A GENERATING FUNCTIONS PROOF OF A CURIOUS IDENTITY
Alois Panholzer
Department of Algebra and Computer Mathematics, Technical University Vienna, Wiedner
Hauptstrasse 8–10, A–1040 Vienna, Austria
Alois.Panholzer@tuwien.ac.at
Helmut Prodinger1
School of Mathematics, University of the Witwatersrand, P. O. Wits, 2050 Johannesburg, South Africa
helmut@staff.ms.wits.ac.za
http://www.wits.ac.za/helmut/index.htm
Received: 2/17/02, Accepted: 3/4/02, Published: 3/19/02
Abstract
We give an alternative proof of an identity that appeared recently in Integers. It is
shorter than the original one and uses generating functions.
In the paper [2] that appeared a few days ago the identity
Sm := (x + m + 1)
m
X
µ
i
(−1)
i=0
x+y+i
m−i
¶µ
¶ X
¶
µ ¶
m µ
y + 2i
x+i
x
i
−
(−4) = (x − m)
i
m−i
m
i=0
(the main result) was proved using double recursions. Here we show this result using
generating functions. This proof is perhaps more pleasant and shorter than the original
one. The relevant functions can be found in [1, p. 201ff.]. We need
µ
¶
¡
¢r X tk + r
r
Bt (z) =
zk ,
k
tk
+
r
k≥0
¢r
¡
X µtk + r¶
Bt (z)
zk ,
¡
¢−1 =
k
1 − t + t Bt (z)
k≥0
and in particular
B−1 (z) =
1
1+
√
1 + 4z
2
and
B2 (z) =
1−
√
1 − 4z
.
2z
Supported by The John Knopfmacher Centre for Applicable Analysis and Number Theory
2
Now
m
X
µ
i
(−1)
i=0
Further,
x+y+i
m−i
¶µ
X µx + y + m − i¶
m
= [z ]
i≥0
i
z ·
i
X µy + 2i¶
i≥0
i
¢x+y+m
¢y
¡
B2 (−z)
m B−1 (z)
= [z ]
¡
¢−1 ·
¡
¢−1
2 − B−1 (z)
−1 + 2 B2 (−z)
¢x+m+1
1 ¡
= [z m ]
.
B−1 (z)
1 + 4z
¡
¶
m µ
X
x+i
i=0
¶
y + 2i
i
m−i
i
m
(−4) = (−4)
¶
m µ
X
x+m−i
i=0
(− 14 )i
i
X µx + m − i¶
1
i
1 − z i≥0
µ
¶
1 X x+m−i i
m
= [z ]
z
i
1 + 4z i≥0
= (−4)m [z m ]
= [z m ]
(− z4 )i
¡
¢x+m+1
1
B
(z)
.
−1
(1 + 4z)3/2
We use the substitution z = u/(1 − u)2 and get:
¶
µ
¡
¢x+m+1
1
x+m+1
m
Sm = [z ]
B−1 (z)
−
3/2
1 + 4z
(1 + 4z)
µ
¶
I
¢x+m+1
¡
1
x+m+1
1
1
=
(z)
dz
B
−
−1
2πi
z m+1
1 + 4z
(1 + 4z)3/2
I
1
(1 − u)2m+2 (1 + u)(2 + x + m) − 2 1 + u
=
du
2πi
um+1
(1 + u)3 (1 − u)x+m−1 (1 − u)3
(1 − u)m−x ((1 + u)(2 + x + m) − 2)
= [um ]
(1 + u)2
(1 − u)m−x
(1 − u)m−x
− 2[um ]
= (2 + x + m)[um ]
.
1+u
(1 + u)2
Expanding the terms leads now to
¶
m µ
(1 − u)m−x X x − k − 1
[u ]
(−1)k
=
m
−
k
1+u
k=0
m
and
µ
¶
m
(1 − u)m−x X
x−k−1
[u ]
=
(k + 1)
(−1)k .
2
(1 + u)
m−k
k=0
m
(−z)i
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 2 (2002), #A06
This gives
3
µ
¶
m
X
£
¤ x−k−1
(x − k) + (m − k)
(−1)k
Sm =
m
−
k
k=0
¶ µ
¶¸
m ·µ
X
x−k
x−k−1
+
(−1)k
= (x − m)
m
−
k
m
−
k
−
1
k=0
µ ¶
x
= (x − m)
,
m
as desired (the last sum is telescoping).
References
[1] R. L. Graham, D. E. Knuth, and O. Patashnik. Concrete Mathematics (Second Edition). Addison
Wesley, 1994.
[2] Z.-W. Sun. A curious identity involving binomial coefficients. Integers, pages A4, 8 pp. (electronic),
2002.