Download Differences between consecutive numbers of quadruples Take 4

Differences between consecutive numbers of quadruples
Take 4 integers in a row (like 2 6 4 13). Take the (magnitude) of
their differences of consecutive integers (where the last number is subtracted from the first): 4 2 9 11. Continue to do that, and observe –
(2 7 2 7; 5 5 5 5) they eventually become all equal.
Problem: Can for every number of steps it should take until the differences are all equal a suitable quadruple of four numbers be found?
If yes, prove it, and, if possible, exhibit the quadruple for the given
number of steps . If no, for which number of steps can we find no such
quadruple?
1st Example: A sequence of 4 Steps:
1234
1113
0022
0202
2222
2nd Example: 4 Steps
1 3 7 11
2 4 4 10
2068
2626
4444
3rd Example: 6 Steps
5 3 0 11
2 3 11 6
1854
7313
4224
2020
2222
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→ Maximum of the four numbers becomes smaller or stays equal at
each step.
Let’s call the number of steps (4, 4, and 6 above) of these sequences
their length. The question is: Given any sequence, can we always find
another longer sequence?
Given a quadruple a = (a1, a2 , a3, a4 ), let us focus on finding an increasing quadruple
b = (b1, b2, b3 , b4 )
meaning b1 ≤ b2 ≤ b3 ≤ b4, which has a as its quadruple of differences:
a1 = b2 − b1 , a2 = b3 − b2, a3 = b4 − b3, and a4 = b4 − b1 .
If we add the first of these three equations, we get:
a1 + a2 + a3 = b4 − b1 = a4 .
(1)
Let’s see if this is fulfilled, for let’s say:
a = (1, 2, 3, 4).
Hmm, not fulfilled: 4 6= 1 + 2 + 3. However, if we reduced every entry
by one:a′ = a − (1, 1, 1, 1), then
a′ = (0, 1, 2, 3)
and indeed, a′4 = 3 = 0 + 1 + 2.
Interesting: the sequence of differences of a′ will be the same as that
of a. It therefore has the same length. So, let’s try to find the b for
the new a′ : Let’s pick any number for b1, let’s say: 0.
Then b2 has to be a′1 , since the difference between b2 and b1 = 0 has
to be a′1 . So, also, b2 = 0.
After that: pick b3 such that its difference with b2 is a′2 = 1. So:
b3 = a′2 + b2 = a′2 + a′1 = 1 + 0 = 1. finally, b4 must be chosen such
that its difference with b3 = 1 is a′3 = 2, so b4 = 3.
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Let’s check if it is consistent with b4 − b1 = a′4 : Yes, 3 − 0 = 3, Check!.
This last thing works out because we made sure that (1) is fulfilled.
So, the new sequence starts with b = (0, 0, 1, 3) and has length 5.
Trying the same trick again, leads us to adding to each of the four
numbers 1, to end up at the quadrupel
c = (1, 1, 2, 4).
Nicely, this quadruple fulfills c4 = c1 + c2 c3 . So, (taking a discrete
’antiderivative’): d1 = 0, d2 = c1 = 1, d3 = c1 + c2 = 2, and d4 = c4 = 4
gives us a quadruple of length l(d) = 6, where
d = (0, 1, 2, 4).
Finally, however, the end of easy-street seems to have been reached:
Observe that, while d4 = 4 > 0 + 1 + 2 = d1 + d2 + d3 , we are not able
to fi this by a shift: Adding one to every coordinate of d gives us
d′ = (1, 2, 3, 5)
which gives that now d′4 = 5 < 6 = 1 + 2 + 3 = d′1 + d′2 + d′3 . We shifted
past our goal, but we can’t shift by less than 1.
We anticipate, that shifting the integers alone is not enough, but that
multiplication may help.
If multiplying each element of b with z ∈ Z, a quadruple emerges that
has the corresponding multiple differences of coordinates. Repeating
the process of taking differences gets us a sequence of equal entries
which are multiples of the initial terminal quadruple:
(10 10 20 40)
Therefore this new sequence has the same length!
So, while we couldn’t continue to extend the sequence backwards by
shifting the entries of the last quadruple, we could pass to another
sequence (by multiplication) where the shift operation works again:
So, let us multiply d by 2, to obtain e:
e = (0, 2, 4, 8).
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If we add 1 to the entries of this qudruple, we get e′ with:
e′ = (1, 3, 5, 9).
Indeed: e′4 = 9 = 1 + 3 + 5 = e′1 + e′2 + e′3 ! So now we can continue to
form the quadruple of which this will be the quadruple of differences:
f1 = 0, f2 = e′1 = 1, f3 = e′1 + e′2 = 4, f4 = e′4 = 9, which is the last
quadruple of a sequence of length 6.
Note that through consecutive multiplications and shifts the order of
the first coordinate being the smallest and the fourth being the largest
isn’t ever changed. Therefore, the condition of the last entry having
to be the sum of the first three will always be sufficient for finding a
sequence with length increased by one.
Here is the link to a program in R and some of its results, computing
sequences of arbitrary length;
http://web.cs.du.edu/~sobieczk/mathclub/programs/
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