Some Variations of Two Combinatorial Identities

M.J. Kronenburg

arXiv:1705.05675·math.CO·May 17, 2017

Some Variations of Two Combinatorial Identities

M.J. Kronenburg

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TL;DR

This paper introduces new variations of existing combinatorial identities, proves them using integral representations, and shows how some known identities are special cases of these new results.

Contribution

It presents novel variations of two known combinatorial identities and demonstrates their proofs via integral representations, expanding the understanding of these identities.

Findings

01

New variations of combinatorial identities are established.

02

Some existing identities are shown as special cases of the new ones.

03

The integral representation method is effectively used for proofs.

Abstract

Given two combinatorial identities proved earlier, a new set of variations of these combinatorial identities is listed and proved with the integral representation method. Some identities from literature are shown to be special cases of these new identities.

Equations72

k = 0 \sum n (k a) (n - k b) (c k) (d n - k) = (n - c - d a + b - c - d) (c a) (d b)

k = 0 \sum n (k a) (n - k b) (c k) (d n - k) = (n - c - d a + b - c - d) (c a) (d b)

k = 0 \sum a (k a) (m + k b) (c k) (d m + k) = (m + a - d a + b - c - d) (c a) (d b)

k = 0 \sum a (k a) (m + k b) (c k) (d m + k) = (m + a - d a + b - c - d) (c a) (d b)

k = 0 \sum n (k - p c + d - b) (n - k b) (c k) (d n - k) = (n - d - p n - b) (c + d + p - n p) (d b)

k = 0 \sum n (k - p c + d - b) (n - k b) (c k) (d n - k) = (n - d - p n - b) (c + d + p - n p) (d b)

k = 0 \sum a (k a) (p + k c + d - a) (c k) (d n - k) = (a + p n + a + p - c - d) (d - a - p n - a) (c a)

k = 0 \sum a (k a) (p + k c + d - a) (c k) (d n - k) = (a + p n + a + p - c - d) (d - a - p n - a) (c a)

k = 0 \sum a (k a) (p - k c + d - a) (c k) (d n - k) = (c + d - p n - p) (p - c n - a) (c a)

k = 0 \sum a (k a) (p - k c + d - a) (c k) (d n - k) = (c + d - p n - p) (p - c n - a) (c a)

k = 0 \sum n (k a) (n - k b) (a + b - d p + k) (d n - k) = (n - d n + p - b) (a + b - n p) (d b)

k = 0 \sum n (k a) (n - k b) (a + b - d p + k) (d n - k) = (n - d n + p - b) (a + b - n p) (d b)

k = 0 \sum n (k a) (n - k b) (a + b - d p - k) (d n - k) = (a + b - n d + p - n) (n - d p - a) (d b)

k = 0 \sum n (k a) (n - k b) (a + b - d p - k) (d n - k) = (a + b - n d + p - n) (n - d p - a) (d b)

k = 0 \sum n (k a) (n - k b) (c k) (a + b - c p + k) = (n - c n + p - b) (a + b - n p + c) (c a)

k = 0 \sum n (k a) (n - k b) (c k) (a + b - c p + k) = (n - c n + p - b) (a + b - n p + c) (c a)

k = 0 \sum n (k a) (n - k b) (c k) (a + b - c p - k) = (a + b - n p - n) (n - c p - a) (c a)

k = 0 \sum n (k a) (n - k b) (c k) (a + b - c p - k) = (a + b - n p - n) (n - c p - a) (c a)

k = 0 \sum a + p (k - p a) (m + k b) (c k) (m m + k) = (a + p - c a + b - c - m) (c b - m) (m b)

k = 0 \sum a + p (k - p a) (m + k b) (c k) (m m + k) = (a + p - c a + b - c - m) (c b - m) (m b)

k = 0 \sum a (k a) (p + k c + d - a) (c k) (d m + k) = (d - a - p m - p) (a + p m + c) (c a)

k = 0 \sum a (k a) (p + k c + d - a) (c k) (d m + k) = (d - a - p m - p) (a + p m + c) (c a)

k = 0 \sum a (k a) (p - k c + d - a) (c k) (d m + k) = (p - c m + a + p - c - d) (c + d - p m + c) (c a)

k = 0 \sum a (k a) (p - k c + d - a) (c k) (d m + k) = (p - c m + a + p - c - d) (c + d - p m + c) (c a)

k = 0 \sum a (k a) (m + k b) (a + b - d p + k) (d m + k) = (b - m d + p - m) (m + a - d p) (d b)

k = 0 \sum a (k a) (m + k b) (a + b - d p + k) (d m + k) = (b - m d + p - m) (m + a - d p) (d b)

k = 0 \sum a (k a) (m + k b) (a + b - d p - k) (d m + k) = (m + a - d m + p - b) (b - m p - a) (d b)

k = 0 \sum a (k a) (m + k b) (a + b - d p - k) (d m + k) = (m + a - d m + p - b) (b - m p - a) (d b)

k = 0 \sum a (k a) (m + k b) (c k) (a + b - c p + k) = (b - c - m p - m) (m + a p + c) (c a)

k = 0 \sum a (k a) (m + k b) (c k) (a + b - c p + k) = (b - c - m p - m) (m + a p + c) (c a)

k = 0 \sum a (k a) (m + k b) (c k) (a + b - c p - k) = (m + a m + p - b) (b - m - c p - a) (c a)

k = 0 \sum a (k a) (m + k b) (c k) (a + b - c p - k) = (m + a m + p - b) (b - m - c p - a) (c a)

k = 0 \sum n (k - p a) (c k) (n - d - k b - d)

k = 0 \sum n (k - p a) (c k) (n - d - k b - d)

= = = k = 0 \sum \infty Res_{x} \frac{( 1 + x ) ^{a}}{x ^{k - p + 1}} Res_{y} \frac{( 1 + y ) ^{k}}{y ^{c + 1}} Res_{z} \frac{( 1 + z ) ^{b - d}}{z ^{n - d - k + 1}} Res_{x} Res_{y} Res_{z} \frac{( 1 + x ) ^{a} ( 1 + z ) ^{b - d}}{x ^{- p + 1} y ^{c + 1} z ^{n - d - 1}} k = 0 \sum \infty (\frac{( 1 + y ) z}{x})^{k} Res_{y} Res_{z} Res_{x} \frac{( 1 + x ) ^{a} ( 1 + z ) ^{b - d} x ^{p}}{y ^{c + 1} z ^{n - d + 1} ( x - ( 1 + y ) z )} Res_{y} Res_{z} \frac{( 1 + ( 1 + y ) z ) ^{a} ( 1 + y ) ^{p} ( 1 + z ) ^{b - d}}{y ^{c + 1} z ^{n - d - p + 1}}

= = = k = 0 \sum \infty Res_{x} \frac{( 1 + x ) ^{a}}{x ^{k - p + 1}} Res_{y} \frac{( 1 + y ) ^{k}}{y ^{c + 1}} Res_{z} \frac{( 1 + z ) ^{b - d}}{z ^{n - d - k + 1}} Res_{x} Res_{y} Res_{z} \frac{( 1 + x ) ^{a} ( 1 + z ) ^{b - d}}{x ^{- p + 1} y ^{c + 1} z ^{n - d - 1}} k = 0 \sum \infty (\frac{( 1 + y ) z}{x})^{k} Res_{y} Res_{z} Res_{x} \frac{( 1 + x ) ^{a} ( 1 + z ) ^{b - d} x ^{p}}{y ^{c + 1} z ^{n - d + 1} ( x - ( 1 + y ) z )} Res_{y} Res_{z} \frac{( 1 + ( 1 + y ) z ) ^{a} ( 1 + y ) ^{p} ( 1 + z ) ^{b - d}}{y ^{c + 1} z ^{n - d - p + 1}}

(1 + (1 + y) z)^{a} = ((1 + z) + y z)^{a} = k = 0 \sum a (k a) (1 + z)^{k} (y z)^{a - k}

(1 + (1 + y) z)^{a} = ((1 + z) + y z)^{a} = k = 0 \sum a (k a) (1 + z)^{k} (y z)^{a - k}

= k = 0 \sum a (k a) (c - a + k p) (n - d - p - a + k b - d + k) k = 0 \sum a (k a) (c - a + k p) (a + b + p - n b - d + k)

= k = 0 \sum a (k a) (c - a + k p) (n - d - p - a + k b - d + k) k = 0 \sum a (k a) (c - a + k p) (a + b + p - n b - d + k)

k = 0 \sum n (k - c a - c) (p + k b) (d n - k)

k = 0 \sum n (k - c a - c) (p + k b) (d n - k)

= = = = k = 0 \sum \infty Res_{x} \frac{( 1 + x ) ^{a - c}}{x ^{k - c + 1}} Res_{y} \frac{( 1 + y ) ^{b}}{y ^{p + k + 1}} Res_{z} \frac{( 1 + z ) ^{n - k}}{z ^{d + 1}} Res_{x} Res_{y} Res_{z} \frac{( 1 + x ) ^{a - c} ( 1 + y ) ^{b} ( 1 + z ) ^{n}}{x ^{- c + 1} y ^{p + 1} z ^{d + 1}} k = 0 \sum \infty (\frac{1}{x y ( 1 + z )})^{k} Res_{y} Res_{z} Res_{x} \frac{( 1 + x ) ^{a - c} ( 1 + y ) ^{b} ( 1 + z ) ^{n} x ^{c}}{y ^{p + 1} z ^{d + 1} ( x - 1/ ( y ( 1 + z )))} Res_{y} Res_{z} \frac{( 1 + 1/ ( y ( 1 + z )) ) ^{a - c} ( 1 + y ) ^{b} ( 1 + z ) ^{n - c}}{y ^{c + p + 1} z ^{d + 1}} Res_{y} Res_{z} \frac{( 1 + y ( 1 + z ) ) ^{a - c} ( 1 + y ) ^{b} ( 1 + z ) ^{n - a}}{y ^{a + p + 1} z ^{d + 1}}

= = = = k = 0 \sum \infty Res_{x} \frac{( 1 + x ) ^{a - c}}{x ^{k - c + 1}} Res_{y} \frac{( 1 + y ) ^{b}}{y ^{p + k + 1}} Res_{z} \frac{( 1 + z ) ^{n - k}}{z ^{d + 1}} Res_{x} Res_{y} Res_{z} \frac{( 1 + x ) ^{a - c} ( 1 + y ) ^{b} ( 1 + z ) ^{n}}{x ^{- c + 1} y ^{p + 1} z ^{d + 1}} k = 0 \sum \infty (\frac{1}{x y ( 1 + z )})^{k} Res_{y} Res_{z} Res_{x} \frac{( 1 + x ) ^{a - c} ( 1 + y ) ^{b} ( 1 + z ) ^{n} x ^{c}}{y ^{p + 1} z ^{d + 1} ( x - 1/ ( y ( 1 + z )))} Res_{y} Res_{z} \frac{( 1 + 1/ ( y ( 1 + z )) ) ^{a - c} ( 1 + y ) ^{b} ( 1 + z ) ^{n - c}}{y ^{c + p + 1} z ^{d + 1}} Res_{y} Res_{z} \frac{( 1 + y ( 1 + z ) ) ^{a - c} ( 1 + y ) ^{b} ( 1 + z ) ^{n - a}}{y ^{a + p + 1} z ^{d + 1}}

(1 + y (1 + z))^{a - c} = ((1 + y) + y z)^{a - c} = k = 0 \sum a - c (k a - c) (1 + y)^{k} (y z)^{a - c - k}

(1 + y (1 + z))^{a - c} = ((1 + y) + y z)^{a - c} = k = 0 \sum a - c (k a - c) (1 + y)^{k} (y z)^{a - c - k}

= k = 0 \sum a - c (k a - c) (c + p + k b + k) (c + d - a + k n - a) k = 0 \sum a - c (k a - c) (b - c - p b + k) (c + d - a + k n - a)

= k = 0 \sum a - c (k a - c) (c + p + k b + k) (c + d - a + k n - a) k = 0 \sum a - c (k a - c) (b - c - p b + k) (c + d - a + k n - a)

k = 0 \sum a (k a) (k b) (a + b p + k) = (a p) (b p)

k = 0 \sum a (k a) (k b) (a + b p + k) = (a p) (b p)

k = 0 \sum n (k m - x + y) (n - k n + x - y) (m + n x + k) = (m x) (n y)

k = 0 \sum n (k m - x + y) (n - k n + x - y) (m + n x + k) = (m x) (n y)

k = 0 \sum a (k a) (k - d b) (a + b p + k) = (a - d p) (b + d p + d)

k = 0 \sum a (k a) (k - d b) (a + b p + k) = (a - d p) (b + d p + d)

k = 0 \sum n (k a) (n - k b) (a + b p + k) = (a + b - n p) (n p - b + n)

k = 0 \sum n (k a) (n - k b) (a + b p + k) = (a + b - n p) (n p - b + n)

k = 0 \sum a (k a) (k b) (a + b a + b + x + k) = (a a + b + x) (b a + b + x)

k = 0 \sum a (k a) (k b) (a + b a + b + x + k) = (a a + b + x) (b a + b + x)

k = 0 \sum a (k a) (k b) (a + b a + b + x - k) = (a x + a) (b x + b)

k = 0 \sum a (k a) (k b) (a + b a + b + x - k) = (a x + a) (b x + b)

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TopicsAdvanced Combinatorial Mathematics · Advanced Mathematical Theories · Mathematics and Applications

Full text

Some Variations of Two Combinatorial Identities

M.J. Kronenburg

Abstract

Given two combinatorial identities proved earlier, a new set of variations of these combinatorial identities is listed and proved with the integral representation method. Some identities from literature are shown to be special cases of these new identities.

Keywords: binomial coefficient, combinatorial identities.

MSC 2010: 05A10, 05A19

1 Two Combinatorial Identities

In an earlier paper [9], the following two combinatorial identities were proved:

[TABLE]

In this paper a list of identities that are variations of these two identities is provided, and it is shown how they are proved with the integral representation method.

2 A Set of New Combinatorial Identities

The following are variations of the first identity (1.1):

[TABLE]

The following are variations of the second identity (1.2):

[TABLE]

3 Proof of the New Combinatorial Identities

The new combinatorial identities listed above can all be proved with the integral representation method [2, 3]. Two of these identities are proved below, and the other identities have similar proofs.

For identity (2.1), applying the trinomial revision identity [6, 7, 8], the left side of the identity simplifies to:

[TABLE]

Using the integral representation method [2, 3], this becomes:

[TABLE]

Now the following is used:

[TABLE]

Collecting the residues, the following expression results:

[TABLE]

This expression is recognized as (1.2) with $c=0$ provided that $m=c-a=b-d$ , or equivalently $a=c+d-b$ . The parameters of (1.2) thus become $a\leftarrow c+d-b$ , $b\leftarrow p$ , $c\leftarrow 0$ , $d\leftarrow c+d+p-n$ and $m\leftarrow b-d$ , and substituting these parameters in the right side of (1.2) gives the result.

For identity (2.2), applying the trinomial revision identity [6, 7, 8], the left side of the identity simplifies to:

[TABLE]

Using the integral representation method [2, 3], this becomes:

[TABLE]

Now the following is used:

[TABLE]

Collecting the residues, the following expression results:

[TABLE]

This expression is recognized as (1.2) with $c=0$ provided that $m=b=c+d-a$ . The parameters of (1.2) thus become $a\leftarrow a-c$ , $b\leftarrow n-a$ , $c\leftarrow 0$ , $d\leftarrow d-p-a$ and $m\leftarrow c+d-a$ , and substituting these parameters in the right side of (1.2) gives the result.

4 Some Identities from Literature

Some identities from literature are special cases of the identities above.

T.S. Nanjundiah gave the following two identities [1, 4, 10, 11]:

[TABLE]

The first equation is equation (2.11) with $m=d=0$ , and the second equation is equation (2.4) with $a=m-x+y$ , $b=n+x-y$ , $d=0$ and $p=x$ .

M.T.L. Bizley gave the following two identities [1, 4]:

[TABLE]

The first equation is equation (2.9) with $c=0$ , $d=a+b$ , $m=p$ and $p=-d$ , and the second equation is equation (2.4) with $d=0$ .

H.W. Gould gave the following identity [4]:

[TABLE]

This equation is equation (2.11) with $m=d=0$ and $p=a+b+x$ .

J. Surányi gave the following identity [4, 11, 12]:

[TABLE]

This equation is equation (2.12) with $m=d=0$ and $p=a+b+x$ .

L. Takács gave the following identity [12, 14]:

[TABLE]

This equation is equation (2.4) with $b=m-a$ and $d=0$ .

J. Riordan gave the following identity [4]:

[TABLE]

This equation is equation (2.5) with $a=n$ , $b=m$ , $d=0$ and $p=x+n$ .

R.P. Stanley gave the following two identities [4, 5, 13]:

[TABLE]

These two identities can be shown to be equivalent to (2.5) with $d=0$ using the following two symmetry identities [6, 7, 8]:

[TABLE]

In the first identity, applying the first symmetry identity to the first and third binomial coefficients on the left and to the second binomial coefficient on the right, and then replacing $q$ by $-q-1$ , results in the following identity:

[TABLE]

When $q<p$ , interchanging $p$ with $q$ and $a$ with $b$ , which does not change Stanley’s identities, makes $q>p$ . This equation is equation (2.5) with $a\leftarrow q-p$ , $b\leftarrow p$ , $d\leftarrow 0$ , $n\leftarrow a$ and $p\leftarrow q+b$ .

In the second identity, applying the second symmetry identity to the second binomial coefficient on the left and the first symmetry identity to the first binomial coefficient on the right, and then replacing $p$ with $-p-1$ , results in the same identity.

Bibliography14

The reference list from the paper itself. Each links out to its DOI / PubMed record.

1[1] M.T.L. Bizley, A Generalization of Nanjundiah’s Identity, Amer. Math. Monthly 77 (1970) 863-865.
2[2] R.V. Churchill, J.W. Brown, Complex Variables and Applications , Mc Graw-Hill, 1984.
3[3] G.P. Egorychev, Integral Representation and the Computation of Combinatorial Sums , Translations of Mathematical Monographs, 59, Amer. Math. Soc., 1984.
4[4] H.W. Gould, Combinatorial Identities , rev. ed., Morgantown, 1972.
5[5] H.W. Gould, A New Symmetrical Combinatorial Identity, J. Combin. Theory Ser. A 13 (1972) 278-286.
6[6] R.L. Graham, D.E. Knuth, O. Patashnik, Concrete Mathematics, A Foundation for Computer Science , 2nd ed., Addison-Wesley, 1994.
7[7] D.E. Knuth, The Art of Computer Programming, Volume 1: Fundamental Algorithms , 3rd ed., Addison-Wesley, 1997.
8[8] M.J. Kronenburg, The Binomial Coefficient for Negative Arguments, ar Xiv:1105.3689 [math.CO]