the Add Term button. The program is set up so that
the user need not enter the initial terms in ascending order.
•
Changing an Initial Term: The user selects the term in the Initial Terms list that
he wants to change, enters the new value of the selected term into the Current
Term box, and clicks the Update Term button. The program automatically places
the new term in its appropriate place in the list.
•
Removing Terms from the List: The user selects the term that he wants to
remove from the list of initial terms and clicks the Remove Term button to
remove the term from the list. Additionally, the user can remove all of the terms
from the list by clicking the Clear All Initial Terms button.
•
Calculation Range: The program only displays the terms of the sequence that are
less than the number in this box.
•
Display Numbers with Multiple Representations: The user can toggle this
option on or off. This indicates if any number in the range of calculation can be
written as a sum of distinct previous terms of the sequence in more than one way.
The user can set minimum number of representations that a number must have to
be displayed.
•
Calculate Sequence: Calculates the sequence beginning with the numbers in the
Initial Terms list, and displays them in the text frame below. Each sequence that
the user generates is added to the text box after the previously calculated
sequences. There are also buttons available to clear the output display and to
print the outputs directly from the program.
Phase III: Investigating Other Weakly Complete Sequences
Now that I had a program to efficiently generate series of sequences, I could now
investigate different types of sequences. The first set of sequences that I looked at were
the sequences beginning with {1, 2, n,...}. These sequences are given in Figure 11.
Using the same investigation techniques that I had used earlier, I found that for n ≥ 5, the
sequence is

S = {1, 2, n, n + 4}   {S i } ,
i 0
x2
where Si =
{s
j 0
i, j
 n  3
} , and si,j = axi – 2 (x – 2) + 4j, where x = 
, and a = n + 2x + 4.
 4 
Though I have not yet written a proof of this conjecture, the techniques for proving it should be
similar to those used to prove the {1, n,...} formula.
Next, I generated sequences beginning with {2, n,...}, and these are shown in Figure 12.
Notice that while each of these sequences still has groups of similar numbers, there is an unusual
pattern in each of these subsets – some numbers are 1 apart from each other, some are 3 apart,
and if n is congruent to 3 modulo 4, there are some numbers that are 4 apart from each other. If
we calculate the averages of the terms in each subset of a sequence, we find that these averages
form a geometric sequence; i.e., there is a constant ratio of the average of one subset to the
average of the previous subset. The sequences seem to have the following pattern:

n2

i 0
j 0
y 0
S  {2, n, n  1}   S i , where S i  {si , j }  {si , 4 y  2 , si , 4 y 3 } , and si,,j = axi – b + j
However, I was unable to determine formulae for a, x, or b.
I then decided to investigate the {n, n + 1,...} sequences. These are shown in Figure 13.
Once again, I compared the averages of each subset of a sequence and found that they formed a
geometric sequence in which the ratio was equal to n. Using quadratic curve-fitting with the aid
of the MATLAB software package, I was able to determine a value for a in terms of n. The

n2
i 0
j 0
formula that I found was: S = {n, n + 1}   {S i } , where S i  {si , j } , and si , j  an i  b  j ,
where a 
3n  2
n2
and b 
. The proof is given in Figure 16.
2
2
Using similar techniques, I found the formulae for the following sequences:

n 1
i 1
j 0
 Start with n, n + 2: S = {n, n  2, n  3, n  4,...2n  1}   S i , where S i  {si , j }  {si ,1 } ,
and si , j  an i 
n n2
3n  n  6
. This works for n ≥ 2.
 j , where a 
2(n  1)
2(n  1)
2
2

n 1
i 1
j 0
 Start with n, n + 3: S = {n, n  3, n  4,...2n  2}   S i , where S i  {si , j }  {si , 2 } ,
and si , j  an i 
n2  n  4
3n 2  3n  10
. This works for n ≥ 4.
 j , where a 
2(n  1)
2(n  1)

n 1
i 1
j 0
 Start with n, n + 4: S = {n, n  4, n  5...2n  3}   S i , where S i  {si , j }  {si ,3 } ,
and si , j  an i 
n n6
3n  5n  14
. This works for n ≥ 5.
 j , where a 
2(n  1)
2(n  1)
2
2

n 1
i 1
j 0
 Start with n, n + 4: S = {n, n  5, n  6...2n  4}   S i , where S i  {si , j }  {si , 4 } ,
and si , j  an i 
n2  n  8
3n 2  7n  18
. This works for n ≥ 6.
 j , where a 
2(n  1)
2(n  1)
The sequences themselves are shown in Figures 15 through 18. Notice that these formulae seem
to hold only when n > m. When n < m or n = m, these formulae are no longer valid, and the
patterns when n < m and when n = m are different as well. With this in mind, I decided to
categorize all sequences that begin with n and n + m in terms of the value of n compared to m.
The three categories are alpha sequences (denoted by  (n, m) ), in which n > m; beta sequences
(denoted by  (n) ), in which n = m; and gamma sequences (denoted by  ( n, m) ), in which n <
m. After analyzing the patterns in formulae that I had obtained for specific alpha sequences, I
determined that the general formula for an alpha sequence is:

 (n, m)  {n, n  m, n  m  1, n  m  2,...,2n  m  1}   S i ,
i 1
n 1
where S i   si , j  si ,m1 , and si , j  an i  b  j ,
j 0
where a 
3n 2  (2m  3)n  4m  2
n 2  n  2(m  1)
, and b 
2(n  1)
2(n  1)
I also investigated the set of beta sequences (i.e., the sequences that start with {n, 2n}), which are
shown in Figure 19, and determined that the formula for a beta sequence is:

 (n)  {n,2n,2n  1,2n  2,...,3n  1,4n}   S i ,
i 1
n2
where S i   si , j , and si , j  an i  b  j , where a 
j 0
n2
5n 2  n  6
, and b 
2
2(n  1)
The final week of my research this summer was spent in writing the proof for this formula,
which is given in Figure 20.
Conclusions and Future Research
Throughout my investigation of weakly complete sequences, I found that the formula for
a particular sequence tends to follow a specific paradigm. A sequence beginning with {n, n + m}
starts with the initial terms, followed by an infinite series of subsets containing similar numbers.
Between the initial terms and the infinite series, there may exist other terms that do not belong to
a particular subset, and the values of these terms depend on n and m. Each subset Si of the
sequence is an indexed set of terms si, j. The range of Si and the terms that are omitted from it are
dependent upon the values of n and m. The formula si, j = axi – b + j seems to be consistent for
all sequences, but the values of a, x, and b are all determined by the values of n and m.
We have found and proven a formula for sequences beginning with {1, n,...}, and our
categorization of sequences with two given initial terms into alpha, beta, and gamma sequences
should set the tone for future research on weakly complete sequences. We have found and
proven a formula for all beta sequences, and we have a conjecture for the general formula for all
alpha sequences that needs only to be proven. However, though we did investigate some gamma
sequences and conjectured on parts of the formulae for some, the patterns in the formulae for
gamma sequences have so far eluded us. The subsets of these sequences increase in size as both
n and m increase, which makes many calculations involving these sequences rather tedious.
However, anyone with access to powerful computing machinery might be able to process these
sequences more efficiently and perhaps determine what the pattern is, if there even is one.
Finally, our research for this summer focused only on sequences that had two given
initial terms. Weakly complete sequences with three or more initial terms may also be of interest
to future researchers. The ultimate goal of future research on this topic is to find a single
unifying formula that can determine the pattern for any weakly complete sequence.
Figure 1. Sequences for 1, n for n from 1 to 10.
Start with 1, 3
Get 1, 3, 5, 7, 14, 28, 56, 112, 224, 448, 896, 1792, 3584, 7168, ...
Start with 1, 4
Get 1, 4, 6, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, ...
Start with 1, 5
Get 1, 5, 7, 9, 11, 29, 31, 89, 91, 269, 271, 809, 811, 2429, 2431, 7289, 7291, ...
Start with 1, 6
Get 1, 6, 8, 10, 12, 32, 34, 98, 100, 296, 298, 890, 892, 2672, 2674, 8018, 8020, ...
Start with 1, 7
Get 1, 7, 9, 11, 13, 15, 50, 52, 54, 206, 208, 210, 830, 832, 834, 3326, 3328, 3330, ...
Start with 1, 8
Get 1, 8, 10, 12, 14, 16, 54, 56, 58, 222, 224, 226, 894, 896, 898, 3582, 3584, 3586, ...
Start with 1, 9
Get 1, 9, 11, 13, 15, 17, 19, 77, 79, 81, 83, 397, 399, 401, 403, 1997, 1999, 2001, 2003, 9997,
9999, ...
Start with 1, 10
Get 1, 10, 12, 14, 16, 18, 20, 82, 84, 86, 88, 422, 424, 426, 428, 2122, 2124, 2126, 2128, ...
Figure 2. Proof for the {1, 6,...} Sequence
Theorem: Consider the sequence 1, 6, 8,..., which is defined by setting the first two terms equal
to 1 and 6, and then demanding that every integer n ≥ 7 must be a sum of distinct terms of the
sequence. Then this sequence is given by:
1, 6, 8, 10, 12, 32, 34, 98, 100, 296, 298...
i.e.,
1, 6, 8, 11 ∙ 30 ± 1, 11 ∙ 31 ± 1, 11 ∙ 32 ± 1, 11 ∙ 33 ± 1, ....
Proof: By definition, 1 and 6 are the first two terms of the sequence, and we can find that the
next three terms are 8, 10, and 12. Note that 10 = 11 ∙ 30 – 1, and 12 = 11 ∙ 30 + 1; this is our
base case for proof by induction. Now assume that the sequence begins with the terms
1, 6, 8, 11 ∙ 30 ± 1,..., 11 ∙ 3k ± 1,
where k ≥ 0. We now need to show that the next two terms of the sequence are 11 ∙ 3k+1 + 1 and
3k+1 – 1.
First, we need to show that every integer less than 11 ∙ 3k+1 – 1 can be represented as a
sum of distinct terms in the sequence. By definition of the sequence, we know that every integer
up to 11 ∙ 3k + 1 can be represented, so now we only need to prove that any integer n with
11 ∙ 3k + 2 ≤ n ≤ 11 ∙ 3k+1 – 2 can be represented. Let B be the set whose elements are the terms
of the sequence that are strictly less than 11 ∙ 3k – 1. Then by definition of the sequence, the
elements of B represent 1 and every integer between 6 and 11 ∙ 3k – 2 inclusive. By adding 11 ∙
3k – 1 to each of these numbers, we find that our sequence represents the number 11 ∙ 3k and all
numbers between 11 ∙ 3k + 5 and (11 ∙ 3k – 2) + (11 ∙ 3k – 1) = 2 ∙ 11 ∙ 3k – 3. Similarly, by
adding 11 ∙ 3k + 1 to each number represented by B, we find that we can represent 11 ∙ 3k + 2 and
all the numbers from 11 ∙ 3k + 7 to (11 ∙ 3k – 2) + (11 ∙ 3k + 1) = 2 ∙ 11 ∙ 3k – 1.
Now if we note that (11 ∙ 3k – 1) + (11 ∙ 3k + 1) = 2 ∙ 11 ∙ 3k, we find that by adding both
of these terms to each number represented by B, we can represent the number 2 ∙ 11 ∙ 3k + 1 and
all of the numbers from 2 ∙ 11 ∙ 3k + 6 to (11 ∙ 3k – 2) + (2 ∙ 11 ∙ 3k) = 3 ∙ 11 ∙ 3k – 2 = 11 ∙ 3k+1 – 2.
Putting all of this information together, we have representations for all numbers up to 11 ∙ 3k+1 –
2, with the possible exceptions of
11 ∙ 3k + 3
11 ∙ 3k + 4
2 ∙ 11 ∙ 3k + 2
2 ∙ 11 ∙ 3k + 4
2 ∙ 11 ∙ 3k + 5.
2 ∙ 11 ∙ 3k + 3
Note that
 b = 1 + 6 + 8 + (11 ∙ 30 – 1) + (11 ∙ 30 + 1) +...+ (11 ∙ 3k–1 – 1) + (11 ∙ 3k–1 + 1)
bB
= 15 + 2 ∙ 11 ∙
k 1
3
i
i 0
3k  1
2
= 11 + 4 + 11 ∙ (3k – 1)
= 4 + 11 ∙ (1 + 3k – 1)
= 11 ∙ 3k + 4.
= 15 + 2 ∙ 11 ∙
Now note that we have the following sums:
11 ∙ 3k + 3 =
b
bB {1}
11 ∙ 3k + 4 =
b
bB
2 ∙ 11 ∙ 3 + 2 = (11 ∙ 3k – 1) +
k
b
bB {1}
b
2 ∙ 11 ∙ 3k + 4 = (11 ∙ 3k + 1) +  b
2 ∙ 11 ∙ 3k + 3 = (11 ∙ 3k – 1) +
bB
bB {1}
2 ∙ 11 ∙ 3 + 5 = (11 ∙ 3 + 1) +  b .
k
k
bB
This shows that our sequence represents every number less than 11 ∙ 3k + 1 – 1.
Next, we need to show that 11 ∙ 3k + 1 – 1 is not represented by the terms of the sequence
up to 11 ∙ 3k + 1. First, note that since
11 ∙ 3k + 1 – 1 > 11 ∙ 3k + 4 =  b ,
bB
any such representation must use one or both of the numbers 11 ∙ 3k – 1 and 11 ∙ 3k + 1. If we try
to use only 11 ∙ 3k – 1, we find that the elements of B must represent
(11 ∙ 3k+1 – 1) – (11 ∙ 3k – 1) = 2 ∙ 11 ∙ 3k,
and this number is too large to be represented by only the elements of B. Similarly, if we try to
use only 11 ∙ 3k + 1, then the elements of B must represent
(11 ∙ 3k+1 – 1) – (11 ∙ 3k + 1) = 2 ∙ 11 ∙ 3k – 2,
and this is also too large to be represented with only the elements of B. If we try using both of
these elements, then the elements of B must represent
(11 ∙ 3k+1 – 1) – (11 ∙ 3k + 1) – (11 ∙ 3k - 1) = 11 ∙ 3k – 1.
But since 11 ∙ 3k – 1 is an element of the sequence, it cannot be represented as a sum of elements
of B. Therefore, 11 ∙ 3k+1 – 1 cannot be represented as a sum of previous terms in the sequence,
and must therefore be the next term.
Since 11 ∙ 3k+1 = (11 ∙ 3k+1 – 1) + 1, it is obvious that 11 ∙ 3k+1 is not in the sequence. All
that remains is to show that 11 ∙ 3k+1 + 1 cannot be represented as a sum of any of the previous
terms of the sequence. Since 11 ∙ 3k+1 + 1 = (11 ∙ 3k+1 – 1) + 2, and the number 2 is not in the
sequence, we know that we cannot use 11 ∙ 3k+1 – 1 in our representation. Following the same
reasoning that we used for 11 ∙ 3k+1 – 1, we find that any representation of 11 ∙ 3k+1 + 1 must use
both 11 ∙ 3k – 1 and 11 ∙ 3k + 1, and therefore the difference
(11 ∙ 3k+1 + 1) – [(11 ∙ 3k + 1) + (11 ∙ 3k - 1)] = 11 ∙ 3k + 1
would have to be represented using only the elements of B. Again, because 11 ∙ 3k + 1 is in the
sequence, we know by definition that it cannot be a sum of any distinct elements of B.
Therefore, 11 ∙ 3k + 1 is the next term in the sequence, and thus by induction we have established
the pattern for the sequence as stated in the theorem. QED.
Figure 3. Proof for {1, 4} Sequence
Theorem: Consider the sequence 1, 4, 6,..., which is defined by setting the first two terms equal to 1 and
4, and then demanding that every integer n ≥ 5 be a sum of distinct terms of the sequence. Then the
sequence is given by:
1, 4, 6, 8, 16, 32, 64, ...,
i.e.,
1, 4, 6, 8 ∙ 20, 8 ∙ 21, 8 ∙ 22, 8 ∙ 23, ....
Proof: By definition, 1 and 4 are the first two terms of the sequence. Then it is easy to find that the next
two terms are 6 and 8. Note that 8 = 8 ∙ 20; this is our base case for proof by induction. Now assume that
the sequence begins with the terms:
1, 4, 6, 8 ∙ 20, ..., 8 ∙ 2k,
where k ≥ 0. We need to show that the next term of the sequence is 8 ∙ 2k+1.
First, we need to show that every number less than 8 ∙ 2k+1 – 1 can be represented as a sum of
previous terms in the sequence. By definition of the sequence, we know that every integer up to 8 ∙ 2k can
be represented, so now we just need to prove that any integer n such that 8 ∙ 2k + 1 ≤ n ≤ 8 ∙ 2k+1 – 1 can be
represented. Let B be the set whose elements are the terms of the sequence strictly less than 8 ∙ 2k. Then
by definition of the sequence, the elements of B represent the number 1 and every integer from 4 to 8 ∙ 2k
– 1 inclusive. By adding 8 ∙ 2k to each of these numbers, we find that our sequence represents the number
8 ∙ 2k + 1 and all of the numbers from 8 ∙ 2k + 4 to (8 ∙ 2k – 1) + 8 ∙ 2k = 8 ∙ 2k+1 – 1. Now we have
representations for all numbers less than 8 ∙ 2k+1 except 8 ∙ 2k + 2 and 8 ∙ 2k + 3. Note that
b = 1 + 4 + 6 + 8 ∙ 2
0
+ 8 ∙ 21 + 8 ∙ 22 + ... + 8 ∙ 2k – 1
bB
k 1
= 11 + 8
2
i
i 0
= 3 + 8 + 8 (2k – 1)
= 3 + 8 (1 + 2k – 1)
= 8 ∙ 2k + 3.
Then we have 8 ∙ 2k + 3 =
 b and 8 ∙ 2
bB
k
+2=
b .
This shows that our sequence represents every
bB {1}
number less than 8 ∙ 2k+1.
Now we need to show that 8 ∙ 2k+1 cannot be represented by the terms of to the sequence up to 8 ∙
b , any representation of 8 ∙ 2k+1 must use the term 8 ∙ 2k. Then the
2k. First, note that since 8 ∙ 2k+1 >

bB
difference 8 ∙ 2k+1 – 8 ∙ 2k = 8 ∙ 2k must be represented by the elements of B. But this is impossible,
because 8 ∙ 2k is an element of the sequence, and thus by definition cannot be represented by the elements
of B. Therefore, 8 ∙ 2k+1 cannot be represented as a sum of previous terms in the sequence, and hence
must be the next term of the sequence. Thus by induction, we have established the pattern for the
sequence as stated in the theorem. QED.
Figure 4: Proof for {1, 7} Sequence
Theorem: Consider the sequence 1, 7, 9, ..., which is defined by setting the first two terms equal
to 1 and 7, and then demanding that every integer n ≥ 8 be a sum of distinct terms of the
sequence. Then this sequence is given by:
1, 7, 9, 11, 13, 15, 50, 52, 54, 206, 208, 210, ...,
i.e.,
1, 7, 9, 13 ∙ 40 + c, 13 ∙ 41 + c, 13 ∙ 42 + c, ...,
where c {-2, 0, 2}.
Proof: By definition, the first two terms of the sequence are 1 and 7, and we can then figure out
that the next four terms are 9, 11, 13, and 15. Note that 11 = 13 ∙ 40 – 2, 13 = 13 ∙ 40, and
15 = 13 ∙ 40 + 2. Now assume that the sequence begins with the terms
1, 7, 9, 13 ∙ 40 + c, ..., 13 ∙ 4k + c,
where k ≥ 0. We need to show that the next three terms of the sequence are 13 ∙ 4k+1 – 2, 13 ∙
4k+1, and 13 ∙ 4k+1 + 2.
First, we need to show that every integer less than 13 ∙ 4k+1 – 2 can be represented as a
sum of distinct terms in the sequence. By definition of the sequence, we know that every integer
up to 13 ∙ 4k + 2 can be represented, so we just need to prove that every integer n for
13 ∙ 4k + 3 ≤ n ≤ 13 ∙ 4k+1 – 3 inclusive has a representation. Let B be the set whose elements are
the terms of the sequence strictly less than 13 ∙ 4k – 2. Then by definition of the sequence, the
elements of B represent 1 and every integer between 7 and 13 ∙ 4k – 3 inclusive, and we will
define the set P (B) to be the set whose elements are these numbers. By adding 13 ∙ 4k – 2 to
each of these numbers, we can represent the number 13 ∙ 4k – 1 and every number from 13 ∙ 4k +
5 to (13 ∙ 4k – 3) + (13 ∙ 4k – 2) = 2 ∙ 13 ∙ 4k – 5. Similarly, by adding 13 ∙ 4k to each element of P
(B), we can represent 13 ∙ 4k + 1 and every integer from 13 ∙ 4k + 7 to (13 ∙ 4k – 3) + 13 ∙ 4k = 2 ∙
13 ∙ 4k – 3. We can also add 13 ∙ 4k + 2 to each element of P(B) and find that we can represent
13 ∙ 4k + 3 and every number from 13 ∙ 4k + 9 to (13 ∙ 4k – 3) + (13 ∙ 4k + 2) = 2 ∙ 13 ∙ 4k – 1.
Now if we note that (13 ∙ 4k – 2) + 13 ∙ 4k = 2 ∙ 13 ∙ 4k – 2, we see that by adding both of
these terms to the elements of P(B), we can represent 2 ∙ 13 ∙ 4k – 1 and all numbers from 2 ∙ 13 ∙
4k + 5 to (13 ∙ 4k – 3) + (2 ∙ 13 ∙ 4k – 2) = 3 ∙ 13 ∙ 4k – 5. Similarly, we can add (13 ∙ 4k – 2) + (13
∙ 4k + 2) = 2 ∙ 13 ∙ 4k to each element of P(B) and thus represent 2 ∙ 13 ∙ 4k + 1 and every number
from 2 ∙ 13 ∙ 4k + 7 to (13 ∙ 4k – 3) + (2 ∙ 13 ∙ 4k) = 3 ∙ 13 ∙ 4k – 3, and we can add (13 ∙ 4k) + (13 ∙
4k + 2) = 2 ∙ 13 ∙ 4k + 2 to each element of P(B) and represent 2 ∙ 13 ∙ 4k + 3 and every number
from 2 ∙ 13 ∙ 4k + 9 to (13 ∙ 4k – 3) + (2 ∙ 13 ∙ 4k + 2) = 3 ∙ 13 ∙ 4k – 1. Finally, if we note that (13 ∙
4k – 2) + (13 ∙ 4k) + (13 ∙ 4k + 2) = 3 ∙ 13 ∙ 4k, we can add all three terms to the elements of P(B)
and represent 3 ∙ 13 ∙ 4k + 1 and every number from 3 ∙ 13 ∙ 4k + 7 to (13 ∙ 4k – 3) + (3 ∙ 13 ∙ 4k) =
4 ∙ 13 ∙ 4k – 3 = 13 ∙ 4k+1 – 3. Putting all of this information together, we find that we can
represent all numbers up to 13 ∙ 4k+1 – 3, with the possible exceptions of 13 ∙ 4k + 4, 2 ∙ 13 ∙ 4k +
6
4, and
 (3  13  4
k
i).
i 2
Note that
 b = 1 + 7 + 9 + (13 ∙ 40 – 2) + 13 ∙ 40 + (13 ∙ 40) +...+ (13 ∙ 4k–1 – 2) + 13 ∙ 4k–1 + (13 ∙ 4k–1 + 2)
bB
= 17 + 3 ∙ 13
k 1
4
i
i 0
4k 1
3
k
= 4 + 13 + 13 (4 – 1)
= 4 + 13 (1 + 4k – 1)
= 13 ∙ 4k + 4.
= 17 + 3 ∙ 13 ∙
Now note that we have the following sums:
13 ∙ 4k + 4 =
b
bB
2 ∙ 13 ∙ 4 + 4 = (13 ∙ 4k) +
k
b
bB
3 ∙ 13 ∙ 4 + 2 = (13 ∙ 4 – 2) + (13 ∙ 4k) +
k
k
b
bB
b
3 ∙ 13 ∙ 4 + 3 = (13 ∙ 4 – 2) + (13 ∙ 4 + 2) +
k
k
k
bB {1}
b
b
3 ∙ 13 ∙ 4k + 4 = (13 ∙ 4k – 2) + (13 ∙ 4k + 2) +
bB
3 ∙ 13 ∙ 4 + 5 = (13 ∙ 4 ) + (13 ∙ 4 + 2) +
k
k
k
bB {1}
3 ∙ 13 ∙ 4 + 6 = (13 ∙ 4 ) + (13 ∙ 4 + 2) +  b .
k
k
k
bB
This shows that our sequence represents every number less than 13 ∙ 4k+1 – 2.
Next, we need to show that 13 ∙ 4k+1 – 2 is not represented by the terms of the sequence
up to 13 ∙ 4k + 2. First, note that since
13 ∙ 4k+1 – 2 > 13 ∙ 4k + 4 =  b ,
bB
any such representation must one or more of the numbers 13 ∙ 4k – 2, 13 ∙ 4k, or 13 ∙ 4k + 2. If we
try to use only 13 ∙ 4k + 2, the largest of these, we find that the elements of B must represent
(13 ∙ 4k+1 – 2) – (13 ∙ 4k + 2) = 3 ∙ 13 ∙ 4k – 4,
and this number is too large to be represented by only the elements of B, and if we try to use
either of the smaller elements by itself, the difference will also be too large. The largest sum that
we can make with a pair of the above elements is (13 ∙ 4k) + (13 ∙ 4k + 2) = 2 ∙ 13 ∙ 4k + 2, so if
we use both of these elements in our representation, the difference
(13 ∙ 4k+1 – 2) – (2 ∙ 13 ∙ 4k + 2) = 2 ∙ 13 ∙ 4k – 4
is once again too large to be represented using only the elements of B. Our last option is to try
using all three of the above elements, and we find that the elements of B must represent
(13 ∙ 4k+1 – 2) – [(13 ∙ 4k – 2) + (13 ∙ 4k) + (13 ∙ 4k + 2)] = 13 ∙ 4k – 2.
But since 13 ∙ 4k – 2 is an element of the sequence, we know by definition that it cannot be
represented as sum of the elements of B. Therefore, 13 ∙ 4k+1 – 2 cannot be represented as a sum
of previous terms in the sequence, and must therefore be the next term.
Since 13 ∙ 4k+1 – 1 = (13 ∙ 4k+1 – 2) + 1, it is obvious that 13 ∙ 4k+1 is not in the sequence.
Our next step is to show that 13 ∙ 4k+1 cannot be represented as a sum of any of the previous
terms in the sequence. Since 13 ∙ 4k+1 = (13 ∙ 4k+1 – 2) + 2, and the number 2 is not in the
sequence, we cannot use 13 ∙ 4k+1 – 2 in our representation. Following the same reasoning that
we used for 13 ∙ 4k+1 – 2, we find that any representation of 13 ∙ 4k+1 must use 13 ∙ 4k – 2, 13 ∙ 4k,
and 13 ∙ 4k + 2, and therefore the difference
(13 ∙ 4k+1) – [(13 ∙ 4k – 2) + (13 ∙ 4k) + (13 ∙ 4k + 2)] = 13 ∙ 4k
would have to be represented using only the elements of B. But this is impossible, since 13 ∙ 4k
is in the sequence, and thus by definition cannot be represented by the elements of B. Hence, we
cannot represent 13 ∙ 4k+1 as a sum of previous terms of the sequence, so this must be the next
term in the sequence.
Since 13 ∙ 4k+1 + 1 = (13 ∙ 4k+1) + 1, we know that 13 ∙ 4k+1 + 1 is not in the sequence. Our
final step here is to prove that 13 ∙ 4k+1 + 2 cannot be represented as a sum of previous terms of
the sequence. Since 13 ∙ 4k+1 + 2 = (13 ∙ 4k+1 – 2) + 4 = (13 ∙ 4k+1) + 2, and the numbers 2 and 4
are not in the sequence, we cannot use 13 ∙ 4k – 2 or 13 ∙ 4k+1 in our representation of 13 ∙ 4k+1 +
2. Using the same reasoning as before, we know our representation must use 13 ∙ 4k – 2, 13 ∙ 4k,
and 13 ∙ 4k + 2. The difference
(13 ∙ 4k+1 + 2) – [(13 ∙ 4k – 2) + (13 ∙ 4k) + (13 ∙ 4k + 2)] = 13 ∙ 4k + 2
is already in the sequence, so it cannot be represented as a sum of the elements of B. Therefore,
13 ∙ 4k+1 + 2 cannot be represented as a sum of previous terms, and must therefore be the next
term in the sequence. Thus we have shown by induction that the sequence follows the pattern
given in the theorem. QED.
Figure 5. Proof for {1, 9} Sequence
Theorem: Define the sequence S = {1, 9, 11, ...} by setting the first two terms equal to 1 and 9,
and then demanding that every integer x ≥ 10 be a sum of distinct terms of S. Then this sequence
is given by:
S = {1, 9, 11, 13, 15, 17, 19, 77, 79, 81, 83, 397, 399, 401, 403, 1997, 1999, 2001, 2003, ...}

= {1, 9, 11}   {S i } ,
i 0
3
where Si =
{s } , sj = 16  5  3  2 j .
i
j
j 0
Proof: By definition, 1 and 9 are the first two terms of the sequence, and we can then figure out
that the next five terms of the sequence are 11, 13, 15, 17, and 19. Note that {13, 15, 17, 19} =
S0. Define the set
Px  S 

S
i
,
i  x 1
and assume that the sequence begins with Pk. We need to show the next terms of the sequence
are the elements of Sk+1.
First, we need to show that every integer less than 16 ∙ 5k+1 – 3 can be represented as a
sum of distinct terms in the sequence. By definition, we know that every integer up to 16 ∙ 5k + 3
can be represented, so we just need to prove that every integer x for 16 ∙ 5k + 4 ≤ x ≤ 16 ∙ 5k+1 – 4
inclusive has a representation. Define the set R (f) as the set of all integers that can be
represented as a sum of distinct elements of the set f. Then by definition of the sequence, we
have
R( Pk 1 )  {1}  {x : x  [9, 16 ∙ 5k – 4]}.
We will now attempt to represent other numbers by adding one or more elements of Sk to each of
the elements of R (Pk–1). We’ll start by adding one element of Sk to each element of R (Pk–1).
Since
3
S k  {16  5 k  3  2 j} ,
j 0
by adding 1 to each element of Sk, we can represent
3
{s
j 0
3
j
 1}  {16  5 k  2  2 j} = {16 ∙ 5k – 2, 16 ∙ 5k, 16 ∙ 5k + 2, 16 ∙ 5k + 4}.
j 0
The smallest element of Sk is 16 ∙ 5k – 3, and when we add this term to each of the remaining
elements of R (Pk–1), we find that we can represent every number from (16 ∙ 5k – 3) + 9 = 16 ∙ 5k
+ 6 to (16 ∙ 5k – 3) + (16 ∙ 5k – 4) = 2 ∙ 16 ∙ 5k – 7. Similarly, by adding 16 ∙ 5k + 3, the largest
element of Sk, to each of the elements, we find that we can represent every number from 16 ∙ 5k +
12 to (16 ∙ 5k + 3) + (16 ∙ 5k – 4) = 2 ∙ 16 ∙ 5k – 1.
Next we will add pairs of elements of Sk to each element of R (Pk–1). We know by
definition of the sequence that the sum of any two elements of Sk is equal to (16 ∙ 5k – 3 + 2 j1) +
(16 ∙ 5k – 3 + 2 j2) = 2 ∙ 16 ∙ 5k – 6 + 2 (j1 + j2), where j1, j2  J = {0, 1, 2, 3} and j1  j2 . If we
let j1 = 0, we can represent the sum j1 + j2 as 1, 2, or 3 if we let j2 equal any of the remaining
numbers in J. Similarly, if we let j1 = 3 and let j2 equal any of the remaining numbers in J, we
can represent 3, 4, or 5 as the sum j1 + j2. Thus we have shown that j1 + j2 can equal any integer
from 1 to 5 inclusive, and by substitution, we find that the sum of any two elements of Sk is equal
to 2 ∙ 16 ∙ 5k plus any even integer from -4 to 4 inclusive. For the sake of simplicity, we will call
the set of all of these possible sums C2. Then by adding 1 to each element of C2, we can
represent 2 ∙ 16 ∙ 5k plus any odd integer from -3 to 5 inclusive. By adding 2 ∙ 16 ∙ 5k – 4, the
smallest element of C2, to each of the remaining elements of R (Pk–1), we can represent every
number from (2 ∙ 16 ∙ 5k – 4) + 9 = 2 ∙ 16 ∙ 5k + 5 to (2 ∙ 16 ∙ 5k – 4) + (16 ∙ 5k – 4) = 3 ∙ 16 ∙ 5k –
8. Similarly, by adding 2 ∙ 16 ∙ 5k + 4, the largest element of C2, to each remaining element of R
(Pk–1), we can represent every number from 2 ∙ 16 ∙ 5k + 13 to (2 ∙ 16 ∙ 5k + 4) + (16 ∙ 5k – 4) = (3
∙ 16 ∙ 5k.
Next, we will add sets of three elements of Sk to each element of R (Pk–1). By definition,
we know that the sum of any three terms of Sk is equal to (16 ∙ 5k – 3 + 2 j1) + (16 ∙ 5k – 3 + 2 j2)
+ (16 ∙ 5k – 3 + 2 j3) = 3 ∙ 16 – 9 + 2 (j1 + j2 + j3), for distinct j1, j2, j3  {0, 1, 2, 3}. If we let j1 =
0, j2 = 1, and j3 = 2 or 3, we find that j1 + j2 + j3 = 3 or 4. If we let j1 = 3, j2 = 2, and j3 = 0 or 1,
we find that j1 + j2 + j3 = 5 or 6. So j1 + j2 + j3 is equal to 3, 4, 5, or 6. Then by substitution, we
find that the sum of any three terms of Sk is equal to 3 ∙ 16 ∙ 5k plus any odd integer from -3 to 3
inclusive, and we will call the set of all these possible sums C3. If we add 1 to each element of
C3, we can represent 3 ∙ 16 ∙ 5k plus every even integer from -2 to 4 inclusive. By adding 3 ∙ 16 ∙
5k – 3, the smallest element of C3, to each remaining element of R (Pk–1), we have a
representation for every number from (3 ∙ 16 ∙ 5k – 3) + 9 = 3 ∙ 16 ∙ 5k + 6 to (3 ∙ 16 ∙ 5k – 3) + (16
∙ 5k – 4) = 4 ∙ 16 ∙ 5k – 7. Similarly, by adding 3 ∙ 16 ∙ 5k + 3, the largest element of C3, to each of
the remaining elements of R (Pk–1), we have a representation for 3 ∙ 16 ∙ 5k + 12 to (3 ∙ 16 ∙ 5k +
3) + (16 ∙ 5k – 4) = 4 ∙ 16 ∙ 5k – 1.
Finally, if we add all four terms (the sum of which is 4 ∙ 16 ∙ 5k) to each element of R (Pk–
k
k
k
k
1), we can represent 4 ∙ 16 ∙ 5 + 1 and all numbers from 4 ∙ 16 ∙ 5 + 9 to (4 ∙ 16 ∙ 5 ) + (16 ∙ 5 –
k
k+1
4) = 5 ∙ 16 ∙ 5 – 4 = 16 ∙ 5 – 4. Putting all of the above information together, we have
representations for every number less than 16 ∙ 5k+1 – 3 except 16 ∙ 5k + 5, 3 ∙ 16 ∙ 5k + 5, and
8
 4  16  5
k
i.
i 2
Note that
 p = 1 + 9 + 11 + 4 ∙ 16 ∙ 50 + 4 ∙ 16 ∙ 51 + ... + 4 ∙ 16 ∙ 5k–1
pPk 1
k 1
= 21 + 4 ∙ 16
5
i
i 0
5k  1
= 5 + 16 + 4 ∙ 16 ∙
4
= 5 + 16 + 16 (5k – 1)
= 5 + 16 (1 + 5k – 1)
= 16 ∙ 5k + 5.
Now note that we have the following sums:
16 ∙ 5k + 5 =  p
pPk 1
3 ∙ 16 ∙ 5 + 5 = 2 ∙ 16 ∙ 5k +
k
p
pPk 1
4 ∙ 16 ∙ 5k + 2 = 3 ∙ 16 ∙ 5k – 3 +
p
pPk 1
4 ∙ 16 ∙ 5k + 3 = 3 ∙ 16 ∙ 5k – 1 +
p
pPk 1 {1}
4 ∙ 16 ∙ 5k + 4 = 3 ∙ 16 ∙ 5k – 1 +
p
pPk 1
4 ∙ 16 ∙ 5k + 5 = 3 ∙ 16 ∙ 5k + 1 +
p
pPk 1 {1}
4 ∙ 16 ∙ 5k + 6 = 3 ∙ 16 ∙ 5k + 1 +
p
pPk 1
4 ∙ 16 ∙ 5k + 7 = 3 ∙ 16 ∙ 5k + 3 +
p
pPk 1 {1}
4 ∙ 16 ∙ 5k + 8 = 3 ∙ 16 ∙ 5k + 3 +
 p.
pPk 1
This shows that our sequence represents every number less than 16 ∙ 5k+1 – 4.
Our next step is to show that each element of Sk+1 cannot be represented by the terms of
the sequence up to 16 ∙ 5k + 4. For each element sj  Sk+1 that we can prove is in the sequence,
we can determine immediately that sj + 1 is not in the sequence. Note that since 16 ∙ 5k+1 – 3, the
smallest element of Sk+1, is greater than  p , any representation of an element of Sk+1 must use
pPk 1
at least one element of Sk. If we use the smallest element of Sk+1, 16 ∙ 5k+1 – 3, and the greatest
element of C3, 3 ∙ 16 ∙ 5k + 3, we find that the difference
(16 ∙ 5k+1 – 3) – (3 ∙ 16 ∙ 5k + 3) = 2 ∙ 16 ∙ 5k – 6
is too large to represent using only the elements of Pk–1, so if we try using a greater element of
Sk+1 and / or a smaller combination of elements of Sk, the difference will still be too large to be
represented using only the elements of Pk–1. Therefore, our only option for any element of Sk+1 is
to use all four elements of Sk in our representation. Then, for the jth element of Sk+1, our
difference is
(16 ∙ 5k+1 – 3 + 2 j) – (4 ∙ 16 ∙ 5k) = 16 ∙ 5k – 3 + 2 j
would have to be represented by the elements of Pk–1. But this is impossible, because (16 ∙ 5k – 3
+ 2 j)  Sk, and by definition cannot be represented as a sum of the elements of Pk–1. So we
know right away that 16 ∙ 5k+1 – 3 cannot be represented as a sum of previous terms of the
sequence, and must therefore be the next term. Since each element sj of Sk+1 is equal to 16 ∙ 5k+1
– 3 + 2 j for j  {0, 1, 2, 3}, then the difference sj – s1 = (16 ∙ 5k+1 – 3 + 2 j) – (16 ∙ 5k+1 – 3) = 2 j.
Because 0 ≤ j ≤ 3, we have 0 ≤ 2 j ≤ 6. With the exception of 1, no number in this interval is in
1
the sequence. However, the difference can’t be 1, because that would imply that j = , which is
2
k+1
not an integer. Therefore, the difference between any element of Sk+1 and 16 ∙ 5 – 3 is not in
the sequence, and thus we cannot use 16 ∙ 5k+1 – 3 in a representation of any element of Sk+1.
Therefore, each element of Sk+1 cannot be written as a sum of previous terms of the sequence,
and is therefore in the sequence. Thus we have shown by induction that the sequence follows the
formula stated in the theorem. QED.
Figure 6. General Proof for {1, n,...} Sequence
Theorem: Define the sequence S = {1, n, ...} by setting the first two terms equal to 1 and n ≥ 4,
and then demanding that every integer greater than n be a sum of distinct terms of S. Then this
sequence is given by

S = {1, n, n + 2}   {S i } ,
i 0
 n  1
where Si = {s j } , and sj = axi – b + 2j, where x = 
, a = n + x + 2, and b = x – 2.
 2 
j 0
b
Proof: By definition, 1 and n are the first two terms in the sequence, and then it is obvious that n
+ 1 is not in the sequence, but n + 2 is; and since n + 3 = (n + 2) + 1, we also know that n + 3 is
not in the sequence. Our first step is to prove that the elements of
b
S0 =
{a  b  2 j}
j 0
are the next terms of S, i.e., that each element of S0 cannot be represented as a sum of the
previous terms of S. We know that for each term sj  S0 that we can prove is in the sequence, sj
+ 1 is not in the sequence. For any two distinct elements of S0, the difference is
s j 2  s j1  (a  b  2 j 2 )  (a  b  2 j1 )  2( j 2  j1 ) ,
where 0 ≤ j1 < j2 ≤ b. Then by algebra, we have 0 < j2 – j1 ≤ b – j1. Since b = x – 2 and x ≤
we then have
0  j 2  j1  b  j1  b  x  2 
n 1
,
2
n 1
n3 n
2
 ,
2
2
2
and then by algebra, we have 0 < 2 (j2 – j1) < n. So the difference of any two elements of S0 must
be strictly between 0 and n. The only number in this interval that is represented by the elements
of S is 1. But because j1 and j2 are integers, their difference must also be an integer, and
therefore the difference of any two elements of S0 must be even, so the difference cannot be 1.
Therefore, we can’t use any elements of S0 to represent sj  S0, so we must use at least one
element of {1, n, n + 2}. The smallest element of S0 is s0 = a – b = n + x + 2 – (x – 2) = n + 4,
and the largest element of S0 is sb = a – b + 2 b = a + b = n + x + 2 + (x – 2) = n + 2 x. Since 1 +
(n + 2) = n + 3, this is not enough to represent any element of S0. The next smallest sum we can
use is n + (n + 2) = 2n + 2. If n is even, then n = 2y for some integer y, and by definition
1
 2 y  1 
x
 y    y.

2
 2  
Then by substitution, n + 2x = n + 2y = n + n = 2n < 2n + 2. If n is odd, then n = 2y + 1 for some
integer y, and then
2y  2
x
  y  1  y  1 .
 2 
Then by substitution, n + 2x = n + 2 (y + 1) = n + 2y + 2 = n + (2y + 1) + 1 = 2n + 1 < 2n + 2.
Either way, sb = n + 2x < n + (n + 2), and so the sum of n and n + 2 is too large to represent any
element of S0. Therefore, the elements of S0 cannot be represented as sums of previous terms of
S, and so these must be the next terms of the sequence.
Now define the set Px  S 

S
i
, and assume that the sequence begins with Pk. We
i  x 1
now need to show that the next terms of the sequence are the elements of Sk+1.
First, we need to show that every integer less than the smallest element of Sk+1 (which is
ax – b) can be represented as a sum of distinct terms in the sequence. By definition, we know
that every integer up to the largest element of Sk (which is axk + b) inclusive can be represented,
so we just need to prove that every integer z for axk + b + 1 ≤ z ≤ axk+1 – b – 1 has a
representation. Define the set R (f) as the set of all integers that can be represented as a sum of
distinct elements of the set f. Then by definition of the sequence, we have
k+1
R (Pk–1)  {1}  {z: z  [n, axk – b – 1]}.
We will now attempt to represent other numbers by adding one or more elements of Sk to each of
the elements of R (Pk–1). Let Cm = {c j } be the set whose elements are all of the possible sums
of m distinct terms of Sk. If we add m terms of Sk, we get maxk – mb + 2 (j1 + j2 + ... + jm) for
some distinct ji  [0, b]. The smallest element of Cm is
maxk – mb + 2 (0 + 1 + 2 + ... m – 1) = maxk – mb + 2 ∙
m( m  1)
2
= maxk – mb + m2 – m,
and the largest term of Cm is
maxk – mb + 2 [b + (b – 1) + (b – 2) + ... + (b – [m – 1])] = maxk – mb + 2 [mb –
m( m  1)
]
2
= maxk – mb + 2mb – m (m – 1)
= maxk + mb – m2 + m.
m
Since we can represent every integer from the smallest possible value of
j
i 1
i
to the largest
m
possible value of  ji , we know that the set Cm represents every integer of the same parity from
i 1
maxk – mb + m2 – m to maxk + mb – m2 + m. By adding 1 (the first element of the sequence) to
each element of Cm, we can also represent every integer of the same parity from maxk – mb + m2
– m + 1 to maxk + mb – m2 + m + 1. Putting these two intervals together, we have a
representation for every number from maxk – mb + m2 – m to maxk + mb – m2 + m + 1, and we
will call this interval Am. We can also add each element c j  C m to each of the remaining
elements of R (Pk–1) to represent each number from cj + n to cj + axk – b – 1. For every cj, we
have cj+1 = cj + 2, and then we have
(cj + axk – b – 1) – (cj+1 + n) = cj + axk – b – 1 – (cj + 2) – n
= cj + (n + x + 2) xk – x + 2 – 1 – cj – 2 – n
= nxk + xk+1 + 2xk – x – 1 – n
= n (xk – 1) + x (xk – 1) + 2xk – 1
= (n + x) (xk – 1) + 2xk – 1.
We know that n + x > 0, and that xk – 1 and 2xk will both increase as k increases. If k = 0, then
we have (n + x) (1 – 1) + 2 – 1 = 1 > 0. Therefore, the difference (cj + axk – b – 1) – (cj+1 + n) is
always positive, so the interval of representation for each cj intersects the intervals of
representation for both the next and previous elements of Cm, so we can combine all of these
intervals and find that we have a representation for every number from maxk – mb + m2 – m + n
to (maxk + mb – m2 + m) + (axk – b – 1) = (m + 1) axk + (m – 1) b – m2 + m – 1, and we’ll call this
interval Bm.
If we add only one element of Sk (i.e., the elements of C1) to each element of R (Pk–1), by
the formulae proven above, we have representations for every number from axk – b to axk + b + 1
and from axk – b + n to 2axk – 1. Note that axk + b + 1 = axk + x – 1 and that axk – b + n = axk – x
+ 2 + n. If n is even, then we have
axk – b + n = axk – y + 2 + 2y
= axk + y + 2
= axk + x + 2.
If n is odd, then we have
axk – b + n = axk – (y + 1) + 2 + 2y + 1
= axk + y + 2
= axk + x + 1.
Putting this information together, we find that we are missing a representation for axk + x, and if
n is even, we are also missing a representation for axk + x + 1. Note that
b
 s   (ax
sSi
i
 b  2 j)
j 0
= (b + 1) (axi – b) + 2
b
j
j 0
b(b  1)
2
i
= (b + 1) (ax – b) + b (b + 1)
= (b + 1) (axi – b) + 2 ∙
= (b + 1) (axi – b + b)
= (b + 1) (axi)
= (x – 1) (axi),
and then
 p = 1 + n + (n + 2) +
pPk 1
= 2n + 3 + a (x – 1)
k 1
s
i  0 sS i
k 1
x
i
i 0
 xk 1

= 2n + 3 + a (x – 1) 
 x 1 
= 2n + 3 + a (xk – 1)
= 2n + 3 + (n + x + 2) (xk – 1).
If n is odd, then we have
 p = 2 (2y + 1) + 3 + (2y + 1 + y + 1 + 2) (xk – 1)
pPk 1
= 4y + 5 + (3y + 4) (xk – 1)
= y + 1 + (3y + 4) + (3y + 4) (xk – 1)
= y + 1 + (3y + 4) (1 + xk – 1)
= axk + x.
If n is even, then we have
 p = 2 (2y) + 3 + (2y + y + 2) (xk – 1)
pPk 1
= 4y + 3 + (3y + 2) (xk – 1)
= y + 1 + (3y + 2) + (3y + 2) (xk – 1)
= y + 1 + (3y + 2) (1 + xk – 1)
= axk + x + 1.
Now if n is odd, we have axk + x =
=
 p ; if n is even, we have axk + x =  p and axk + x + 1
pPk 1
 p.
pPk 1 {1}
pPk 1
Now if we add b terms of Sk, we have
Ab = [(maxk – mb + m2 – m), (maxk + mb – m2 + m + 1)]
= [(baxk – b2 + b2 – b), (baxk + b2 – b2 + b + 1)]
= [(baxk – x + 2), (baxk + x – 1)]
and
Bb = [(maxk – mb + m2 – m + n), ([m+1] axk + [m – 1] b – m2 + m – 1)]
= [(baxk – b2 + b2 – b + n), ([b + 1] axk + [b – 1] b – b2 + b – 1)]
= [(baxk – x + 2 + n), ([x – 1] axk – 1)].
Therefore, we have representations for every number from baxk – x + 2 to baxk + x – 1 and from
baxk – x + 2 + n to (x – 1) axk – 1. If n is even, then we have
baxk – x + 2 + n = baxk – y + 2 + 2y
= baxk + y + 2
= baxk + x + 2,
and if n is odd, we have
baxk – x + 2 + n = baxk – (y + 1) + 2 + 2y + 1
= baxk + y + 2
= baxk + x + 1.
Putting all of this information together, we find that we are missing a representation for baxk + x,
and if n is even, we are also missing a representation for baxk + x + 1. We know that (b – 1) axk
is an element of Cb–1 and is therefore represented by our sequence. Then if n is odd, we have
baxk + x = [(b – 1) axk] +  p ; if n is even, we have baxk + x = [(b – 1) axk] +  p and baxk +
pPk 1
x + 1 = [(b – 1) ax ] +
k
pPk 1 {1}
 p.
pPk 1
If we add all of the terms of Sk, we get (x – 1) axk. Then if we add this to each element of
R (Pk–1), we can represent (x – 1) axk + 1 and every number from (x – 1) axk + n to [(x – 1) axk] +
(axk – b – 1) = axk+1 – b – 1. We still need representations for every number from (x – 1) axk + 2
to (x – 1) axk + n – 1 inclusive.
If n is odd, then
 p = axk + x.
If we add this to each element of Cb, we have the set
pPk 1

b
 bax
j 0
k

b
b2j 

 p
pPk 1



  bax k  b  2 j  ax k  x
j 0
b


  x  1ax k  x  2  2 j  x
j 0
 ( x  1)ax
b
=
k
 2  2 j,
j 0
which represents every number of the same parity from (x – 1) axk + 2 to
(x – 1) axk + 2b + 2 = (x – 1) axk + 2 (x – 2) + 2
= (x – 1) axk + 2x – 2
= (x – 1) axk + 2 (y + 1) – 2
= (x – 1) axk + 2y
= (x – 1) axk + n – 1.
By adding
 p to each element of Cb, we can also represent every number of the same parity
pPk 1 {1}
from (x – 1) axk + 1 to (x – 1) axk + n – 2.
If n is even, then
 p = axk + x + 1.
If we add this to every element of Cb, we have the
pPk 1
set
b

 bax
j 0

k
b2j 

 p   bax
pPk 1

k

 b  2 j  ax k  x  1
b


  ( x  1)ax k  x  2  2 j  x  1
j 0
b


  ( x  1)ax k  3  2 j ,
j 0
which represents every number of the same parity from (x – 1) axk + 3 to
(x – 1) axk + 3 + 2b = (x – 1) axk + 2 (x – 2) + 3
= (x – 1) axk + 2x – 1
= (x – 1) axk + 2y – 1
= (x – 1) axk + n – 1.
By adding
 p to each of the elements of Cb, we can also represent every number of the same
pPk 1 {1}
parity from (x – 1) axk + 2 to (x – 1) axk + n – 2.
If n ≥ 9, then b ≥ 3, and we need to show the numbers that can be represented by adding
m terms of Sk for 1 < m < b. If we subtract the smallest element of Bm from the largest element
of Am, we get (maxk + mb – m2 + m + 1) – (maxk – mb + m2 – m + n) = 2mb – 2m2 + 2m + 1 – n.
We will have a continuous representation if the intervals intersect or are consecutive (i.e., only 1
apart from each other), and this will happen if 2mb – 2m2 + 2m + 1 – n ≥ -1, or if 2mb – 2m2 +
2m + 2 – n ≥ 0. If m = 1, then we have
2mb – 2m2 + 2m + 2 – n = 2b – 2 + 2 + 2 – n
= 2b + 2 – n
= 2(x – 2) + 2 – n
= 2x – 2 – n.
n 1
, by algebra we have 2x ≤ n + 1, and then 2x – 2 – n ≤ -1. Therefore, the
2
difference equation is negative for m = 1. If m = 2, then we have 2mb – 2m2 + 2m + 2 – n = 4b –
n
 n  1
n 1
 2      2   2 . Then by
8 + 4 + 2 – n = 4b – 2 – n. Note that b = x – 2 = 

2
 2 
2 2
n

algebra, we have 4b – 2 – n ≥ 4   2  – 2 – n = 2n – 8 – 2 – n = n – 10. So the difference is
2

non-negative if n ≥ 10. If n = 9, then b = 3, and 4b – 2 – n = 4∙3 – 2 – 9 = 1 > 0, so the difference
is also non-negative for n = 9. Since the difference is negative for m = 1 and non-negative for m
= 2, then the difference equation has a root for some 1 < m ≤ 2. If m = b, then the difference is
Since x ≤
2b2 – 2b2 + 2b + 2 – n = 2x – 2 – n, which we already proved is negative. If m = b – 1, then the
difference is
2mb – 2m2 + 2m + 2 – n = 2 (b – 1) b – 2 (b – 1)2 + 2 (b – 1) + 2 – n
= 2b2 – 2b – 2b2 + 4b – 2 + 2b – 2 + 2 – n
= 4b – 2 – n,
which we already proved is non-negative for n ≥ 9. Since the difference is negative for m = b
and non-negative for m = b – 1, then the difference equation has a root for some b – 1 ≤ m < b.
Since 2mb – 2m2 + 2m + 2 – n is a quadratic equation, we know that it has no more than two
roots, so there are no roots between 2 and b – 1. Therefore, the difference of the largest term of
A minus the smallest term of B will remain non-negative for all m from 2 to b – 1 inclusive, and
therefore A and B intersect for these values of m, and so we can represent every number from
maxk – mb + m2 – m to (m + 1) axk + (m – 1) b – m2 + m – 1 inclusive.
Now we must prove that there are no gaps between the intervals of representation for
consecutive values of m. If we add m + 1 terms, the smallest possible value of Cm+1 is (m + 1)
axk – (m + 1) b + (m + 1)2 – (m + 1) = (m + 1) axk – (m + 1) b + m2 + 2m + 1 – m – 1 = (m + 1)
axk – (m + 1) b + m2 + m. If we subtract the smallest element of Am+1 from the largest element of
Bm, we get
[(m + 1) axk + (m – 1) b – m2 + m – 1] – [(m + 1) axk – (m +1) b + m2 + m]
= [(m – 1) + (m + 1)] b – 2m2 – 1
= 2mb – 2m2 – 1.
We will have continuous representation over consecutive values of m if this difference is greater
than or equal to -1, i.e., if 2mb – 2m2 ≥ 0. We can factor this inequality into 2m (b – m) ≥ 0.
This is true if 0 ≤ m ≤ b, which by definition is true. Therefore, there are no gaps between the
intervals of representation for consecutive values of m, and thus we have shown that every
number less than the smallest element of Sk+1 can be represented as a sum of distinct elements of
the sequence.
Our next step is to show that each element of Sk+1 cannot be represented by the terms of
the sequence up to axk + b. For each element sj  Sk+1 that we can prove is in S, we know that sj
+ 1 is not in the sequence. For any two distinct elements of Sk+1, the difference is
s j2  s j1  (ax k 1  b  2 j2 )  (ax k 1  b  2 j2 )  2( j2  j1 ) ,
where 0 ≤ j1 < j2 ≤ b. But we already proved in the first paragraph that the difference 2 (j2 – j1) is
not represented in the sequence, so we cannot use any elements of Sk+1 to represent other
elements of the same set. Now note that if we subtract the sum of the elements of Pk–1 from axk+1
– b, the smallest element of Sk+1, the smallest that the difference can be is
(ax k 1  b) 
 p  (ax
k 1
 b)  (ax k  x  1)
pPk 1
= axk+1 – b – axk – x – 1
= xak – x + 2 – axk – x – 1
= (x – 1) axk – 2x + 1
= x [(x – 1) axk–1 – 2] + 1.
We know that x ≥ 2, so x – 1 ≥ 1. Since a = n + x + 2 and n ≥ 4, by substitution we have a ≥ 8.
1
Then if k = 0, we have (x – 1) ax-1 ≥ 1 ∙ 8 ∙ = 4 > 2. Therefore, both factors x and [(x – 1) axk–1
2
– 2] are both positive, and since both will remain positive for larger values of x and k, we know
that the difference (ax k 1  b)   p will be greater than zero. Thus each element of Sk+1 is too
pPk 1
large to represent by just the elements of Pk–1, and hence any representation of sj  Sk+1 must use
at least one element of Sk.
b
Since Sk =
{s } , we know that Sk has b + 1 elements.
j
Consider the set Cb, the set
j 0
whose elements are the possible sums of all but one term of Sk. The largest element of Cb is the
largest possible sum of all but one term of Sk, i.e., the sum of all but the smallest term of Sk. This
sum is equal to
b
 s   (ax k  b  2 j)
sS k { s0 }
j 1
b
 b(ax k  b)  2 j
j 1
 b(b  1) 
 b(ax k  b)  2

 2 
 b(ax k  b)  b(b  1)
 b(ax k  b  b  1)
 b(ax k  1)
 bax k  b .
If we subtract the largest element of Cb from the smallest element of Sk+1, we get
(axk+1 – b) – (baxk + b) = axk+1 – b – baxk – b
= xaxk – (x – 2) axk – 2 (x – 2)
= (x – x + 2) axk – 2x + 4
= 2axk – 2x + 4.
Then if n is even, we have
[(axk+1 – b) – (baxk + b)] 
 p = (2axk – 2x + 4) – (axk + x + 1)
pPk 1
 ax k  3x  3
 (n  x  2) x k  3x  3
 (2 y  y  2) x k  3 y  3
 (3 y  2) x k  3 y  3
As k increases, (3y + 2) xk also increases. If k = 0, we have (3y + 2) – 3y + 3 = 5 > 0. If n is odd,
we have
[(axk+1 – b) – (baxk + b)]   p = (2axk – 2x + 4) – (axk + x)
pPk 1
 ax k  3x  4
 (n  x  2) x k  3x  4
 (2 y  1  y  1  2) x k  3( y  1)  4
 (3 y  4) x k  3 y  1
As k increases, (3y + 4) xk also increases. If k = 0, we have (3y + 4) – 3y + 1 = 5 > 0. In either
case, the difference of (axk+1 – b) – (baxk + b) –  p is greater than zero. Therefore, if we
pPk 1
subtract the largest element of Cb from the smallest element of Sk+1, the difference is too large to
represent with just the elements of Pk–1. If we use a smaller element of Cb and/or a larger
element of Sk+1, the difference will still be too large to represent using only the elements of Pk–1.
Therefore, any representation of an element of Sk+1 must use every element of Sk. We have
shown earlier that the sum of all of the terms of Sk is (x – 1) axk. By definition, the jth element of
Sk+1 is axk+1 – b + 2j, so when we subtract all of the elements of Sk from any element of Sk+1, we
get
(axk+1 – b + 2j) – (x – 1) axk = xaxk – b + 2j – (x – 1) axk
= (x – x + 1) axk – b + 2j
= axk – b + 2j.
But (ax k  b  2 j )  S k , and by definition of the sequence, this difference cannot be represented
as a sum of the elements of Pk–1. Therefore, the elements of Sk+1 cannot be represented as sums
of previous terms of the sequence, and hence must be the next terms of the sequence. Thus we
have shown by induction that the sequence follows the formula stated in the theorem. QED.
Figure 7. Java Code for Sequence Program v. 1.0
/* Sequence Program (Version 1.0)
* By Mike Paul
* Calculates sequence for 1,n,...
* Input: n
* Output: Terms of the sequence
*/
import java.util.Scanner;
class Sequence1
{
static public void main(String [] args)
{
Scanner in = new Scanner(System.in);
System.out.print("Enter n: ");
int n = in.nextInt();
int values[] = new int[22001];
int new_values[] = new int[22001];
System.out.print("1, " + n + ", ");
values[1] = 1;
values[n] = 1;
values[n + 1] = 1;
for(int i = n; i < 11001; i++)
{
if (values[i] == 0)
{
System.out.print(i + ", ");
for (int j = 1; j < 11001; j++)
{
new_values[j] = values[j];
}
for (int k = 1; k < 11001; k++)
{
if (new_values[k] == 1)
values[i + k] = 1;
}
values[i] = 1;
}
}
}
}
Figure 8. Sequence Program v. 2.2
/* Sequence Program (Version 2.2)
* By Mike Paul
* Input: Number of initial terms, range, initial terms,
representation counts (y/n)
* Output: Sequence, numbers with multiple representations (if
desired)
* Added: User can turn off representation counts
*/
import java.util.Scanner;
class Sequence_2_2
{
static public void main (String [] args)
{
Scanner in = new Scanner (System.in);
System.out.println ("How many initial terms? ");
int terms = in.nextInt();
System.out.println ("How far do you want to calculate? ");
int range = in.nextInt();
int values[] = new int[2 * range + 1];
int new_values[] = new int[2 * range + 1];
int n[] = new int[terms];
int sum = 0;
for (int i = 0; i < terms; i++)
{
System.out.println ("Enter term # " + i + ": ");
n[i] = in.nextInt();
sum = sum + n[i];
}
System.out.println ("Display represention counts?");
System.out.println ("1) Yes
2) No");
int option = in.nextInt();
System.out.println();
System.out.println ("The sequence is: ");
for (int i = 0; i < terms; i++)
{
for (int j = 1; j <= sum; j++)
{
new_values[j] = values[j];
}
for (int j = 1; j <= sum; j++)
{
if (new_values[j] > 0)
values[j + n[i]]++;
}
values[n[i]]++;
System.out.print (n[i] + ", ");
}
for (int i = n[terms - 1]; i < range + 1; i++)
{
if (values[i] == 0)
{
System.out.print (i + ", ");
for (int j = 1; j <= range; j++)
{
new_values[j] = values[j];
}
for (int j = 1; j <= range; j++)
{
if (new_values[j] > 0)
values[i + j]++;
}
values[i]++;
sum = sum + i;
}
}
if (option == 1)
{
int highest_sum_count = 1;
for (int i = 1; i <= range; i++)
{
if (values[i] > highest_sum_count)
highest_sum_count = values[i];
}
int number_count;
System.out.println();
if (highest_sum_count > 1)
{
for (int i = 2; i <= highest_sum_count; i++)
{
number_count = 0;
System.out.println();
for (int j = 1; j <= range; j++)
{
if (values[j] == i)
number_count++;
}
if (number_count > 0)
{
System.out.println ("There are " +
number_count + " numbers in this range with " + i + " possible
representations: ");
for (int j = 1; j <= range; j++)
{
if (values[j] == i)
System.out.print (j + ", ");
}
System.out.println();
}
}
}
else
System.out.println ("There are no numbers in this
range with multiple representations.");
}
}
}
Figure 9. Sequence Program v. 3.12
1 /* Sequence Program (Version 3.12)
2 * Mike Paul
3 * Input: Initial terms, range of calculation, show multiple
representations?(y/n)
4 * Output: Sequence, numbers with multiple representations
(if desired)
5 * Modifications:
6 *
1) Added a button to print the results on paper,
changed the font for the display to
7 *
12-pt. Times New Roman (which is what we've been
using previous printouts).
8 *
2) Current Term spinner now automatically changes to
match the value of the currently
9 *
selected term in the list.
10 *
3) Catches most instances of user entering a limit of
calculation over 3,500,000.
11 */
12
13 import javax.swing.*;
14 import java.awt.event.*;
15 import java.awt.*;
16 import java.awt.print.*;
17 import javax.swing.event.*;
18
19 public class Sequence3_12 extends JFrame implements
ActionListener, ListSelectionListener, ChangeListener
20 {
21
public static void main (String [] args)
22
{
23
new Sequence3_12();
24
}
25
26
JTextArea results;
27
DefaultListModel initial;
28
JList initialTerms;
29
JButton removeTerm, clearTerms, update, addTerm,
calculate, clearDisplay, printResults;
30
JSpinner currentTerm, nextTerm, limit, repCount;
31
JCheckBox multRep;
32
JScrollPane initialScroll, resultScroll;
33
final int MAX_LIMIT = 3500000;
34
35
int range, terms, addedTerm, updatedTerm, rCount;
36
37
public Sequence3_12()
38
{
39
this.setTitle("Sequence Program");
40
this.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
41
JPanel mainPanel = new JPanel();
42
mainPanel.setLayout(new GridBagLayout());
43
44
// List of initial terms
45
46
JPanel initialPanel = new JPanel();
47
initialPanel.setLayout(new GridLayout(1,0));
48
initial = new DefaultListModel();
49
initialTerms = new JList(initial);
50
initialTerms.setVisibleRowCount(10);
51
initialTerms.setSelectionMode(ListSelectionModel.SINGLE_SELECTIO
N);
52
initialScroll = new JScrollPane(initialTerms,
JScrollPane.VERTICAL_SCROLLBAR_ALWAYS,
53
JScrollPane.HORIZONTAL_SCROLLBAR_NEVER);
54
initialPanel.add(initialScroll);
55
initialPanel.setBorder(BorderFactory.createTitledBorder("Initial
Terms"));
56
addItem(mainPanel, initialPanel, 0, 0, 1, 1,
GridBagConstraints.WEST,
57
GridBagConstraints.BOTH);
58
initialPanel.setVisible(true);
59
60
// Components to modify or remove terms
61
62
Box changeBox = Box.createVerticalBox();
63
removeTerm = new JButton("Remove Term");
64
clearTerms = new JButton("Clear all initial terms");
65
changeBox.add(removeTerm);
66
changeBox.add(Box.createVerticalStrut(5));
67
changeBox.add(clearTerms);
68
changeBox.add(Box.createVerticalStrut(10));
69
Box currentBox = Box.createVerticalBox();
70
currentTerm = new JSpinner(new SpinnerNumberModel(1,
1, 10000, 1));
71
update = new JButton("Update Term");
72
currentBox.add(currentTerm);
73
currentBox.add(Box.createVerticalStrut(10));
74
currentBox.add(update);
75
currentBox.setBorder(BorderFactory.createTitledBorder("Current
Term"));
76
changeBox.add(currentBox);
77
addItem(mainPanel, changeBox, 1, 0, 1, 1,
GridBagConstraints.WEST,
78
GridBagConstraints.NONE);
79
80
// Components to add terms
81
82
Box nextBox = Box.createHorizontalBox();
83
nextBox.add(new JLabel("Next term: "));
84
nextBox.add(Box.createHorizontalStrut(5));
85
nextTerm = new JSpinner(new SpinnerNumberModel(1, 1,
10000, 1));
86
nextBox.add(nextTerm);
87
nextBox.add(Box.createHorizontalStrut(5));
88
addTerm = new JButton("Add Term");
89
nextBox.add(addTerm);
90
addItem(mainPanel, nextBox, 0, 1, 2, 1,
GridBagConstraints.WEST,
91
GridBagConstraints.NONE);
92
93
// Spinner to set limit of calculation
94
95
Box limitBox = Box.createHorizontalBox();
96
limitBox.add(new JLabel("Calculate sequence up to "));
97
limitBox.add(Box.createHorizontalStrut(5));
98
limit = new JSpinner(new SpinnerNumberModel(10000, 1,
MAX_LIMIT, 1));
99
limitBox.add(limit);
100
limitBox.add(Box.createHorizontalStrut(5));
101
limitBox.add(new JLabel("(maximum " + MAX_LIMIT +
")"));
102
addItem(mainPanel, limitBox, 0, 3, 2, 1,
GridBagConstraints.WEST,
103
GridBagConstraints.NONE);
104
105
// Checkbox to display numbers with multiple
representations
106
107
Box multRepBox = Box.createHorizontalBox();
108
multRep = new JCheckBox("Display numbers with ");
109
multRepBox.add(multRep);
110
multRepBox.add(Box.createHorizontalStrut(5));
111
repCount = new JSpinner(new SpinnerNumberModel(2, 2,
20, 1));
112
multRepBox.add(repCount);
113
multRepBox.add(Box.createHorizontalStrut(5));
114
multRepBox.add(new JLabel(" or more
representations"));
115
addItem(mainPanel, multRepBox, 0, 4, 2, 1,
GridBagConstraints.WEST,
116
GridBagConstraints.NONE);
117
118
// Button to calculate the sequence
119
120
Box calcBox = Box.createHorizontalBox();
121
calculate = new JButton("Calculate Sequence");
122
calcBox.add(calculate);
123
calcBox.add(Box.createHorizontalGlue());
124
125
// Button to print the results on paper
126
127
printResults = new JButton("Print Results");
128
calcBox.add(printResults);
129
calcBox.add(Box.createHorizontalGlue());
130
131
// Button to clear the results display
132
133
clearDisplay = new JButton("Clear Display");
134
calcBox.add(clearDisplay);
135
addItem(mainPanel, calcBox, 0, 5, 2, 1,
GridBagConstraints.WEST,
136
GridBagConstraints.HORIZONTAL);
137
138
// Text field to display the generated sequence
139
140
results = new JTextArea(20, 50);
141
results.setLineWrap(true);
142
results.setWrapStyleWord(true);
143
results.setEditable(false);
144
results.setFont(new Font("Times New Roman",
Font.PLAIN, 12));
145
resultScroll = new JScrollPane(results,
JScrollPane.VERTICAL_SCROLLBAR_ALWAYS,
146
JScrollPane.HORIZONTAL_SCROLLBAR_NEVER);
147
addItem(mainPanel, resultScroll, 0, 6, 3, 2,
GridBagConstraints.NORTHWEST,
148
GridBagConstraints.BOTH);
149
150
// Put it all together on screen
151
152
this.add(mainPanel);
153
this.pack();
154
this.setVisible(true);
155
156
/* ActionListeners enable the buttons to actually do
stuff when clicked.
157
* addTerm will always be enabled, so it is activated
here.
158
* The remove buttons will not be enabled until the
user adds initial terms to the list.
159
* printResults and clearDisplay will not be enable
until there is text in the results display.
160
*/
161
162
addTerm.addActionListener(this);
163
initialTerms.addListSelectionListener(this);
164
limit.addChangeListener(this);
165
}
166
167
// Position each component on the window
168
169
public void addItem(JPanel p, JComponent c, int x, int y,
int width, int height, int align, int fill)
170
{
171
GridBagConstraints gc = new GridBagConstraints();
172
gc.gridx = x;
173
gc.gridy = y;
174
gc.gridwidth = width;
175
gc.gridheight = height;
176
gc.weightx = 100.0;
177
gc.weighty = 100.0;
178
gc.insets = new Insets(5, 5, 5, 5);
179
gc.anchor = align;
180
gc.fill = fill;
181
p.add(c, gc);
182
}
183
184
// Determine which button has been clicked and execute
the appropriate action
185
186
public void actionPerformed(ActionEvent e)
187
{
188
terms = initial.size();
189
190
if (e.getSource() == addTerm)
191
{
192
addedTerm = spinnerValue(nextTerm);
193
addTermToList(addedTerm, terms);
194
}
195
196
else if (e.getSource() == removeTerm)
197
198
199
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234
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237
238
239
240
241
242
243
{
removeTermFromList();
}
else if (e.getSource() == clearTerms)
{
initial.clear();
turnOffRemoveButtons();
}
else if (e.getSource() == update)
{
updatedTerm = spinnerValue(currentTerm);
int i = initialTerms.getSelectedIndex();
removeTermFromList();
addTermToList(updatedTerm, terms - 1);
}
else if (e.getSource() == calculate)
{
ChangeEvent ce = new ChangeEvent(limit);
if (results.getText().equals(""))
{
turnOnDisplayButtons();
}
boolean calculated = false;
while (!calculated)
{
try
{
this.calculateSequence();
calculated = true;
}
catch (OutOfMemoryError m)
{
limit.setValue(MAX_LIMIT);
}
}
}
else if (e.getSource() == clearDisplay)
{
results.setText("");
turnOffDisplayButtons();
244
245
246
247
248
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250
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252
253
254
255
256
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258
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286
287
288
289
290
}
else if (e.getSource() == printResults)
{
try
{
results.print();
}
catch (PrinterException p)
{
JOptionPane.showMessageDialog(
this,
"Unable to print results.",
"Printing Error",
JOptionPane.ERROR_MESSAGE);
}
}
}
public void valueChanged(ListSelectionEvent s)
{
if (s.getSource() == initialTerms)
{
try
{
int i = initialTerms.getSelectedIndex();
currentTerm.setValue(this.getTerm(i));
}
catch (ArrayIndexOutOfBoundsException a)
{
}
}
}
public void stateChanged(ChangeEvent c)
{
if (c.getSource() == limit)
{
try
{
if (spinnerValue(limit) > MAX_LIMIT)
{
limit.setValue(MAX_LIMIT);
}
}
catch (OutOfMemoryError m)
{
291
292
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297
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304
305
306
307
308
309
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311
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321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
limit.setValue(MAX_LIMIT);
}
}
}
public void calculateSequence()
{
String display = "";
range = spinnerValue(limit);
rCount = spinnerValue(repCount);
int values[] = new int[2 * range + 1];
int newValues[] = new int[2 * range + 1];
int n[] = new int[terms];
int sum = 0;
display += "Start with ";
for (int i = 0; i < terms; i++)
{
n[i] = getTerm(i);
sum += n[i];
display = display + n[i];
if (i < terms - 1)
{
display += ", ";
}
else
{
display = display + "\n" + "Get ";
}
}
for (int i = 0; i < terms; i++)
{
for (int j = 1; j <= sum; j++)
{
newValues[j] = values[j];
}
for (int j = 1; j <= sum; j++)
{
if (newValues[j] > 0)
values[j + n[i]]++;
}
values[n[i]]++;
display += (n[i] + ", ");
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
}
for (int i = n[terms - 1]; i < range + 1; i++)
{
if (values[i] == 0)
{
display += (i + ", ");
for (int j = 1; j <= range; j++)
{
newValues[j] = values[j];
}
for (int j = 1; j <= range; j++)
{
if (newValues[j] > 0)
values[i + j]++;
}
values[i]++;
sum = sum + i;
}
}
display += "...";
if (multRep.isSelected())
{
int highestSumCount = 1;
int numberCount;
for (int i = 1; i <= range; i++)
{
if (values[i] > highestSumCount)
highestSumCount = values[i];
}
if (highestSumCount >= rCount)
{
for (int i = rCount; i <= highestSumCount; i++)
{
numberCount = 0;
for (int j = 1; j <= range; j++)
{
if (values[j] == i)
numberCount++;
385
}
386
387
if (numberCount > 0)
388
{
389
display += "\n \n";
390
display += ("There are " + numberCount + "
numbers in this range with " + i + " possible representations:
\n");
391
int counter = 0;
392
393
for (int j = 1; j <= range; j++)
394
{
395
if (values[j] == i)
396
{
397
display += j;
398
counter++;
399
400
if (counter != numberCount)
401
{
402
display += ", ";
403
}
404
}
405
}
406
}
407
}
408
}
409
else
410
{
411
display += "\n \n";
412
display += ("There are no numbers in this range
with " + rCount + " or more representations.");
413
}
414
}
415
416
display += "\n \n";
417
results.append(display);
418
}
419
420
public int spinnerValue(JSpinner spinner)
421
{
422
String s = spinner.getValue().toString();
423
return Integer.parseInt(s);
424
}
425
426
public int getTerm(int i)
427
{
428
String s = initial.get(i).toString();
429
430
431
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
return Integer.parseInt(s);
}
public void addTermToList(int addedTerm, int terms)
{
if (terms == 0)
{
initial.addElement(addedTerm);
turnOnRemoveButtons();
initialTerms.setSelectedIndex(0);
}
else
{
if (addedTerm > getTerm(terms - 1))
{
initial.addElement(addedTerm);
initialTerms.setSelectedIndex(terms);
}
else
{
for (int i = 0; i < terms; i++)
{
if (addedTerm < getTerm(i))
{
initial.add(i, addedTerm);
initialTerms.setSelectedIndex(i);
break;
}
else if (addedTerm == getTerm(i))
{
break;
}
}
}
}
}
public void removeTermFromList()
{
int i = initialTerms.getSelectedIndex();
initial.remove(i);
if (i == terms - 1)
{
initialTerms.setSelectedIndex(i - 1);
}
476
477
478
479
480
481
482
483
484
485
486
487
488
489
490
491
492
493
494
495
496
497
498
499
500
501
502
503
504
505
506
507
508
509
510
511
512
513
514 }
else
{
initialTerms.setSelectedIndex(i);
}
if (terms == 1)
{
turnOffRemoveButtons();
}
}
public void turnOffRemoveButtons()
{
removeTerm.removeActionListener(this);
clearTerms.removeActionListener(this);
update.removeActionListener(this);
calculate.removeActionListener(this);
}
public void turnOnRemoveButtons()
{
removeTerm.addActionListener(this);
clearTerms.addActionListener(this);
update.addActionListener(this);
calculate.addActionListener(this);
}
public void turnOffDisplayButtons()
{
clearDisplay.removeActionListener(this);
printResults.removeActionListener(this);
}
public void turnOnDisplayButtons()
{
clearDisplay.addActionListener(this);
printResults.addActionListener(this);
}
Figure 10. Screen Shot of Sequence Program v. 3.12
Figure 11. Sequences beginning with {1, 2, n,...}
Start with 1, 2, 5
Get 1, 2, 5, 9, 13, 26, 52, 104, 208, 416, 832, 1664, 3328, 6656, ...
Start with 1, 2, 6
Get 1, 2, 6, 10, 14, 28, 56, 112, 224, 448, 896, 1792, 3584, 7168, ...
Start with 1, 2, 7
Get 1, 2, 7, 11, 15, 30, 60, 120, 240, 480, 960, 1920, 3840, 7680, ...
Start with 1, 2, 8
Get 1, 2, 8, 12, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, ...
Start with 1, 2, 9
Get 1, 2, 9, 13, 17, 21, 55, 59, 169, 173, 511, 515, 1537, 1541, 4615, 4619, ...
Start with 1, 2, 10
Get 1, 2, 10, 14, 18, 22, 58, 62, 178, 182, 538, 542, 1618, 1622, 4858, 4862, ...
Start with 1, 2, 11
Get 1, 2, 11, 15, 19, 23, 61, 65, 187, 191, 565, 569, 1699, 1703, 5101, 5105, ...
Start with 1, 2, 12
Get 1, 2, 12, 16, 20, 24, 64, 68, 196, 200, 592, 596, 1780, 1784, 5344, 5348, ...
Start with 1, 2, 13
Get 1, 2, 13, 17, 21, 25, 29, 96, 100, 104, 396, 400, 404, 1596, 1600, 1604, 6396, 6400, 6404, ...
Start with 1, 2, 14
Get 1, 2, 14, 18, 22, 26, 30, 100, 104, 108, 412, 416, 420, 1660, 1664, 1668, 6652, 6656, 6660, ...
Start with 1, 2, 15
Get 1, 2, 15, 19, 23, 27, 31, 104, 108, 112, 428, 432, 436, 1724, 1728, 1732, 6908, 6912, 6916, ...
Start with 1, 2, 16
Get 1, 2, 16, 20, 24, 28, 32, 108, 112, 116, 444, 448, 452, 1788, 1792, 1796, 7164, 7168, 7172, ...
Start with 1, 2, 17
Get 1, 2, 17, 21, 25, 29, 33, 37, 149, 153, 157, 161, 769, 773, 777, 781, 3869, 3873, 3877, 3881,
...
Start with 1, 2, 18
Get 1, 2, 18, 22, 26, 30, 34, 38, 154, 158, 162, 166, 794, 798, 802, 806, 3994, 3998, 4002, 4006,
...
Figure 12. Sequences beginning with {2, n,...}
Start with 2, 3
Get 2, 3, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192
Start with 2, 4
Get 2, 4, 5, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192
Start with 2, 5
Get 2, 5, 6, 9, 10, 28, 29, 85, 86, 256, 257, 769, 770, 2308, 2309, 6925, 6926
Start with 2, 6
Get 2, 6, 7, 10, 11, 14, 45, 46, 49, 185, 186, 189, 745, 746, 749, 2985, 2986, 2989
Start with 2, 7
Get 2, 7, 8, 11, 12, 16, 50, 51, 55, 206, 207, 211, 830, 831, 835, 3326, 3327, 3331
Start with 2, 8
Get 2, 8, 9, 12, 13, 16, 53, 54, 57, 217, 218, 221, 873, 874, 877, 3497, 3498, 3501
Start with 2, 9
Get 2, 9, 10, 13, 14, 17, 18, 75, 76, 79, 80, 385, 386, 389, 390, 1935, 1936, 1939, 1940, 9685, 9686, 9689,
9690
Start with 2, 10
Get 2, 10, 11, 14, 15, 18, 19, 22, 102, 103, 106, 107, 110, 630, 631, 634, 635, 638, 3798, 3799, 3802,
3803, 3806
Start with 2, 11
Get 2, 11, 12, 15, 16, 19, 20, 24, 109, 110, 113, 114, 118, 673, 674, 677, 678, 682, 4057, 4058, 4061,
4062, 4066
Start with 2, 12
Get 2, 12, 13, 16, 17, 20, 21, 24, 114, 115, 118, 119, 122, 702, 703, 706, 707, 710, 4230, 4231, 4234,
4235, 4238
Start with 2, 13
Get 2, 13, 14, 17, 18, 21, 22, 25, 26, 146, 147, 150, 151, 154, 155, 1049, 1050, 1053, 1054, 1057, 1058,
7370, 7371, 7374, 7375, 7378, 7379
Start 2, 14
Get 2, 14, 15, 18, 19, 22, 23, 26, 27, 30, 183, 184, 187, 188, 191, 192, 195, 1503, 1504, 1507, 1508, 1511,
1512, 1515, 12063, 12064, 12067, 12068, 12071, 12072, 12075
Start with 2, 15
Get 2, 15, 16, 19, 20, 23, 24, 27, 28, 32, 192, 193, 196, 197, 200, 201, 205, 1576, 1577, 1580, 1581, 1584,
1585, 1589, 12648, 12649, 12652, 12653, 12656, 12657, 12661
Figure 13. Sequences Beginning with {n, n + 1}
Start with 1, 2
Get 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192
Start with 2, 3
Get 2, 3, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192
Start with 3, 4
Get 3, 4, 5, 6, 16, 17, 49, 50, 148, 149, 445, 446, 1336, 1337, 4009, 4010
Start with 4, 5
Get 4, 5, 6, 7, 8, 27, 28, 29, 111, 112, 113, 447, 448, 449, 1791, 1792, 1793, 7167, 7168, 7169
Start with 5, 6
Get 5, 6, 7, 8, 9, 10, 41, 42, 43, 44, 211, 212, 213, 214, 1061, 1062, 1063, 1064, 5311, 5312,
5313, 5314
Start with 6, 7
Get 6, 7, 8, 9, 10, 11, 12, 58, 59, 60, 61, 62, 358, 359, 360, 361, 362, 2158, 2159, 2160, 2161,
2162
Start with 7, 8
Get 7, 8, 9, 10, 11, 12, 13, 14, 78, 79, 80, 81, 82, 83, 561, 562, 563, 564, 565, 566, 3942, 3943,
3944, 3945, 3946, 3947
Start with 8, 9
Get 8, 9, 10, 11, 12, 13, 14, 15, 16, 101, 102, 103, 104, 105, 106, 107, 829, 830, 831, 832, 833,
834, 835, 6653, 6654, 6655, 6656, 6657, 6658, 6659
Start with 9, 10
Get 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 127, 128, 129, 130, 131, 132, 133, 134, 1171, 1172,
1173, 1174, 1175, 1176, 1177, 1178
Start with 10, 11
Get 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 156, 157, 158, 159, 160, 161, 162, 163, 164, 1596,
1597, 1598, 1599, 1600, 1601, 1602, 1603, 1604
Start with 11, 12
Get 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 188, 189, 190, 191, 192, 193, 194, 195, 196,
197, 2113, 2114, 2115, 2116, 2117, 2118, 2119, 2120, 2121, 2122
Start with 12, 13
Get 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 223, 224, 225, 226, 227, 228, 229, 230, 231,
232, 233, 2731, 2732, 2733, 2734, 2735, 2736, 2737, 2738, 2739, 2740, 2741
Figure 14. Proof for {n, n + 1,...} Sequences
Theorem: Define the sequence S = {n, n + 1, ...} by setting the first two terms equal to n and n +
1, where n ≥ 2, and then demanding that every integer greater than n + 1 be a sum of distinct
terms of S. Then this sequence is given by:

S = {n, n + 1}   {S i } ,
i 0
n2
3n  2
 j , where a 
where S i  {si , j } , and si , j  an i 
.
2
2
j 0
Proof: By definition, n and n + 1 are the first two terms of the sequence. Our first step is to
n2
n2
 j} are in the next terms of the sequence, i.e., that
prove that the elements of S0 = {a 
2
j 0
each element of S0 cannot be represented as a sum of previous terms of S. For any two distinct
elements of S0, the difference is:
n2
n2
s 0, j2  s 0, j1  (a 
 j 2 )  (a 
 j1 )  j 2  j1 ,
2
2
where 0 ≤ j1 < j2 ≤ n – 2 < n. Then by algebra, we have 0 < j2 – j1 ≤ n – 2 – j1 < n. Therefore, the
difference is an integer less than n, which by definition is not in the sequence. Therefore, we
cannot represent an element of S0 using other elements of S0. We know that the elements of S0
are the integers from
n2
3n  2 n  2
s 0, 0  a 
0

n2
2
2
2
n2
3n  2 n  2
 (n  2) 

 2n .
to s 0,n 2  a 
2
2
2
n2
Since the sum of the first two terms of S is 2n + 1, we know that each integer from n + 2 to 2n
cannot be represented as a sum of either of the first two terms in the sequence. Therefore, the
elements of S0 cannot be represented as sums of the previous terms of S, and must therefore be
the next terms of the sequence.
Now define the set Px  S 

{S } , and assume that the sequence begins with Pk for
i
i  x 1
some k ≥ 0. We now need to show that the next terms of the sequence are the elements of Sk+1.
First, we need to show that every integer less than the smallest element of Sk+1, which
n2
is an k 1 
, can be represented as a sum of distinct terms of S. By definition, we know that
2
n2
every integer up to and including the largest element of Sk, an k 
, can be represented, so
2
n2
n
n2
 1  an k  to an k 1 
1
we just need to prove that every integer from an k 
2
2
2
n
 an k 1  can be represented. Define the set R (f) as the set of all integers that can be
2
represented as a sum of distinct elements of the set f. Then by definition of the sequence, we
n
have R (Pk–1)  {z: z  [n, an k  ]}. We will now attempt to represent other numbers by
2
adding one or more elements of Sk to each element of R (Pk–1). Let Cm be the set whose elements
are all of the possible sums of m distinct terms of Sk. If we add m terms of Sk, we get
n2
man k  m
  ( j1  j 2  ...  j m ) for distinct ji  [0, n – 2]. The smallest element of Cm is:
 2 
n2
 n  2  m(m  1)
k
man k  m
  (0  1  2  ...  m  1)  man  m

2
 2 
 2 
 m 1 n  2 
 man k  m


2 
 2
 m 1 n  2 
 man k  m

2


 m  n  1
= man k  m
,
2


and the largest term of Cm is:
n2
man k  m
  (n  2)  (n  2  1)  (n  2  2)  ...  (n  2  [m  1]) 
 2 
m(m  1)
n2
 man k  m
  m(n  2) 
2
 2 
n2
n2
 m 1
 man k  m
  2m
  m

 2 
 2 
 2 
n2
 m 1
 man k  m
  m

 2 
 2 
 n  2  m 1
 man k  m

2


 n  m 1
 man k  m
.
2


m
Since
j
i 1
i
can represent every integer from its minimum value to its maximum value, we know
 m  n  1
 n  m 1 
k
that Cm represents every integer from man k  m
 to man  m
 . By adding
2
2




each element cj  of Cm to each element of R (Pk–1), we can also represent every number from cj
n
+ n to c j  an k  . Since the elements of Cm are consecutive, the intervals of representation for
2
each element of Cm intersect, so we can combine these intervals and find that we have a
 m  n 1
 n  m 1
k
representation for every number from man k  m
  n to man  m

2
2




n
 n  m 1 n
 an k   (m  1)an k  m
  , and we will call this interval Bm.
2
2

 2
If we add only one element of Sk (i.e., the elements of C1) to each element of R( Pk 1 ) , by
n2
the formulae proven above, we have representations for every number from an k 
to
2
n2
n2
n 11 n
an k 
 n to 2an k 
  2an k  1 . Note that
and from an k 
2
2
2
2
n

2
n
2 2n
an k 
 n  an k   
2
2 2 2
n
2
 an k 
2
n

24
 an k 
2
n

2
 an k 
2.
2
n2
 1 . Note that
Then we find that we find that we are missing a representation for an k 
2
n2
n2
s   (an i 
 j)

2
sSi
j 0
(n  1)( n  2) (n  2)( n  1)
 (n  1)an i 

2
2
i
 (n  1)an ,
and then
k 1
 p  n  (n  1)   s
pPk 1
i 0 sSi
k 1
 2n  1   (n  1)an i
i 0
k 1
 2n  1  (n  1)a  n i
i 0
 nk 1

 2n  1  (n  1)a
n

1


k
 2n  1  a (n  1)
 2n  1  an k  a
4n  2 3n  2


 an k
2
2
n
 an k 
2
n 2 2
 an k   
2 2 2
n2
 an k 
 1.
2
Now if n ≥ 4, we can add n – 2 terms of Sk; this is the same as adding all but one term of
Sk, so we can obtain the elements of Cn–2 by subtracting each element of Sk from the sum of all of
the elements of Sk:
n2
n2
n2

 n2 

C n 2   (n  1)an k  (an k 
 j )   (n  2)an k 
 j
2
2
 j 0 

j 0 
By evaluating the values of this set, we find that we have representations for every number from
n2
n2
(n  2)an k 
to (n  2)an k 
. If we add each element of Cn–2 to each element
2
2
n2
n2
 n  (n  2)an k 
2
of R( Pk 1 ) , we can represent every number from (n  2)an k 
2
2
n2
n
to (n  2)an k 
+ an k   (n  1)an k  1. However, we are missing a representation
2
2
n

2
 1 . We know that (n  3)an k is an element of C n 3 and is therefore
for (n  2)an k 
2
n2
represented by the sequence. Then we have (n  2)an k 
 1  (n  3)an k   p .
2
pPk 1
If we add all of the terms of Sk, we get (n  1)an k . Then if we add this to each element
n
of R( Pk 1 ) , we can represent every number from (n  1)an k  n to (n  1)an k  an k 
2
n
n
 nan k   an k 1  . We still need representations for every number from (n  1)an k  1
2
2

 n2 
n2
n
k
 j  an k  
to (n  1)an  n  1. Note that  c   p    (n  2)an k 
2
2
pPk 1 
cC n  2 
j 0 
n2
  (n  1)an k  n  1  j. Therefore, by adding
j 0
 p to each element of C
pPk 1
n2
, we can
represent every number from (n  1)an k  1 to (n  1)an k  n  1.
If n ≥ 5, then n – 2 ≥ 3, and we need to show the numbers that can be represented by
adding m terms of Sk for 1 < m < n – 2. If we subtract the smallest element of Bm from the largest
element of Cm, we get

 n  m  1  
 m  n  1 
 n  m 1
 m  n  1
k
k
  man  m
  n  m
  m
n
man  m
2
2
2
2

 

 





m
 (n  m  1  m  n  1)  n
2
m
 (2n  2m  2)  n
2
 mn  m 2  m  n .
We will have a continuous representation if the intervals Cm and Bm intersect or are consecutive
(i.e., only 1 apart from each other), and this will happen if mn – m2 – m – n ≥ -1, or if mn – m2 –
m – n + 1 ≥ 0. If m = 1, then we have mn – m2 – m – n + 1 = n – 1 – 1 – n + 1 = -1 < 0, so the
difference equation is negative for m = 1. If m = 2, then we have mn – m2 – m – n + 1 = 2n – 4 –
2 – n + 1 = n – 5, so the difference is non-negative for n ≥ 5. Since the difference is negative for
m = 1 and non-negative for m = 2, then the difference equation has a root for some 1 < m ≤ 2. If
m = n – 2, then the difference is (n – 2) n – n – (n – 2)2 – (n – 2) + 1 = n2 – 2n – n – n2 + 4n – 4 –
n + 2 + 1 = -1 < 0, so the difference is negative for m = n – 2. If m = n – 3, then the difference is
(n – 3) n – n – (n – 3)2 – (n – 3) + 1 = n2 – 3n – n – n2 + 6n – 9 – n + 3 + 1 = n – 5, which is nonnegative for n ≥ 5. Since the difference is negative for m = n – 2 and non-negative for m = n – 3,
then the difference equation has a root for some n – 3 ≤ m < n – 2. Since mn – m2 – m – n + 1 is
a quadratic equation in m, we know that it has no more than two roots, so there are no roots
between 2 and n – 3. Therefore, the difference of the largest term of Cm minus the largest term
of Bm will be non-negative for all m from 2 to n – 3 inclusive, and therefore Cm and Bm intersect
 m  n  1
for these values of m, and so we can represent every number from man k  m
 to
2


 n  m 1 n
(m  1)an k  m
  inclusive.
2

 2
Now if n ≥ 3, we must prove that there are no gaps between the intervals of
representation for consecutive values of m. If we add m + 1 terms, the smallest possible value of
mn 2
Cm is (m  1)an k  (m  1)
 . If we subtract the smallest element of Cm+1 from the
2


largest element of Bm, the difference is

 n  m 1 n  
 m  n  2 
k
k
    (m  1)an  (m  1)

(m  1)an  m
2
2

 2 



 n  m 1 n
mn 2
 m
   (m  1)

2
2

 2


mn  m 2  m  n  m(m  n  2)  (m  n  2)
2
2
2
mn  m  m  n  m  mn  2m  m  n  2

2
2
2mn  2m  4m  2

2
2
 mn  m  2m  1 .

We will have continuous representation over consecutive values of m if this difference is greater
than or equal to -1, i.e., if mn – m2 – 2m ≥ 0. We can factor this inequality into m (n – m – 2) ≥ 0.
This is true if 0 ≤ m ≤ n – 2, which by definition is true because Si has n – 1 elements. Therefore,
there are no gaps between the intervals of representation for consecutive values of m, and thus
we have shown that every number less than the smallest element of Sk+1 can be represented as a
sum of distinct elements of the sequence.
Our next step is to show that each element of Sk+1 cannot be represented by the terms of
n2
the sequence up to an k 
. For any two distinct elements of Sk+1, the difference is
2
n2
 k 1 n  2
 

 j 2    an k 1 
 j1   j 2  j1 , where 0 ≤ j1 < j2 ≤ n – 2 < n. But we
 an 
2
2

 

already proved in the first paragraph that the difference j2 – j1 is not represented in the sequence,
so we cannot represent an element of Sk+1 using other elements of Sk+1. First, note that since we
have
n2
n
an k 1 
 j > an k    p ,
2
2 pPk 1
any representation of an element of Sk+1 must use at least one element of Sk. Consider the
set C n 2 , the set whose elements are the possible sums of all but one term of Sk. If we subtract
the largest element of C n 2 from the smallest element of Sk+1, we get
n  2
n2
n2
 k 1 n  2  
k
k
k
an  2   (n  2)an  2   nan  2  (n  2)an  2
 [n  (n  2)]an k  (n  2)
 2an k  n  2 .
n  2 
n  2
n


 (n  2)an k 
  p  2an k  n  2   an k  
Then we have an k 1 


2  
2  pPk 1
2


 an k 
3n
2
2
3n
3n
3n  2 3n 4
 2 increases. If k = 0, we have a 
2

 = 3 > 0.
2
2
2
2 2
Therefore, if we subtract the largest element of C n 2 from the smallest element of Sk+1, the
difference is too large to represent with just the elements of Pk–1. If we use a larger element of
Sk+1 and/or a smaller element of C n 2 , the difference will still be too large to represent using only
the elements of Pk–1. Therefore, any representation of an element of Sk+1 must use every element
of Sk. We have shown earlier that the sum of all of the terms of Sk is (n – 1) ank. By definition,
n2
 j , so when we subtract all of the elements of Sk from any
the jth element of Sk+1 is an k 1 
2
element of Sk+1, we get
n2
 k 1 n  2

 j   (n  1)an k  nan k 
 j  (n  1)an k
 an 
2
2


n2
 [n  (n  1)]an k 
j
2
n2
 an k 
 j.
2
n2


 j   S k , and by definition of the sequence, this difference cannot be
But  an k 
2


represented as a sum of the elements of Pk–1. Therefore, the elements of Sk+1 cannot be
represented as sums of previous terms of the sequence, and hence must be the next terms of the
sequence. Thus we have shown by induction that the sequence follows the formula stated in the
theorem. QED.
As k increases, an k 
Figure 15. Sequences Beginning with {n, n + 2}
Start with 1, 3
Get 1, 3, 5, 7, 14, 28, 56, 112, 224, 448, 896, 1792, 3584, 7168, 14336, ...
Start with 2, 4
Get 2, 4, 5, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, ...
Start with 3, 5
Get 3, 5, 6, 7, 17, 19, 53, 55, 161, 163, 485, 487, 1457, 1459, 4373, 4375, 13121, 13123, ...
Start with 4, 6
Get 4, 6, 7, 8, 9, 29, 31, 32, 121, 123, 124, 489, 491, 492, 1961, 1963, 1964, 7849, 7851, 7852, ...
Start with 5, 7
Get 5, 7, 8, 9, 10, 11, 44, 46, 47, 48, 229, 231, 232, 233, 1154, 1156, 1157, 1158, 5779, 5781, 5782,
5783, ...
Start with 6, 8
Get 6, 8, 9, 10, 11, 12, 13, 62, 64, 65, 66, 67, 386, 388, 389, 390, 391, 2330, 2332, 2333, 2334, 2335,
13994, 13996, 13997, 13998, 13999, ...
Start with 7, 9
Get 7, 9, 10, 11, 12, 13, 14, 15, 83, 85, 86, 87, 88, 89, 601, 603, 604, 605, 606, 607, 4227, 4229,
4230, 4231, 4232, 4233, ...
Start with 8, 10
Get 8, 10, 11, 12, 13, 14, 15, 16, 17, 107, 109, 110, 111, 112, 113, 114, 883, 885, 886, 887, 888, 889,
890, 7091, 7093, 7094, 7095, 7096, 7097, 7098, ...
Start with 9, 11
Get 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 134, 136, 137, 138, 139, 140, 141, 142, 1241, 1243, 1244,
1245, 1246, 1247, 1248, 1249, 11204, 11206, 11207, 11208, 11209, 11210, 11211, 11212, ...
Start with 10, 12
Get 10, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 164, 166, 167, 168, 169, 170, 171, 172, 173, 1684,
1686, 1687, 1688, 1689, 1690, 1691, 1692, 1693, 16884, 16886, 16887, 16888, 16889, 16890,
16891, 16892, 16893, ...
Start with 11, 13
Get 11, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 197, 199, 200, 201, 202, 203, 204, 205, 206, 207,
2221, 2223, 2224, 2225, 2226, 2227, 2228, 2229, 2230, 2231, ...
Start with 12, 14
Get 12, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 233, 235, 236, 237, 238, 239, 240, 241, 242,
243, 244, 2861, 2863, 2864, 2865, 2866, 2867, 2868, 2869, 2870, 2871, 2872, ...
Figure 16. Sequences beginning with {n, n + 3}
Start with 1, 4
Get 1, 4, 6, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, ...
Start with 2, 5
Get 2, 5, 6, 9, 10, 28, 29, 85, 86, 256, 257, 769, 770, 2308, 2309, 6925, 6926, ...
Start with 3, 6
Get 3, 6, 7, 8, 12, 31, 32, 94, 95, 283, 284, 850, 851, 2551, 2552, 7654, 7655, ...
Start with 4, 7
Get 4, 7, 8, 9, 10, 32, 33, 35, 132, 133, 135, 532, 533, 535, 2132, 2133, 2135, 8532, 8533, 8535,
...
Start with 5, 8
Get 5, 8, 9, 10, 11, 12, 48, 49, 51, 52, 248, 249, 251, 252, 1248, 1249, 1251, 1252, 6248, 6249,
6251, 6252, ...
Start with 6, 9
Get 6, 9, 10, 11, 12, 13, 14, 67, 68, 70, 71, 72, 415, 416, 418, 419, 420, 2503, 2504, 2506, 2507,
2508, ...
Start with 7, 10
Get 7, 10, 11, 12, 13, 14, 15, 16, 89, 90, 92, 93, 94, 95, 642, 643, 645, 646, 647, 648, 4513, 4514,
4516, 4517, 4518, 4519, ...
Start with 8, 11
Get 8, 11, 12, 13, 14, 15, 16, 17, 18, 114, 115, 117, 118, 119, 120, 121, 938, 939, 941, 942, 943,
944, 945, 7530, 7531, 7533, 7534, 7535, 7536, 7537, ...
Start with 9, 12
Get 9, 12, 13, 14, 15, 16, 17, 18, 19, 20, 142, 143, 145, 146, 147, 148, 149, 150, 1312, 1313,
1315, 1316, 1317, 1318, 1319, 1320, ...
Start with 10, 13
Get 10, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 173, 174, 176, 177, 178, 179, 180, 181, 182, 1773,
1774, 1776, 1777, 1778, 1779, 1780, 1781, 1782, ...
Start with 11, 14
Get 11, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 207, 208, 210, 211, 212, 213, 214, 215, 216,
217, 2330, 2331, 2333, 2334, 2335, 2336, 2337, 2338, 2339, 2340, ...
Start with 12, 15
Get 12, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 244, 245, 247, 248, 249, 250, 251, 252, 253,
254, 255, 2992, 2993, 2995, 2996, 2997, 2998, 2999, 3000, 3001, 3002, 3003, ...
Figure 17. Sequences beginning with {n, n + 4}
Start with 1, 5
Get 1, 5, 7, 9, 11, 29, 31, 89, 91, 269, 271, 809, 811, 2429, 2431, 7289, 7291, ...
Start with 2, 6
Get 2, 6, 7, 10, 11, 14, 45, 46, 49, 185, 186, 189, 745, 746, 749, 2985, 2986, 2989, ...
Start with 3, 7
Get 3, 7, 8, 9, 13, 14, 48, 49, 50, 195, 196, 197, 783, 784, 785, 3135, 3136, 3137, ...
Start with 4, 8
Get 4, 8, 9, 10, 11, 16, 51, 52, 53, 207, 208, 209, 831, 832, 833, 3327, 3328, 3329, ...
Start with 5, 9
Get 5, 9, 10, 11, 12, 13, 52, 53, 54, 56, 267, 268, 269, 271, 1342, 1343, 1344, 1346, 6717, 6718,
6719, 6721, ...
Start with 6, 10
Get 6, 10, 11, 12, 13, 14, 15, 72, 73, 74, 76, 77, 444, 445, 446, 448, 449, 2676, 2677, 2678, 2680,
2681, ...
Start with 7, 11
Get 7, 11, 12, 13, 14, 15, 16, 17, 95, 96, 97, 99, 100, 101, 683, 684, 685, 687, 688, 689, 4799,
4800, 4801, 4803, 4804, 4805, ...
Start with 8, 12
Get 8, 12, 13, 14, 15, 16, 17, 18, 19, 121, 122, 123, 125, 126, 127, 128, 993, 994, 995, 997, 998,
999, 1000, 7969, 7970, 7971, 7973, 7974, 7975, 7976, ...
Start with 9, 13
Get 9, 13, 14, 15, 16, 17, 18, 19, 20, 21, 150, 151, 152, 154, 155, 156, 157, 158, 1383, 1384,
1385, 1387, 1388, 1389, 1390, 1391, ...
Start with 10, 14
Get 10, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 182, 183, 184, 186, 187, 188, 189, 190, 191, 1862,
1863, 1864, 1866, 1867, 1868, 1869, 1870, 1871, ...
Start with 11, 15
Get 11, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 217, 218, 219, 221, 222, 223, 224, 225, 226,
227, 2439, 2440, 2441, 2443, 2444, 2445, 2446, 2447, 2448, 2449, ...
Start with 12, 16
Get 12, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 255, 256, 257, 259, 260, 261, 262, 263, 264,
265, 266, 3123, 3124, 3125, 3127, 3128, 3129, 3130, 3131, 3132, 3133, 3134, ...
Figure 18. Sequences beginning with {n, n + 5}
Start with 1, 6
Get 1, 6, 8, 10, 12, 32, 34, 98, 100, 296, 298, 890, 892, 2672, 2674, 8018, 8020, ...
Start with 2, 7
Get 2, 7, 8, 11, 12, 16, 50, 51, 55, 206, 207, 211, 830, 831, 835, 3326, 3327, 3331, ...
Start with 3, 8
Get 3, 8, 9, 10, 14, 15, 16, 68, 69, 70, 74, 349, 350, 351, 355, 1754, 1755, 1756, 1760, 8779, 8780,
8781, 8785, ...
Start with 4, 9
Get 4, 9, 10, 11, 12, 17, 18, 73, 74, 75, 76, 371, 372, 373, 374, 1861, 1862, 1863, 1864, 9311, 9312,
9313, 9314, ...
Start with 5, 10
Get 5, 10, 11, 12, 13, 14, 20, 76, 77, 78, 79, 386, 387, 388, 389, 1936, 1937, 1938, 1939, 9686, 9687,
9688, 9689, ...
Start with 6, 11
Get 6, 11, 12, 13, 14, 15, 16, 77, 78, 79, 80, 82, 473, 474, 475, 476, 478, 2849, 2850, 2851, 2852,
2854, ...
Start with 7, 12
Get 7, 12, 13, 14, 15, 16, 17, 18, 101, 102, 103, 104, 106, 107, 724, 725, 726, 727, 729, 730, 5085,
5086, 5087, 5088, 5090, 5091, ...
Start with 8, 13
Get 8, 13, 14, 15, 16, 17, 18, 19, 20, 128, 129, 130, 131, 133, 134, 135, 1048, 1049, 1050, 1051,
1053, 1054, 1055, 8408, 8409, 8410, 8411, 8413, 8414, 8415, ...
Start with 9, 14
Get 9, 14, 15, 16, 17, 18, 19, 20, 21, 22, 158, 159, 160, 161, 163, 164, 165, 166, 1454, 1455, 1456,
1457, 1459, 1460, 1461, 1462, ...
Start with 10, 15
Get 10, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 191, 192, 193, 194, 196, 197, 198, 199, 200, 1951,
1952, 1953, 1954, 1956, 1957, 1958, 1959, 1960, ...
Start with 11, 16
Get 11, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 227, 228, 229, 230, 232, 233, 234, 235, 236, 237,
2548, 2549, 2550, 2551, 2553, 2554, 2555, 2556, 2557, 2558, ...
Start with 12, 17
Get 12, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 266, 267, 268, 269, 271, 272, 273, 274, 275,
276, 277, 3254, 3255, 3256, 3257, 3259, 3260, 3261, 3262, 3263, 3264, 3265, ...
Figure 19. Beta Sequences, or sequences beginning with {n, 2n,...}
Start with 1, 2
Get 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, ...
Start with 2, 4
Get 2, 4, 5, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, ...
Start with 3, 6
Get 3, 6, 7, 8, 12, 31, 32, 94, 95, 283, 284, 850, 851, 2551, 2552, 7654, 7655, ...
Start with 4, 8
Get 4, 8, 9, 10, 11, 16, 51, 52, 53, 207, 208, 209, 831, 832, 833, 3327, 3328, 3329, ...
Start with 5, 10
Get 5, 10, 11, 12, 13, 14, 20, 76, 77, 78, 79, 386, 387, 388, 389, 1936, 1937, 1938, 1939, 9686,
9687, 9688, 9689, ...
Start with 6, 12
Get 6, 12, 13, 14, 15, 16, 17, 24, 106, 107, 108, 109, 110, 646, 647, 648, 649, 650, 3886, 3887,
3888, 3889, 3890, ...
Start with 7, 14
Get 7, 14, 15, 16, 17, 18, 19, 20, 28, 141, 142, 143, 144, 145, 146, 1002, 1003, 1004, 1005, 1006,
1007, 7029, 7030, 7031, 7032, 7033, 7034, ...
Start with 8, 16
Get 8, 16, 17, 18, 19, 20, 21, 22, 23, 32, 181, 182, 183, 184, 185, 186, 187, 1469, 1470, 1471,
1472, 1473, 1474, 1475, ...
Start with 9, 18
Get 9, 18, 19, 20, 21, 22, 23, 24, 25, 26, 36, 226, 227, 228, 229, 230, 231, 232, 233, 2062, 2063,
2064, 2065, 2066, 2067, 2068, 2069, ...
Start with 10, 20
Get 10, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 40, 276, 277, 278, 279, 280, 281, 282, 283, 284,
2796, 2797, 2798, 2799, 2800, 2801, 2802, 2803, 2804, ...
Figure 20. Proof of Formula for Beta Sequences
Theorem: Define the sequence S = {n, 2n,...} by setting the first two terms equal to n and 2n for
n ≥ 2, and then requiring that every integer greater than 2n be a sum of distinct terms of S. Then
this sequence is given by

S = {n, 2n, 2n + 1, 2n + 2, ..., 3n – 1, 4n}   S i
i 0
n2
where S i  {si , j } , and si , j  an i 
j 0
n2
5n  6n
 j , where a 
.
2
2
2
Proof: By definition, n and 2n are the first two terms of the sequence, and their sum is 3n, so any
number between 2n and 3n cannot be represented using only these two terms. For any two
distinct numbers 2n + r1 and 2n + r2, the difference is (2n + r2) – (2n + r1) = r2 – r1, where 0 ≤ r1
< r2 < n. Then by algebra, we have 0 < r2 – r1 < n – r1 ≤ n, so the difference is less than n. But
no number less than n exists in the sequence, so we cannot represent any number from 2n to 3n –
1 using any other such numbers. Therefore, the numbers from 2n + 1 to 3n – 1 cannot be
represented by adding previous terms in the sequence, and must therefore be the next terms of S.
Then by adding n to any number from 2n to 3n – 1, we can represent every number from 3n to 4n
– 1, so these numbers are not in the sequence. We now need to show that 4n cannot be
represented as a sum of any of the previous terms of the sequence. If we try adding n and 2n + r
for 0 ≤ r ≤ n – 1, the sum is at most 4n – 1, which is too small to represent 4n. If we try adding
any 2n + r terms, the sum is at least 2n + (2n + 1) = 4n + 1, which is too large to represent 4n.
Any sum of three terms is at least n + 2n + (2n + 1) = 5n + 1, which is also too large to represent
4n. Therefore, we cannot represent 4n as a sum of previous terms of the sequence, and thus 4n is
the next term.
n2
n
 1 = a  can
2
2
be represented by adding distinct terms of S. Let Dm be the set of all numbers that can be
represented by adding m distinct numbers in the interval [2n, 3n – 1] = [2n, 2n + n – 1] for any m
such that 1 ≤ m ≤ n, and let Em be the set of all numbers equal to n plus any element of Dm. If we
add m terms of the above interval, we get (2n + r1) + (2n + r2) + ... + (2n + m) = 2mn + r1 + r2 +
m(m  1)
... + rm. Then the smallest element of Dm is 2mn  0  1  2  ...  m  1  2mn 
,
2
m(m  1)
and the largest element of Dm is 2mn  (n  1)  (n  2)  ...  (n  m)  2mn  mn 
2
m(m  1)
 3mn 
. Since r1 + r2 + ... + rm can represent every integer from its minimum value to
2
m(m  1)
its maximum value, we know that Dm represents every number from 2mn 
to
2
m(m  1)
m(m  1)
3mn 
 n to
, and then Em represents every number from 2mn 
2
2
m(m  1)
3mn 
 n.
2
We next need to show that every number from 2n to s0,0 – 1 = a 
Lemma 1: For any value of m such that 1 ≤ m ≤ n – 1, there is continuous representation
from Dm to Em (i.e., Dm and Em either intersect or are consecutive).
Proof: We need to show that for the above values of m, when we subtract the smallest
element of Em from the largest element of Dm, the difference is greater than or equal to -1.
So we will have continuous representation if
m(m  1)  
m(m  1)


 2mn 
 n  1 .
3mn 

2
2
 

Then by algebra, we have
m2  m
m2  m
3mn 
 2mn 
 n  1
2
2
m2 m m2 m
mn 
 
  n 1  0
2
2
2
2
2
mn  m  n  1  0
n(m  1)  (m 2  1)  0
n(m  1)  (m  1)( m  1)  0
(n  1  m)( m  1)  0
This inequality is true if 1 ≤ m ≤ n – 1. QED.
Lemma 2: For any value of m such that 2 ≤ m ≤ n – 3, there is continuous representation
from Dm  E m to Dm1  Em1 .


m2  m
m2  m
Proof: By Lemma 1, we have Dm  Em = 2mn 
 n . We
, 3mn 
2
2


need to show that for the above values of m, when we subtract the smallest element of
Dm1  Em1 from the smallest element of Dm  Em , the difference is greater than or
equal to -1. This difference is equal to
m(m  1)
m(m  1) 
m(m  1)
m(m  1)

 
 n  2n(m  1) 
 3mn 
 n  2n(m  1) 
3mn 

2
2
2
2
 

= 3mn – m (m + 1) + n – 2mn – 2n
= mn – m2 – m – n.
We will have continuous representation if mn – m2 – m – n ≥ -1, i.e., if mn – m2 – m – n +
1 ≥ 0. If m = 1, then the left side of this inequality is n – 1 – 1 – n + 1 = -1, which is
negative. If m = n – 2, then we have (n – 2) n – (n – 2)2 – (n – 2) – n + 1 = n2 – 2n – n2 +
4n – 4 – n + 2 – n + 1 = -1, which is also negative. If m = 2, we have 2n – 4 – 2 – n + 1 =
n – 5. This is non-negative if n ≥ 5*. If m = n – 3, we have (n – 3) n – (n – 3)2 – (n – 3) –
n + 1 = n2 – 3n – n2 + 6n – 9 – n + 3 – n + 1 = n – 5, which we know is non-negative for n
≥ 5*. So the difference is negative for m = 1 and non-negative for m = 2, we know that
the difference equation has a root for some m such that 1 < m ≤ 2; and since the
difference is negative for m = n – 2 and non-negative for m = n – 3, the difference
equation also has a root for some m such that n – 3 ≤ m < n – 2. Since the difference
equation is a quadratic equation in m, we know that it has only two roots, and thus the
equation is non-negative for any m such that 2 ≤ m ≤ n – 3. QED.
* If n ≤ 4, then n – 3 ≤ 1. If n = 4 and m ≥ 2, then the difference equation is still negative
because m ≥ 2 = n – 2. Similarly, if n = 3 and m ≥ 2, then m ≥ 2 = n – 1 > n – 2, so the
difference equation is still negative. Therefore, any values of n less than 5 render this
difference equation extraneous.
Combining the information from both of these lemmas, we have three separate intervals
n2
of continuous representation: D1  E1 ,
 (D
m
 E m ) , and Dn1  En1 . Given the above
m2
formula for the interval Dm  E m , by substitution we have the following intervals:

5n 2  7n 
(
D

E
)

4
n

1
,
 1
 m m 
2
m2


2
2
 5n  7n
5n  3n 
Dn1  En1  
 1,

2
2


n2
D1  E1  [2n,4n  1]
Note that if n = 2, then n – 2 = 0, and n – 1 = 1, so there is only one interval, D1  E1 . If n = 3,
then n – 2 = 1 and n – 1 = 2, so the middle interval is rendered extraneous. Now let Fm be the
interval whose elements are derived by adding 4n to each element of Dm  E m . Then we have
the following intervals:

5n 2  n 
F

8
n

1
,
 1
 m 
2
m2


2
2
 5n  n
5n  5n 

 1,

2
 2

n2
F1 = [6n, 8n – 1]
Fn1
5n 2  5n 5n 2  6n n 2 2 5n 2  6n n  2
n2

   

 1  an 0 
 1 , which is one
Note that
2
2
2 2 2
2
2
2
n 1
less than the smallest element of S0. Note that
 (D
m
 E m ) represents every number from 2n
m 1
5n 2  3n
5n 2  7 n
except 4n (which is one of the terms in the sequence) and
. However, we
2
2
5n 2  7n n2
can show that
  Fm . If we subtract the smallest element in this interval, 8n + 1,
2
m2
to
5n 2  7 n
5n 2  23n
, we get
 1 , which is positive when n ≥ 5. The derivative of this
2
2
10n  23
difference is
, which has a critical value of n = 2.3 and is positive when n ≥ 3.
2
Therefore, because the difference is positive and increasing for n = 5, we know that it is positive
5n 2  7 n
5n 2  n
for all n ≥ 5. Now if we subtract
from the largest element of the interval,
 1,
2
2
5n 2  7 n 5 n 2  n
we get 8n – 1, which is positive for all n ≥ 5. Since 8n + 1 ≤
≤
 1 for n ≥ 5,
2
2
n2
5n 2  7n n2
we know that
  Fm . If n = 2, the interval  ( Dm  E m ) does not exist, and
2
m2
m2
2
5n  7n
5n 2  7 n
= 12 = 4n, which is
 3 , which by definition is not in the sequence. If n = 3,
2
2
5n 2  7 n
one of the terms of the sequence. If n = 4, then
= 26 and F1 = [24, 31], so
2
5n 2  7n
 F1 and therefore still has a representation.
2
n 1
n
5n 2  n
Now notice that  Fm represents every number from 6n to a - except 8n and
.
2
2
m 0
from
5n 2  n
, and we will find a representation
2
for this number later. If n = 4, then 8n = 32 and En–1 = E3 = [31, 34], so 8n  En–1 and therefore
If n = 2, then 8n = 16 = s0,0. If n = 3, then 8n = 24 =
n2
has a representation. If n ≥ 5, we can show that 8n   ( Dm  E m ) . If we subtract the smallest
2
element of this interval, 4n + 1, from 8n, we get 4n – 1, which is positive for all n ≥ 5. If we
5n 2  7n
5n 2  23n
 1 , we get
subtract 8n from the largest element of this interval,
 1 , which
2
2
we have already shown to be positive for all n ≥ 5.
If we add every term in the sequence from 2n to 3n – 1, we get
n 1
n(n  1)
( 2 n  i )  2n 2 

2
i 0
4n 2 n 2  n

2
2
2
5n  n

.
2
5n 2  n
Then if we add n to this sum, we get
. So now we have representations for every
2
n
number from 2n to s0,0 – 1 = a - .
2

We now need to show that the elements of S0 cannot be represented as sums of the
previous terms of the sequence. For any two distinct elements of S0, the difference is
n2
n2
s 0, j2  s 0, j1  (a 
 j 2 )  (a 
 j1 )  j 2  j1 ,
2
2
where 0 ≤ j1 < j2 ≤ n – 2 < n. Then by algebra, we have 0 < j2 – j1 ≤ n – 2 – j1 < n, so the
difference of any two elements of S0 must be strictly between 0 and n. But there are no such
terms in the sequence, so we cannot represent an element of S0 using previous elements of S0,
and we must therefore use at least one of the previous elements of the sequence. Recall that the
largest element Fn–1, i.e., the largest number that can be represented by adding n and 4n to all but
n
n2
 1 , which is too small to represent any
one term in the interval [2n, 3n – 1], is a  = a 
2
2
element of S0. Therefore, any representation of an element of S0 must use every term from 2n to
5n 2  n
3n – 1. Recall that this sum is
, which is too small to represent any element of S0, and if
2
n 1
5n 2  n
we add n to this sum, we get
, which is still too small. If we add 4n to  ( 2n  i ) , we get
2
i 0
n2
5n 2  7n 5n 2  6n n 2 2
 1 , which is too large to represent any element of

   a
2
2
2
2 2 2
S0. Therefore, the elements of S0 cannot be represented as sums of previous terms of the
sequence, and must therefore be the next terms of the sequence.
Now define the set Px  S 

S
i
, and assume that the sequence begins with Pk. We
i  x 1
now need to show that the next terms of the sequence are the elements of Sk+1.
First, we need to show that every number greater than 2n and less than the smallest term
n2
of Sk+1 (which is an k 1 
) can be represented as a sum of distinct term of the sequence. By
2
definition of the sequence, we know that every integer up to the largest element of Sk (which
n2
is an k 
) can be represented, so we just need to prove that every integer from
2
n2
n
n2
n
an k 
 1  an k  to an k 1 
 1  an k 1  can be represented. By definition, the
2
2
2
2
n
set Pk 1 represents the number n and every number from 2n to an k  , and we will call this set
2
of represented numbers R. We will now attempt to represent other numbers by adding one or
more elements of Sk to each element of R. Let Cm be the set whose elements are all of the
possible sums of m distinct elements of Sk. If we add m terms of Sk, we get
n2
man k  m
  j1  j 2  ...  j m for distinct ji  [0, n – 2]. The smallest element of Cm is:
 2 
n2
 n  2  m(m  1)
k
man k  m
  0  1  2  ...m  1  man  m

2
 2 
 2 
 n  2  m 1
 man k  m

2


 n  m 1
 man k  m
,
2


and the largest element of Cm is:
n2
man k  m
  (n  2)  (n  2)  1  (n  2)  2  ...  (n  2)  (m  1)
 2 
m(m  1)
n2
 man k  m
  m(n  2) 
2
 2 
 n  2  m(m  1)
 man k  m

2
 2 
 n  2  m 1
 man k  m

2


 n  m 1
 man k  m
.
2


m
Since
j
i 1
i
can represent every integer from its minimum value to its maximum value, we know
 n  m 1 
 n  m 1 
k
that Cm represents every integer from mank  m
 to man  m
 . By adding
2
2




 n  m 1
n to each element of Cm, we can represent every number from man k  m
  n to
2


 n  m 1
man k  m
  n , and we will call this interval Am. If we add each element cj  Cm to
2


n
each element of R, we can represent every number from 2n + cj to an k   c j . Since the
2
elements of Cm are consecutive integers, these intervals intersect, so we can combine these
 n  m 1
intervals and find that we have representations for every number from man k  m
  2n
2


n
 n  m 1
 n  m 1 n
k
k
to man k  m
  an   (m  1)an  m
  , and we’ll call this interval
2
2
2



 2
Bm.
If add only one element of Sk (i.e., the elements of C1) to each element of R, by the
n2
formulae proven above, we have representations for every number from an k 
to
2
n2
n
n2
n
n2
an k 
 an k   1 , from an k 
 n  an k   1 to an k 
n
2
2
2
2
2
n
n2
n
n2
n
 an k   n  1 , and from an k 
 2n  an k   n  1 to an k 
 an k 
2
2
2
2
2
 2an k  1 . However, we are missing representations for an k 
n2
 s   (an i 
sS i
j 0
n
n
and an k   n . Note that
2
2
n2
 j)
2
 i n  2  n2
 (n  1) an 
 j
2  j 0

 n  2  (n  1)( n  2)
 (n  1)an i  (n  1)

2
 2 
i
 (n  1)an .
Then we have
k 1
 p  n  2n  2n  1  ...  2n  n  1  4n   (n  1)an i
pPk 1
i 0
n 1
k 1
i 0
i 0
 n   (2n  i )  4n  (n  1)a  n i
 n  2n 2 
 nk 1
n(n  1)

 4n  (n  1)a
2
 n 1 
n2  n
 n  2n 
 4n  an k  a
2
2 n 4 n 2 n 2  n 8n 5 n 2  6 n
 an k 




2
2
2
2
2
3n
 an k 
2
n
 an k   n ,
2
2
n
 p .
2 pPk 1 {n}
Now if n ≥ 4, we can add n – 2 terms of Sk, and we have the following intervals of
representation:

 n  n  2 1
 n  n  2  1 
k
C n 2  (n  2)an k  (n  2)
, (n  2)an  (n  2)

2
2





n2
n  2

 (n  2)an k 
, (n  2)an k 
2
2 

n2
n 

 (n  2)an k 
, (n  2)an k   1
2
2 

n2
n2


An  2  (n  2)an k 
 n, (n  2)an k 
 n
2
2


n
n


 (n  2)an k   1, (n  2)an k   n  1
2
2


and then an k 
n2
n2
n

Bn  2  (n  2)an k 
 2n, (n  2)an k 
 an k  
2
2
2

n
 [( n  2)an k   n  1, (n  1)an k  1]
2
n
n
We are missing representations for (n  2)an k  and (n  2)an k   n . We know that
2
2
(n  3)an k  C n 3 and is therefore represented by the sequence. Then we have (n  2)an k 

 p and (n  2)an
pPk 1 {n}
k

n
n
2
n
2
p.
pPk 1
If we add all n – 1 terms of Sk, we get (n – 1) ank. Then if we add this to each element of
R, we can represent (n – 1) ank + n and every number from (n – 1) ank + 2n to (n – 1) ank + ank –
n 1
n
n
 an k 1  . However we are missing representations for  (n  1)an k  j as well as
2
2
j 1

n 1
 (n  1)an
k

 n  j . Note that
j 0

 n2 
n2
n
k
c

p
 j  an k  

   (n  2)an 


2
2
pPk 1 {n}
cC n  2 
j 0 
n2
  (n  1)an k  1  j
j 0
n 1
  (n  1)an k  j,
j 1
and

 n2 
n2
n

k
 j  an k   n
c   p    (n  2)an 

2
2

pPk 1 
cC n  2 
j 0 
n2
  (n  1)an k  n  1  j
j 0
n 1
  (n  1)an k  n  j.
j 1
If n ≥ 5, then n – 2 ≥ 3, and we need to show the numbers that can be represented by
adding m terms of Sk for 1 < m < n – 2. If we subtract the smallest element of Am from the largest
element of Cm, we have

 n  m  1  
 n  m 1 
k
k
  man  m
  n  m(n  m  1)  n
man  m
2
2

 

 

 mn  m 2  m  n .
We will have a continuous representation from Cm to Am if mn – m2 – m – n ≥ -1, i.e., if mn – m2
– m – n + 1 ≥ 0. If m = 1, then we have mn – m2 – m – n + 1 = n – 1 – 1 – n + 1 = -1 < 0, so the
difference equation is negative for m = 1. If m = 2, then we have mn – m2 – m – n + 1 = 2n – 4 –
2 – n + 1 = n – 5, so the difference is non-negative for n ≥ 5. Since the difference is negative for
m = 1 and non-negative for m = 2, then the difference equation has a root for some 1 < m ≤ 2. If
m = n – 2, then the difference is (n – 2) n – n – (n – 2)2 – (n – 2) + 1 = n2 – 2n – n – n2 + 4n – 4 –
n + 2 + 1 = -1 < 0, so the difference is negative for m = n – 2. If m = n – 3, then the difference is
(n – 3) n – n – (n – 3)2 – (n – 3) + 1 = n2 – 3n – n – n2 + 6n – 9 – n + 3 + 1 = n – 5, which is nonnegative for n ≥ 5. Since the difference is negative for m = n – 2 and non-negative for m = n – 3,
then the difference equation has a root for some n – 3 ≤ m < n – 2. Since mn – m2 – m – n + 1 is
a quadratic equation in m, we know that it has no more than two roots, so there are no roots
between 2 and n – 3. Therefore, the difference of the largest term of Cm minus the smallest term
of Am will be non-negative for all m from 2 to n – 3 inclusive, and therefore Cm and Am intersect
for these values of m. Similarly, if we subtract the smallest value of Bm from the largest value of
Am, we have


 n  m 1  
 n  m 1
k
k
  n  man  m
  2n  m(n  m  1)  2n
man  m
2
2

  




 mn  m 2  m  n .
We will have a continuous representation from Am to Bm if mn – m2 – m – n ≥ -1, i.e., if mn – m2
– m – n + 1 ≥ 0. We have already shown that this is true if 2 ≤ m ≤ n – 3. Therefore, the
difference of the largest term of Am minus the smallest term of Bm will be non-negative for all m
from 2 to n – 3 inclusive, and therefore Am and Bm intersect for these values of m, and thus we
 n  m 1
 n  m 1 n
k
can represent every number from man k  m
 to (m  1)an  m

2
2



 2
inclusive.
Now if n ≥ 3, we must prove that there are no gaps between the intervals of
representation for consecutive values of m. If we add m + 1 terms, then the smallest element of
nm2
Cm+1 is (m  1)an k  (m  1)
 . If we subtract the smallest element of Cm+1 from the
2


largest element of Bm, the difference is

 n  m 1 n  
 n  m  2 
k
k
    (m  1)an  (m  1)

(m  1)an  m
2
2

 2 



 n  m 1 n
nm2
 m
   (m  1)

2
2

 2


2
2
mn  m  m  n  mn  m  2m  n  m  2

2
2
2mn  2m  4m  2

2
2
 mn  m  2m  1 .
We will have continuous representation over consecutive values of m if this difference is greater
than or equal to -1, i.e., if mn – m2 – 2m ≥ 0. We can factor this inequality into m (n – 2 – m) ≥ 0.
This is true if 0 ≤ m ≤ n – 2, which by definition is true because there are n – 1 elements in Sk. (If
m = n – 1, we cannot add n elements of Sk, so there is no interval of representation for m = n.)
Therefore, there are no gaps between the intervals of representation for consecutive values of m,
n
and thus we have shown that every number from 2n to an k 1  can be represented as a sum of
2
distinct terms of the sequence.
Our next step is to show that each element of Sk+1 cannot be represented by the terms of
n2
the sequence up to an k 
. For any two distinct elements of Sk+1, the difference is
2
n2
 k 1 n  2
 

 j 2    an k 1 
 j1   j 2  j1 , where 0 ≤ j1 < j2 ≤ n – 2 < n. But we
 an 
2
2

 

already proved earlier that the difference j2 – j1 is not represented in the sequence, so we cannot
represent any element of Sk+1 using previous elements of Sk+1. Now note that if we subtract the
n2
sum of the elements of Pk 1 from an k 1 
, the smallest element of Sk+1, the smallest that
2
the difference can be is
n 2 n 2n
 k 1 n  2   k n

k
 an 
   an   n   (n  1)an    
2  
2
2 2 2 2


k
 (n  1)an  2n  1
 n[( n  1)an k 1  2]  1 .
5n 2  6n
, by substitution we have a ≥ 16. Then if k
2
= 0, we have (n  1)an k 1  1  16  0.5  8  2 . Therefore, both factors n and [( n  1)an k 1  2] are
positive, and since both will remain positive for larger values of n and k, we know that the
n2

difference  an k 
   p will be greater than zero. Thus each element of Sk+1 is too large
2  pPk 1

to represent using only the elements of Pk 1 , and hence any representation of an element of Sk+1
must use at least one element of Sk.
We know that n ≥ 2, so n – 1 ≥ 1. Since a 
Consider the set C n 2 , the set whose elements are the possible sums of all but one term of
n2
 n  n  2 1
k
Sk. The largest element of C n 2 is (n  2)an k  (n  2)
. If we
  (n  2)an 
2
2


subtract this from the smallest element of Sk+1, we get
n  2
n2
n2
 k 1 n  2  
k
 nan k 
 (n  2)an k 
 an 
  (n  2)an 

2  
2 
2
2

k
 2an  n  2 .
Then we have
n  2
n
 k 1 n  2  
k
  p  (2an k  n  2)  (an k   n)
 an 
  (n  2)an 

2  
2  pPk 1
2

2n
n
2n
 
2
2
2
2
5n
 an k 
2
2
5n
5n 2  6n  5n
As k increases, this difference increases. If k = 0, we have a 
2
2
2
2
5n 2  n

 2  0 . Therefore, the difference is always greater than zero, so if we subtract the
2
largest element of C n 2 from the smallest element of Sk+1, the difference is too large to be
 an k 
represented using just the elements of Pk 1 . If we use a smaller element of C n 2 and/or a larger
element of Sk+1, the difference will still be too large. Therefore, any representation of an element
of Sk+1 must use every element of Sk+1. We have shown earlier that the sum of all of the elements
of Sk is (n – 1) ank. If we subtract this from the jth element of Sk+1, we get
n2
 k 1 n  2

 j   (n  1)an k  nan k 
 j  (n  1)an k
 an 
2
2


n2
 (n  n  1)an k 
 j
2
n2
 an k 
 j.
2
n2


 j   S k , and by definition of the sequence, this difference cannot be
But  an k 
2


represented as a sum of the elements of Pk 1 . Therefore, the elements of Sk+1 cannot be
represented as sums of previous terms of the sequence, and hence must be the next terms of the
sequence. Thus we have shown by induction that the sequence follows the formula stated in the
theorem. QED.
Works Cited
Honsberger, Ross. Mathematical Gems III. Washington, D.C.: Mathematical Association of
America, 1985.
Knapp, Michael. Unpublished proof.
Lowe, Doug and Barry Burd. Java All-In-One Desk Reference for Dummies. Hoboken, NJ:
Wiley Publishing, 2007.
Schissel, Eric. “Characterizations of Three Types of Completeness.” Fibonacci Quarterly 27
(1989), 409-419.