A WZ PROOF OF A “CURIOUS” IDENTITY Shalosh B. Ekhad and Mohamud Mohammed ^{1}

Abstract: We give a short WZ-proof of an identity that appeared recently in Integers.

In [1] and [2], the identity,

∑_{i=0}^{m}(x + m + 1)(−1)^{i}\binom{x + y + i}{m − i}\binom{y + 2i}{i} − ∑_{i=0}^{m} \binom{x + i}{m − i}(−4)^{i} = (x − m)\binom{x}{m}...........(∗)

was proved using generating functions and double recursions respectively. Here we cleverly construct the function

g(i) = \frac{−\binom{x}{i}(x^{2} − 2ix − x + i + i^{2})}{2(1 + i)(x + i + 2)(x + i + 1)(−2)^{i}}

with the motives that

∑_{i=0}^{m} \binom{x + i}{m − i}(−4)^{i} = (−2)^{m}(x + m + 1) (\frac{1}{x + 1} + ∑_{i=0}^{m−1} g(i)) ..........(∗∗)

The first sum, call it T (m), on the LHS of (∗) divided by (x+m+1) satisfies the following recurrence equation as found by EKHAD. a_{0}(m)T (m) + a_{1}(m)T (m + 1) + a_{2}(m)T (m + 2) + a_{3}(m)T (m + 3) = 0, where a_{0}(m) = 2(x − m − 1)(x − m − 2); a_{1}(m) = − (x − m − 2) (2 y − x + 5 m + 11), a_{2}(m) = (−yx + 3 ym − 2 xm + 4 m^{2} + 8 y − 5 x + 21 m + 28); a_{3}(m) = (m + 3) (y + m + 3) Now, it is routine(use maple!) to check that the sum of the RHS’s of (∗) and (∗∗) divided by (x + m + 1) satisfies the recurrence equation. Moreover, both sides agree for m = 0, 1, 2. Q.E.D. REFERENCES

[1] Alois Panholzer, and Helmut Prodinger. A generating functions proof of a curious identity. Integers, pages A6, 3 pp. (electronic), 2002.

[2] Zhi-Wei Sun. A curious identity involving binomial coefficients. Integers, pages A4, 8 pp. (electronic), 2002

1 Department of Mathematics, Rutgers University, Piscataway, NJ 08854, USA. mohamudm@math.rutgers.edu

http://www.math.rutgers.edu/~mohamudm/. April 12, 2002.