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DePaul Discoveries
Volume 1 | Issue 1
Article 13
2012
A Generalization of Pascal’s Triangle
Eliya Gwetta
DePaul University, [email protected]
Adrian Pacurar
DePaul University, [email protected]
Elizabeth Mai Smith
DePaul University, [email protected]
Follow this and additional works at: http://via.library.depaul.edu/depaul-disc
Part of the Physical Sciences and Mathematics Commons
Acknowledgements
Faculty Advisors: T. Kyle Petersen, Bridget Tenner
Recommended Citation
Gwetta, Eliya; Pacurar, Adrian; and Smith, Elizabeth Mai (2012) "A Generalization of Pascal’s Triangle," DePaul Discoveries: Vol. 1: Iss.
1, Article 13.
Available at: http://via.library.depaul.edu/depaul-disc/vol1/iss1/13
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Gwetta et al.: A Generalization of Pascal’s Triangle
D E PA U L D I S C O V E R I E S ( 2 O 1 2 )
A Generalization of Pascal’s Triangle
Eliya Gwetta, Adrian Pacurar
and Elizabeth Mai Smith*
Department of Mathematical Sciences
1. Introduction
The branch of mathematics particularly interested in the study of discr
ABSTRACT
called Combinatorics. Enumerative Combinatorics is an area that dea
Combinatorics is a branch of mathematics interested in the study of finite, or countable, sets. In
particular, Enumerative Combinatorics is an area interested inpatterns
counting
many ways
patterns are
created,
such for finite set
canhow
be formed,
and developing
counting
techniques
as counting permutations and combinations. Brenti and Welker, authors of “The Veronese Construction for Formal
an interesting
relationship
between
Eulerian
Power Series and Graded Algebras,” seek an explanation for a illustrates
combinatorial
identity posed
in their
research.
Using numbers and b
techniques practiced in this area of mathematics, we have discovered
that certain
numbers appearing
in their
very important
combinatorial
quantities.
Brentiidentity
and Welker, in [2],
hold properties similar to properties of the well-known binomial coefficients.
analogous to Worpitzky's identity. When this new identity is expanded,
possess properties similar to the properties of the binomial coefficient
coefficients as geonomial coefficients; they represent the bulk of our res
INTRODUCTION
PA S C A L’ S T R I A N G L E A N D B I N O M I A L
2. Pascal's Triangle
C O E F F I C I Eand
N TS Binomial Coefficients
The branch of mathematics particularly interested
in the study of discrete and finite structures is called
Combinatorics. Enumerative Combinatorics is an
A well-known construction in mathematics, called
A well-known construction in mathematics, called Pascal's Triangle
Pascal’s Triangle, contains many beautiful properties.
sets. Worpitzky’s identity [1] illustrates an interesting
The
numbersThe
constructed
here originally
arose from
properties.
numbers constructed
here originally
arose from Hindu
Hindu studies of combinatorics, and will be defined later.
and will
be defined
Before the
we reader
identifyshould
any properties,
the reader
Before
we identify
anylater.
properties,
take
relationship between Eulerian numbers and binomial
some
time them
to discover
them in the
following table.
discover
in the following
table.
area that deals with how many ways patterns can be
formed, and developing counting techniques for finite
coefficients,
two
very
important
combinatorial
quantities. Brenti and Welker, in [2], introduce a new
identity analogous to Worpitzky’s identity. When this
TA B L E 1 : PA S C A L’ S T R I A N G L E
new identity is expanded, the coefficients that arise
possess properties similar to the properties of the
binomial coefficients. We refer to these new coefficients
as geonomial coefficients; they represent the bulk of our
research.
* Author names listed alphabetically
n
0
1
2
3
4
5
6
7

Faculty Advisors: T. Kyle Petersen, Bridget Tenner
Table 1: Pascal's Triangle
k = 0,, n
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1

Department of Mathematical Sciences, Summer 2011
Author contact: [email protected]
The reader has most likely seen these numbers before, whether or not t
— 168 —
Published by Via Sapientiae, 2012
1
DePaul Discoveries, Vol. 1 [2012], Iss. 1, Art. 13
A G E N E R A L I Z AT I O N O F PA S C A L ' S T R I A N G L E
Recall from high school algebra that students were
SY M M E T RY
asked to expand (multiply) the following: (a+b)(a+b).
The most direct observation would be that symmetry lies
A helpful technique usually learned at the time, called
in each row. By cutting the triangle directly through the
“FOIL” (acronym for first, outside, inside, last), made
center, it is clear that the numbers mirror one another.
this computation more straightforward. This method,
The following observation should provide some intuition
however, quickly becomes tedious as the number of
for why this happens: by choosing 0 elements from a set
binomials to multiply increases. To confirm this claim,
of n , there are exactly n elements remaining, so therefore
try to compute the following using the “FOIL” method:
the number of ways to choose 0 is the same as the number
(a+b)(a+b) (a+b)(a+b). Certainly this method will give
of ways to choose n ; there is only one way. Similarly,
a correct solution, but it is an inefficient approach. A
the number of ways to choose 1 element from this set
helpful theorem, named the binomial theorem, provides
is exactly the number of ways to choose n — 1 elements.
a more canny procedure. The following list is easily
The number of ways to choose 2 elements is exactly the
computed by the binomial theorem.
number of ways to choose n — 2 elements from the same
Recall from high school algebra that students were asked to expand (multiply) the following:
set, and so on.
(a  b)(a  b) . A helpful technique usually learned at the time, called ``FOIL" (acronym for first,
(a+b) = 1a+1b
outside, inside, last), made this computation more straightforward. This method, however, quickly
(a+b)(a+b) = 1a2 + 2ab + 1b2
Recall from
school
thattostudents
asked To
to expand
becomes tedious
as thehigh
number
of algebra
binomials
multiplywere
increases.
confirm(multiply)
this claim,the
tryfollowing:
to
(a+b)(a+b) (a+b) = 1a + 3a b + 3ab + 1b
3
2
2
3
A helpful
technique
usually(alearned
``FOIL" (acronym
for first,
(a following
b)(a  b) .using
this
compute the
the ``FOIL"
method:
 b)(a atbthe
)(a time,
 b)(acalled
 b) . Certainly
and (a+b)(a+b) (a+b)(a+b)= 1a + 4a b + 6a b + 4ab + 1b
4
3
2
2
3
4
R OW S U M
Here is a small experiment for the reader: choose any
row, and add up all of the numbers in this row. Observe
outside,
madebut
thisitcomputation
more
straightforward.
This
method,
however,
method will
give ainside,
correctlast),
solution,
is an inefficient
approach.
A helpful
theorem,
named
the quickly
number
binomialsThe
to multiply
increases.
To confirm
thisbyclaim,
binomial becomes
theorem, tedious
providesasa the
more
cannyofprocedure.
following
list is easily
computed
the try to
the sum of all numbers in row three and row four: when
If it is not clear why the previous expansions deserved
n = 3 , the row sum is 1+3+3+1 = 8 ; when n = 4 , the row
some space in this paper, observe the coefficients that
sum is 1+4+6+4+1 =16 . Now notice, 8 = 23 and 16 = 24 ,
binomial compute
theorem. the following using the ``FOIL" method: (a  b)(a  b)(a  b)(a  b) . Certainly this
(a  bbut
) = it1aisan
1b,inefficient approach. A helpful theorem, named the
method will give a correct solution,
and see the exponents here match the value of n that
2
appear in this list,
Triangle.
Recall
from high school algebra that students were asked to expand (multiply) the following:
(a  b)and
(a  b)refer
= 1a 2 back
2ab  1bto
, Pascal’s
binomial theorem, provides a more canny 3procedure.
The2 following
list is easily computed by the
2
3
(
a

b
)
(
a

b
)(
a

b
)
=
1
a

3
a
b

3
ab

1
b
was
chosen.
In
general,
by
selecting
any
single
row, the
The coefficients that arise in the previous
Recall
high
from
school
highalgebra
school
that
algebra
students
that
were
were
to
asked
expand
to(multiply)
expand (multiply)
the first,
following:
the following:
helpful
technique
usually
learned
at
thestudents
time,asked
called
``FOIL"
(acronym
for
(a expansions
b)(a Recall
b) . A from
binomial
andtheorem.
(a  b)(a  b)(a  b)(a  b) = 1a 4  4a 3b  6a 2 b 2  4ab 3  1b 4 .
n
of alltechnique
numbers
in
row
will
be
2 ,forwhere
are well-known as binomial coefficients, which
make
Aathis
helpful
. Atechnique
helpful
usually
learned
usuallyat
learned
thethis
time,
atmethod,
the
called
time,
``FOIL"
calledalways
``FOIL"
(acronym
for first,
first,
(a  blast),
)(aup
(amade
b) b. )(
 sum
b)computation
outside, inside,
more
straightforward.
This
however,
quickly(acronym
(a  b) = 1some
a  1bspace
,
If it is not clear why the previous expansions deserved
in this paper, observe the
n
is
the
row
index.
Wow!
2
2
the numbers in Pascal’s
Consider
a situation
outside,asinside,
outside,
last),
inside,
made
thismade
computation
this
computation
more
straightforward.
moreTo
straightforward.
This
This try
however,
method,
quickly quickly
(a  b)(aTriangle.
 b) = 1a  2ab
 1b , becomes
tedious
the number
of last),
binomials
to
multiply
increases.
confirm
thismethod,
claim,
to however,
coefficients that appear in this list, and refer back to Pascal's Triangle.
(a  b)(a  b)(
) = 1a 3  3the
a 2 b  3number
ab 2  1b 3
in which one wanted
toa  bknow
of
ways
becomes
becomes
tedious
as
tedious
the
number
as
the
of
number
binomials
of
binomials
to
multiply
to
multiply
increases.
increases.
To
confirm
To
this
confirm
claim,
this
tryclaim,
to try to
compute3 the following
using the ``FOIL" method: (a  b)(a  b)(a  b)(a  b) . Certainly
4
2
andin
(athe
 b)previous
(a  b)(aexpansions
 b)(a  b) are
= 1awell-known
 4a 3b  6as
a 2 bbinomial
 4ab coefficients,
 1b 4 .
The coefficients that arise
A LT E R N AT I N G S U M
three cards could be selected from a standard
deck
of
Certainly
this
this
compute
the following
the following
using
``FOIL"
the ``FOIL"
method:
method:
(a  b)(Aa(helpful
ab)(ba)(atheorem,
b)(ba)(ab)named
b. )(
a  bthe
) . Certainly
method will
give a compute
correct
solution,
but the
it isusing
an inefficient
approach.
which makeIf up
numbers
in Pascal's
Triangle.
Consider
a situation
whichinone
to
it isthenot
clear why
the previous
expansions
deserved
someinspace
thiswanted
paper, observe
the
Let’s now consider the alternating sums, where every
fifty-two. This value is 22,100. Only the binomial
most theorem,
patient
method will
method
give
will
a
correct
give
a
solution,
correct
solution,
but
it
is
an
but
inefficient
it
is
an
inefficient
approach.
approach.
A
helpful
A
theorem,
helpful
theorem,
named
the
named
the
provides a more canny procedure. The following list is easily computed by the
know the coefficients
number of ways
three cards
be selected
from
a standard
deck of fifty-two. This
that appear
in thiscould
list, and
refer back
to Pascal's
Triangle.
other number is subtracted from the total. For example,
human being would dare to count this bybinomial
hand.theorem.
These
binomialbinomial
theorem,theorem,
provides provides
a more canny
a more
procedure.
canny procedure.
The following
The following
list is easily
list computed
is easily computed
by the by the
value is 22,100.
Only the most
patient
beingexpansions
would dareare
to well-known
count this by
These
The coefficients
that arise
in human
the previous
as hand.
binomial
coefficients,
the alternating sum of row three is 1 – 3 + 3 – 1 = 0, and
binomial coefficients enumerate combinations, binomial
namely,
binomial
theorem.theorem. (a  b) = 1a  1b,
binomial which
coefficients
enumerate
combinations,
numbera situation
of ways intowhich
choose
make up
the numbers
in Pascal's namely,
Triangle.the
Consider
onekwanted to
the(a alternating
 b)(a  b) = 1a 2 sum
2ab  1of
b 2 , row four is 1 – 4 + 6 – 4 +1= 0. This
the number of ways to choose k elements from a set of
(a  b) =(1aab1)b,= 1a  1b,
know the number of ways three cardscould
n  be selected from a standard deck of fifty-two. This
(a pattern
b)(a  b)(a is
 b)indeed
= 1a 3  3a 2 btrue
23ab 2for
 1b 3every
Every alternating
(a  b)(a(a b)b=)(1aa 
b)2=ab1a21
b 22,ab  1b 2row.
,
understand
exactly
what
n (formally
denoted
elements from
a set of n (formally
denotedas
as   ).).ToTo
understand
exactly what
this is counting,
4
3
2 2
k
and (a  b)(a  b)(a  b)(a  b) = 1a  4a b 36a b 2 34ab 3 2
1b 4 .
 human

(a  b)(a(a b)(ba)(a b)b=)(1aa 
b)3=
a 1ba 3ab
3a 2 b 1b33ab 2  1b 3
value is 22,100. Only the most patient
being would dare to count this by hand. These
this is counting, let’s dissect a smaller example. Say that
row sum, within this famous triangle, is 0.
4
2 32
2
and
(a  band
)(expansions
a(a b)(ba)(a b)(
ba)(a b)b=)(some
1aa4 
b)4space
=a 13ba 
6a4this
ab bpaper,
 46ab
a 2 b3 observe
1b44ab
. 3the
 1b 4 .
let's dissect
a smaller
example. Say
that we wanted
to count namely,
all possible
2clear
from
achoose
If it of
is of
not
the
previous
deserved
in
k
binomial
coefficients
enumerate
combinations,
thecombinations
number
ways
to why
we wanted to count all possible combinations of 2 from a
it is
not
If clear
itinisthis
not
why
clear
the
previous
previous
expansions
expansions
deserved
deserved
some space
some
in space
this paper,
in thisobserve
paper, the
observe the
coefficients If
that
appear
list,
andwhy
refertheback
to Pascal's
Triangle.
 4
set of 4 objects,
denoted
as of
label each
object
set
as {1,2,3,4}
.each
Order
does
 2 n, and
set of
4 aobjects,
denoted
asas inn the
and
labelexactly
object
elements
from
set
(formally
denoted
).,To
understand
what
thisnot
is counting,PA S C A L’ S R EC U R R E N C E
 
k
coefficients
coefficients
appear
this list,
in expansions
and
this refer
list, and
back
refer
to
Pascal's
back to Triangle.
Pascal's
Triangle.
 
The coefficients
thatthat
arise
in that
theinappear
previous
are
well-known
as binomial
coefficients,
in the set as {1,2,3,4}. Order does not matter in this
Pascal’s Triangle features yet another interesting
situation, so we can have {(1,2), (1,3) (1,4), (2,3), (2,4), (3,4)}.
property: if the entries in the previous row are given,
the value appearing in Pascal’s Triangle. The patterns
recurrence relation). For insight, let’s understand exactly
found in Pascal’s Triangle may come as a surprise when
what this means. When n = 3 and k = 2, the binomial
The
coefficients
arisethat
in the
arise
previous
in the
previous
expansions
are well-known
well-known
as binomial
as binomial
coefficients,
coefficients,
let's dissect a smaller example. Say that we wanted to count all possible
combinations
2 from
athat
which
makeThe
up coefficients
theofnumbers
in Pascal's
Triangle.
Consider
a expansions
situation
in are
which
one
wanted
to
which
upmake
the
numbers
up
the could
numbers
in Pascal's
in Triangle.
Pascal's
Triangle.
Consider
Consider
a situation
a situation
in whichinone
which
wanted
one to
wanted to
know the which
numbermake
of
ways
three
cards
be selected
from
a standard
deck of
fifty-two.
This
 4
set
of 4 objects,
as   , and combinations,
label each object in thewhich
set as {1,2,3,4}
. Order does notit is possible to obtain any number in this triangle (a
There
are denoted
six possible
is exactly
 2
know the
know
number
themost
of
number
ways
of
three
ways
cards
three
could
cards
be could
selected
be selected
from
a standard
from
a standard
deck
fifty-two.
deck of fifty-two.
This
This
value is 22,100.
Only
the
patient
human
being
would
dare
to count
this by
hand.ofThese
is value
22,100.
is Only
22,100.
thecombinations,
Only
mostthe
patient
mostnamely,
human
patientbeing
human
would
beingdare
would
to count
dare
this
count
by hand.
this
These
hand. These
k by
binomial value
coefficients
enumerate
the number
of
ways
to to
choose
binomialbinomial
coefficients
coefficients
enumerate
enumerate
combinations,
namely, namely,
the number
the number
of ways ofto ways
choose
to kchoose k
 n  combinations,
denoted as   ).=To
exactlyin
what
thisprevious
is counting, row when n
elements from a set of n (formally
3.understand
Now look
the
coefficient
introduced, but indeed a valid proof is well-established
k 
n
n
  ).+
 ).=To2 +1=
na set
(formally
(formally
denoted
as
Tounderstand
understand
exactly
exactly
thiswhat
is counting,
this is counting,
elements
from a setfrom
of
of n see
3.what
Coincidence?
= 2,
and
thatasdenoted
for each property that will be mentioned. let's dissectelements
 k all
 possible
 k  combinations of 2 from a
a smaller example. Say that we wanted to count
let's dissect
let'sa dissect
smaller
Say that we
Saywanted
that wetowanted
count all
to count
possible
all possible
combinations
combinations
of 2 fromofa 2 from a
 a4example.
smaller example.
set of 4 objects, denoted as   , and label each object in the set as {1,2,3,4}. Order does not
 2
4
4
 
 
 , and
set of 4 objects,
set of 4 denoted
objects, denoted
as   , as
and label
each
label
object
eachinobject
the setinasthe{1,2,3,4}
set as {1,2,3,4}
. Order does
. Order
not does not
 2
 2
— 169 —
http://via.library.depaul.edu/depaul-disc/vol1/iss1/13
2
The coefficients
The coefficients
that
arise in
that
thearise
previous
in theexpansions
previous
expansions
are well-known
are well-known
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binomial coefficients,
The coefficients that arise in the previous
expansions
are well-known
as binomial
coefficients,
which
make
which
up
the
make
numbers
up
theinnumbers
Pascal's
inTriangle.
Pascal's
Consider
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situationainsituation
which one
in which
wantedone
to wanted to
which make up the numbers in Pascal's
Triangle.
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Gwetta et al.: A Generalization of Pascal’s Triangle
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thatas from
 k  ).
k 
k 
 
immediately identifiable is the
symmetry in every row. The reason behind this follows a
In order to obtain
a particular
entry,
add
the
following
let'swedissect
let's
a smaller
aallsmaller
Say
example.
that weSay
wanted
thatofwe
to
count
wanted
count allcombinations
possible combinations
of 2 from aof 2 from a
let's dissect a smaller example. Say that
wanted
todissect
countexample.
possible
combinations
2 from
a alltopossible
similar concept used to explain the symmetry of Pascal’s
two consecutive numbers that are in the previous row:
 4
 4
 4
of 4each
objects,
set of 4denoted
objects,asset
denoted
as label
and
each
and label
object
in the
object
set as
in the
set as. {1,2,3,4}
Order does
. Order
not does not
{1,2,3,4}
set of 4 objects, denoted as   , andsetlabel
doeseach
not
 as
 ,{1,2,3,4}
 2 . , Order
Triangle
. We
will illustrate
the entry directly
above,object
andin the
the entry
immediately
to
 2
 
 2
another reason when we
the left of this value. The following table is modified
discuss the recurrence relation. Other similarities are not
to illustrate the recurrence relation of the binomial
as direct, however, the reader should take a moment to
coefficients.
discover patterns on their own.
Hopefully this triangle is becoming more and more
beautiful to the reader! The following section introduces
a new type of triangle not so well-known in mathematics.
n
0
1
2
3
4
5
6
..
.
k
1
1
1
1
1
1
1
..
.
= 0, . . . , n
1
2 1
3 3 1
4
5
6
6 4
1
10 10 5
1
15 20 15 6 1
Recall from
TA B L E Table
2 : G EO N O2.
MIA
L C O E F F I C I Ecoefficients
N TS W H E N r when
 2
Geonomial
n
0
1
2
3
4
5
6
..
.
k
1
1
1
1
1
1
1
..
.
r=2
= 0, . . . , 2n
1
2
3
4
5
6n
1
3 2 1
6 7 6 3
1
10 16 19 16 10 4
1
Table
Geonomial
coefficients
15
30 2.45
51 45
30 when
15 r 5= 2 1
21
50
90
126
141
126
90 50 21 6 1
k = 0, . . . , 2n
0 1
1 1 1 1
2 1 2 3 2 1
3 1 3 6 7 6 3
1
4 1 4the
10 previous
16 19 16 section
10 4 that
1
Recall from
the coefficients
5 1 5 15 30 45 51 45 30 15 5 1
6 1 section
6 21 50
90
126from
141 126
90 in
50Pascal’s
21the
6 product
1 Triangle
theinprevious
that
the
coefficients
Pascal’s
Triangle
arise
expanding
.. ..
. .
arise from ex
of binomials.
thegeonomial
geonomial
panding the product
of binomials.Interestingly
Interestingly enough,
enough, the
coefficients becom
coefficients become explicit through a similar process
explicit through a similar process of expansion. The following list will illustrate the coeffi
pefully this triangle is becoming more and more beautiful to the reader! The
following The following list will illustrate the
T R I A N G L E O F G EO N O M I A L C O E F F I C I E N TS
Recallof
fromexpansion.
the previous section that the coefficients in Pascal’s Triangle arise from excients connected to the values in Table 2.
connected
to the values
Table
2. coefficients become
n introduces a In
new
type
of triangle
so well-known
in mathematics.
this
section,
we willnot
refer
to the numbers
in the new
panding coefficients
the product of binomials.
Interestingly
enough,inthe
geonomial
explicit through a similar
process of expansion. The following list will illustrate the coeffitriangles as geonomial coefficients. These numbers
(1 + a + a2 )1 = 1a2 + 1a + 1
2 1
3.
Triangle
of
Geonomial
Coefficients
(1aa
) in Table
1a21a1
cients connected
to the values
2.
satisfy a definition stated by Brenti and Welker and
4 4
3 3
2 2a + 1
+
2a2a
+ 3a
3a2 +
(1 (1aa
+ a + a2 )22)2= 1a
1a
2a1
appear in an identity central to the paper [2]. The
(1 + a + a2 )1 = 1a2 + 1a + 1
his section, we will refer to the numbers in the new triangles as geonomial coefficients.
2 3
6
5
4
)

1a
3a
6a

7a336a22 3a1
(1aa
geonomial coefficients count the number of ways we can
6
5
4
(1 + a (1
++aa2 )+3 a=2 )21a
4 3a 3+ 6a2 + 7a + 6a + 3a + 1
= 1a+
+ 2a + 3a + 2a + 1
numbers satisfy
a
definition
stated
by
Brenti
and
Welker
and
appear
in
an
identity
write an integer k as a sum of n digits whose values are
= 1a + 3a + 6a
7a +left-hand
6a + 3a + side
1
(1 + a the
+ a )polynomials
Notice
that
on+the
are
zerocoefficients
and r . Thecount
geonomial
coefficients
l to the paper integers
[2]. Thebetween
geonomial
the number
waysthe
we
can write
Noticeofthat
polynomials
on the left-hand side are raised to some exponent, whic
raised
to
some
exponent
which
happens
to
be
their
row
different
families,
on a parameter
Notice
polynomials
the left-hand
sidetable
are raised
to asome
exponent,
eger k as a sumfall
of ninto
digits
whose values
aredepending
integers between
zero and
r.
Thethat
geonomial
happens
to be
their
rowtheindex
in our on
table.
The next
shows
different
casewhich
of geonomi
index in our table. The next table shows a different case
r  1. If r  1, we recover the entries in Pascal’s Triangle.
happens to be their row index in our table. The next table shows a different case of geonomial
ients fall into different families, depending on a parameter
r ≥ 1. If .r = 1,
we recover
of .geonomial coefficients.
The tables for r  2 and r  3 are illustratedcoefficients
below. The
coefficients
ntries in Pascal’s
Triangle.
= 2 and rcoefficients
= 3 are illustrated below.
patterns
that The
occurtables
with for
ther geonomial
2 3
6
5
4
3
2
TA B L E 3 : G EO N O M I A L C O E F F I C I E N TS W H E N r  3
Table
3. Geonomials coefficients
coefficients when
r = 3r = 3
are true
values of coefficients
r , and the following
cases
Table
3. Geonomials
when
atterns that occur
withfor
theallgeonomial
are true for
all are
values of r, and
the
k
=
0,
.
.
.
,
3n
n
selected for visual reference. For now, let us focus on the = 0, . . . , 3n
ing cases are selected for visual reference. For now, let us focus nonkthe
0mysterious
1
0 1
mysterious similarities between Pascal’s Triangle and
1 1 1 1 1
1 1 1 2 1 1 21 3 4 3 2 1
rities between Pascal’s
Triangle
and the following triangles.
the following
triangles.
2 1 2 3 3 1 34 6 310
4 1 4 10 20
1 3symmetry
35
5 6 1 51015 12
ymmetry. A similar property that is immediately identifiable is3 the
in
4 1 4 ... 10... 20 31
5 the
1 5symmetry
15 35 65
row. The reason behind this follows a similar concept used to explain
of
.. ..
. .
212 12 1 10 6 3 1
31 40 44 40 31 20 10 4 1
12
3 135
1 101 65 35 15 5 1
65 10110135 6155 155
40 44 40 31 20 10 4 1
101 135 155 155 135 101 65 35 15 5 1
’s Triangle . We will illustrate another reason when we discuss the recurrence relation.
er similarities are not as direct, however, the reader should take a moment to discover
ns on their own.
— 170 —
Published by Via Sapientiae, 2012
3
DePaul Discoveries, Vol. 1 [2012], Iss. 1, Art. 13
A G E N E R A L I Z AT I O N O F PA S C A L ' S T R I A N G L E
To be consistent with the format used above, here
A LT E R N AT I N G S U M
are some products of polynomials whose coefficients
The previous section proved that the alternating sum of
represent values from Table 3.
the binomial coefficients will always be 0. For geonomial
coefficients, there are two possible outcomes. When
(1aa2a3)1  1a31a2 1a1
r is an even number, the alternating sum in each row
(1aa2a3)2  1a6 2a53a4 4a33a22a1
(1aa2a3)3  1a9 3a86a7 10a6 12a5 12a410a36a23a1
In general, we were able to prove that the geonomial
coefficients appear in the following expansions:
(1aa2…ar)n , which is a repeated product of a finite
geometric series 1aa2…ar (hence our choice for
the name geonomial). The proof, however, is too technical
to include in this paper, and we invite the reader to email
us for specific details if desired.
is 1. On the other hand, when r is an odd number, the
alternating sum is 0. To test this, consider row three
from Table 2: 13676–311. Now use row three
of Table 3 for confirmation. The alternating sum is
13610121210631 0 . The reader can
prove this property by setting a  –1 in the polynomials
above.
R EC U R R E N C E R E L AT I O N
When
Whenrrisisan
aneven
even
number,
number,
the
alternating
alternating
sum
suminineach
each
row
row
isis1.1. On
Onthe
the
other
otherthe
hand,
hand,when
when
Just
as thethe
binomial
coefficients
are
obtainable
given
rrisisan
anodd
oddnumber,
number,
the
thealternating
alternating
sum
sumisis0.0.Triangle,
To
Totest
testthis,
this,
consider
considerrow
row
three
threefrom
fromTable
Table2:2:
previous
row of Pascal’s
geonomial
coefficients
R OW S U M
11−−33++66−−77++66−−
33++11=
1.Now
Nowuse
use
row
rowthree
threeofof
Table
Table33for
for
confirmation.
confirmation.
The
Thealternating
alternating
have
a=1.
similar
recurrence
relation.
Recall
from Pascal’s
Recall from Pascal’s Triangle that the sum of the numbers
every
can
bereader
obtained
by adding
sum
sumisis11−−33++66−Triangle
−10
10++12
12−−that
12
12++10
10−−66+number
+33−−11==0.
0.The
The
reader
can
canprove
prove
this
thisproperty
propertyby
by
in any single row was 2n (where n represented the row
consecutive
numbers in the previous row. For these
setting
settingaa==−1
−1inintwo
the
thepolynomials
polynomialsabove.
above.
index). Let’s attempt the same experiment! By arbitrarily
new triangles, the number of consecutive values required
selecting row three, when n  3 from Table 2,3.4.
the
row
3.4.
Recurrence
Recurrence
Relation.
Relation. Just
Justasasthe
thebinomial
binomialcoefficients
coefficientsare
areobtainable
obtainablegiven
giventhe
thepreviprevito add is r 1. When r  2 , the number of consecutive
sum is 1367631  27. Remember that
Table
ous
ousrow
rowofofPascal’s
Pascal’sTriangle,
Triangle,geonomial
geonomialcoefficients
coefficientshave
haveaasimilar
similarrecurrence
recurrencerelation.
relation.Recall
Recall
values to add in order to obtain a geonomial coefficient,
5
(21)
. Triangle
2 is the case when r = 2 , and notice that 27  33 
from
from
Pascal’s
Pascal’s
Trianglethat
thatevery
everynumber
numbercan
canbe
beobtained
obtainedby
byadding
addingtwo
twoconsecutive
consecutivenumnumis three. Comparably, when r  3, by adding four
From the same table, let n  5 so we obtainbers
the
sum
bersininthe
theprevious
previousrow.
row.For
Forthese
thesenew
newtriangles,
triangles,the
thenumber
numberofofconsecutive
consecutivevalues
valuesrequired
required
consecutive values in the previous row of the selected
5
15  1530455145301551  (21)
totoadd
addisisr.r++1.1. When
Whenrr==2,2,the
thenumber
numberofofconsecutive
consecutivevalues
valuestotoadd
addininorder
ordertotoobtain
obtainaa
row, a new geonomial coefficient is obtained. Therefore,
Let’s now switch to Table 3 to see if the same
pattern coefficient,isisthree.
geonomial
geonomialcoefficient,
three. Comparably,
Comparably,when
whenrr==3,3,by
byadding
addingfour
fourconsecutive
consecutivevalues
since the
entries
are initially
symmetric,
the
entries
in values
occurs. Add up all values in row two, when n = 2,inso
we previous
get
inthe
theprevious
row
row
of
of
the
the
selected
selected
row,
row,
a
a
new
new
geonomial
geonomial
coefficient
coefficient
is
is
obtained.
obtained.
Therefore,
Therefore,
the following rows will abide to the same property. The
the sum 1+2+3+4+3+2+1 = 16 = 42 = (3+1)2. Also, when
since
sincethe
theentries
entriesare
areinitially
initiallytables
symmetric,
symmetric,
the
theentries
entriesto
ininview
the
thefollowing
following
rows
rowswill
willabide
abidetotothe
the
following
are modified
this property.
n = 5, we have 1+5+15+35+65+101+135+155+155+135
same
sameproperty.
property.The
Thefollowing
followingtables
tablesare
aremodified
modifiedtotoview
viewthis
thisproperty.
property.
+101+65+35+15+5+1=1024=45 =(3+1)5. More generally,
TA B L E 4 : r  4
Table
Table4.4. rr==22
another property of the geonomial coefficients is that
the sum in each row can be obtained by (r1)n, where
r represents a particular case and n is the row index. We
can verify this by revisiting our polynomial and setting
a  1, where we get (1112 )n  3n  (21)n, and
(111213 )n  4n  (31) n, and, more generally,
.
nn
00
11
22
33
44
55
66
.. ..
..
2n
kk==0,0,. . . ., ,2n
11
11 11 11
11 22 33 22
11 33 66 77
10 16
16
11 44 10
15 30
30
11 55 15
21 50
50
11 66 21
.. ..
..
11
66
19
19
45
45
90
90
33
11
16
16 10
10 44 11
51 45
45 30
30 1515 55 11
51
126 141
141 126
126 90
90 50
50 21
21 66 11
126
We
Wehave
haveobtained
obtainedaageneral
generalproof
prooffor
forthese
theserecurrence
recurrencerelations
relationsasaswell,
well, and
andwould
wouldbe
be
thrilled
thrilledfor
forthe
thereader
readertotocontact
contactus
usfor
formore
moredetails.
details.
4.4. Final
FinalDiscussion
Discussion
The
Theidentity
identitypresented
presentedby
byBrenti
Brentiand
andWelker
Welkerisiscombinatorial
combinatorialininnature,
nature,however
howeverasasthey
they
themselves
themselvesstate,
state,there
thereisisno
noknown
knowncombinatorial
combinatorialexplanation
explanationfor
forit.
it.The
TheEulerian
Euleriannumbers
numbers
— 171 —
http://via.library.depaul.edu/depaul-disc/vol1/iss1/13
4
Gwetta et al.: A Generalization of Pascal’s Triangle
D E PA U L D I S C O V E R I E S ( 2 O 1 2 )
TA B L E 5 : G EO N O M I A L C O E F F I C I E N TS W H E N r  3
Table 5. Geonomials coefficients when r = 3
n
0
1
2
3
4
5
..
.
k
1
1
1
1
1
1
..
.
= 0, . . . , 3n
1
2
3
4
5
1
1
3 4 3 2 1
6 10 12 12 10 6
3
1
10 20 31 40 44 40 31 20
10 4 1
15 35 65 101 135 155 155 135 101 65 35 15 5 1
which are contained in this identity have many well-known properties, but the properties
We have obtained a general proof for these recurrence
of the geonomial coefficients are less well-known. We hope that our research will provide
relations as well, and would be thrilled for the reader to
insight for developing a combinatorial explanation of this curious identity.
contact us for more details.
References
[1] D. E. Knuth, The art of computerFprogramming.
Volume
INAL DIS
C U S3.SSorting
I O N and searching. Addison-Wesley Series
The identity presented by Brenti and Welker is
in Computer Science and Information Processing. Addison-Wesley Publishing Co., Reading, Mass.London-Don Mills, Ont., (1973).
combinatorial in nature, however as they themselves
[2] F. Brenti and V. Welker, The Veronese construction for formal power series and graded algebras, Ad-
there
is no
known
combinatorial
vancesstate,
in Applied
Mathematics,
42 (2009),
545–556.
explanation
for it. The Eulerian numbers which are contained in
this identity have many well-known properties, but the
properties of the geonomial coefficients are less wellknown. We hope that our research will provide insight for
developing a combinatorial explanation of this curious
identity.
ACKNOWLEDGEMENTS
I would like to thank my research advisors, Dr. Kyle
Petersen and Dr. Bridget Tenner, for their continuous
support and dedication to enhancing the quality of our
research. I would also like to thank DePaul University
for funding this research project and allowing for us
to have such a unique experience as undergraduate
researchers.
— 172 —
Published by Via Sapientiae, 2012
5
DePaul Discoveries, Vol. 1 [2012], Iss. 1, Art. 13
A G E N E R A L I Z AT I O N O F PA S C A L ' S T R I A N G L E
REFERENCES
[1] D. E. Knuth, The art of computer programming. Volume 3. Sorting
[2] F. Brenti and V. Welker, The Veronese construction for formal power
and searching. Addison-Wesley Series in Computer Science and
series and graded algebras, Advances in Applied Mathematics, 42
Information Processing. Addison-Wesley Publishing Co., Reading,
(2009), 545—556.
Mass.-London-Don Mills, Ont., (1973).
Sanjay Cherala, Stephanie Lyngaas Matt Hartzell and Lihua Jin, chair of the Chemistry Department, enjoy the 9th Annual Natural
Sciences, Mathematics, and Technology Showcase.
— 173 —
http://via.library.depaul.edu/depaul-disc/vol1/iss1/13
6