arXiv:1705.05675v1 [math.CO] 15 May 2017
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:
n X
a
b
k
n−k
a+b−c−d a
b
=
k
n−k
c
d
n−c−d
c
d
(1.1)
k=0
a X
a
k
k=0
b
m+k
k
m+k
a+b−c−d a
b
=
c
d
m+a−d
c
d
(1.2)
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):
n X
c+d−b
b
k
n−k
n−b
p
b
=
k−p
n−k
c
d
n−d−p c+d+p−n d
(2.1)
k=0
a X
a c+d−a k
n−k
k=0
k
p+k
c
d
=
n+a+p−c−d
n−a
a
a+p
d−a−p
c
a X
a c+d−a k
n−k
k=0
k
p−k
c
d
1
=
n−p
n−a a
c+d−p
p−c
c
(2.2)
(2.3)
n X
a
b
p+k
n−k
n+p−b
p
b
=
k
n−k
a+b−d
d
n−d
a+b−n d
(2.4)
n X
a
b
p−k
n−k
d+p−n
p−a
b
=
k
n−k
a+b−d
d
a+b−n
n−d d
(2.5)
n X
a
b
k
p+k
n+p−b
p+c
a
=
k
n−k
c
a+b−c
n−c
a+b−n
c
(2.6)
n X
a
b
k
p−k
p−n
p−a a
=
k
n−k
c
a+b−c
a+b−n n−c
c
(2.7)
k=0
k=0
k=0
k=0
The following are variations of the second identity (1.2):
a+p
X
k=0
a
k−p
b
m+k
k
m+k
a+b−c−m b−m
b
=
c
m
a+p−c
c
m
a X
a c+d−a k
m+k
k
k=0
p+k
c
a X
a c+d−a k
m+k
p−k
k
k=0
a X
a
k=0
k
b
m+k
a X
a
k=0
k
k
a X
a
k=0
3
k
b
m+k
a X
a
k=0
c
=
=
m−p
d−a−p
m+c a
a+p
c
(2.9)
m+a+p−c−d
m+c
a
(2.10)
p−c
c+d−p
c
p+k
m+k
d+p−m
p
b
=
(2.11)
a+b−d
d
b−m
m+a−d d
p−k
m+k
m+p−b
p−a
b
=
a+b−d
d
m+a−d b−m d
(2.12)
k
p+k
p−m
p+c
a
=
c
a+b−c
b−c−m m+a
c
(2.13)
k
p−k
m+p−b
p−a
a
=
c
a+b−c
m+a
b−m−c
c
(2.14)
b
m+k
b
m+k
d
d
(2.8)
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:
n X
a
k
b−d
(3.1)
k−p
c
n−d−k
k=0
2
Using the integral representation method [2, 3], this becomes:
∞
X
(1 + x)a
(1 + y)k
(1 + z)b−d
Res
Res
y
z
xk−p+1
y c+1
z n−d−k+1
k=0
k
∞ (1 + x)a (1 + z)b−d X (1 + y)z
= Resx Resy Resz −p+1 c+1 n−d−1
x
y z
x
Resx
k=0
(3.2)
(1 + x)a (1 + z)b−d xp
= Resy Resz Resx c+1 n−d+1
y z
(x − (1 + y)z)
a
(1 + (1 + y)z) (1 + y)p (1 + z)b−d
= Resy Resz
y c+1 z n−d−p+1
Now the following is used:
(1 + (1 + y)z)a = ((1 + z) + yz)a =
a X
a
k=0
k
(1 + z)k (yz)a−k
Collecting the residues, the following expression results:
a X
a
p
b−d+k
k
c−a+k
n−d−p−a+k
k=0
a
X
a
p
b−d+k
=
k
c−a+k
a+b+p−n
(3.3)
(3.4)
k=0
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 ← c + d − b,
b ← p, c ← 0, d ← c + d + p − n and m ← 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:
n X
a−c
b
n−k
(3.5)
k−c p+k
d
k=0
Using the integral representation method [2, 3], this becomes:
∞
X
(1 + x)a−c
(1 + y)b
(1 + z)n−k
Resy p+k+1 Resz
k−c+1
x
y
z d+1
k=0
k
∞ (1 + x)a−c (1 + y)b (1 + z)n X
1
= Resx Resy Resz
x−c+1 y p+1 z d+1
xy(1 + z)
Resx
k=0
(1 + x)a−c (1 + y)b (1 + z)n xc
= Resy Resz Resx p+1 d+1
y
z
(x − 1/(y(1 + z)))
(1 + 1/(y(1 + z)))a−c (1 + y)b (1 + z)n−c
= Resy Resz
y c+p+1 z d+1
a−c
(1 + y(1 + z)) (1 + y)b (1 + z)n−a
= Resy Resz
y a+p+1 z d+1
3
(3.6)
Now the following is used:
(1 + y(1 + z))a−c = ((1 + y) + yz)a−c =
a−c X
a−c
k=0
k
(1 + y)k (yz)a−c−k
(3.7)
Collecting the residues, the following expression results:
a−c X
a−c
b+k
n−a
k
c+p+k
c+d−a+k
k=0
a−c
X
a−c
b+k
n−a
=
k
b−c−p c+d−a+k
(3.8)
k=0
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 ← a − c, b ← n − a, c ← 0, d ← d − p − a and
m ← 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]:
a X
a
b
p+k
p p
=
k
k
a+b
a
b
(4.1)
k=0
n X
m−x+y
n+x−y
x+k
k
k=0
n−k
m+n
x
y
=
m n
(4.2)
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]:
a X
a
b
p+k
p
p+d
=
(4.3)
k
k−d
a+b
a−d
b+d
k=0
n X
a
b
p+k
p
p−b+n
=
k
n−k
a+b
a+b−n
n
(4.4)
k=0
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]:
a X
a
b
a+b+x+k
a+b+x a+b+x
=
(4.5)
k
k
a+b
a
b
k=0
4
This equation is equation (2.11) with m = d = 0 and p = a + b + x.
J. Surányi gave the following identity [4, 11, 12]:
a X
a
b
a+b+x−k
k=0
k
k
a+b
=
x+a x+b
a
b
(4.6)
This equation is equation (2.12) with m = d = 0 and p = a + b + x.
L. Takács gave the following identity [12, 14]:
n X
a m−a p+k
k=0
n−k
k
m
=
p
m−n
n+a+p−m
n
(4.7)
This equation is equation (2.4) with b = m − a and d = 0.
J. Riordan gave the following identity [4]:
n X
n
m
x+n−k
x
x
=
k
n−k
n+m
m n
(4.8)
k=0
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]:
a X
p+q+k
p
q
p+b q+a
=
k
a−k
b−k
a
b
(4.9)
k=0
a
X
(−1)k
k=0
p+q+1
k
p+a−k
p
q+b−k
q
=
p+a−b q+b−a
(4.10)
a
b
These two identities can be shown to be equivalent to (2.5) with d = 0 using the
following two symmetry identities [6, 7, 8]:
n
−n + k − 1
−k − 1
= (−1)k
= (−1)n−k
(4.11)
k
k
n−k
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:
a X
q−p
p
q+b−k
p+b q+b−a
=
k
a−k
q
a
b
(4.12)
k=0
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 ← q − p, b ← p, d ← 0,
n ← a and p ← 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.
5
References
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77 (1970) 863-865.
[2] R.V. Churchill, J.W. Brown, Complex Variables and Applications, McGraw-Hill,
1984.
[3] G.P. Egorychev, Integral Representation and the Computation of Combinatorial
Sums, Translations of Mathematical Monographs, 59, Amer. Math. Soc., 1984.
[4] H.W. Gould, Combinatorial Identities, rev. ed., Morgantown, 1972.
[5] H.W. Gould, A New Symmetrical Combinatorial Identity, J. Combin. Theory
Ser. A 13 (1972) 278-286.
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for Computer Science, 2nd ed., Addison-Wesley, 1994.
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some Harmonic Number Identities, arXiv:1701.02768 [math.CO]
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65 (1958) 354.
[11] J. Quaintance, H.W. Gould, Combinatorial Identities for Stirling Numbers,
World Scientific, 2016.
[12] L.A. Székely, Common Origin of Cubic Binomial Identities; A Generalization of Surányi’s Proof on Le Jen Shoo’s Formula, J. Combin. Theory Ser. A
40 (1985) 171-174.
[13] R.P. Stanley, Ordered Structures and Partitions, Ph. D. thesis, Harvard University, 1971.
[14] L. Takács, On an identity of Shih-Chieh Chu, Acta Sci. Math. (Szeged)
34 (1973) 383-391.
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