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Let $n, j, j_1 \ge 0$ be integers. Then let $k\ge 2$ and $l=0,\cdots,k-2$. We state a following combinatorial identity: \begin{eqnarray} &&\sum\limits_{s=0}^{l-1} \binom{k+j_1}{l+s+j_1+2} \binom{l+j}{s+1+j} \binom{s}{n} =\\ && \sum\limits_{s=0}^n \binom{k+j_1+j+l-s}{1+j_1+2 l} (-1)^{n+s}\cdot \binom{n-s+j}{j} \binom{l+j}{s} -\\ &&\sum\limits_{s=0}^j \binom{k+j_1}{l+1+j_1-s}(-1)^n \cdot \binom{n+s}{n} \binom{l+j}{j-s} \end{eqnarray} We have discovered this identity by relating the left hand side to the $n$th derivative of the hypergeometric function $F_{2,1}$ at unity and then using the chain rule for differentiation, simple algebra, and the Gauss' formula for $F_{2,1}$ at unity.

Note 1: The left hand side can be seen as a linear combination of positive powers of $k$ with positive coefficients. Here the highest and the lowest powers are $2l+1+j_1$ and $l+j_1+2$ respectively. Interestingly enough the right hand side consists of "border terms" only.Indeed the first sum on the right hand side is related to the highest power and the second sum to all the powers that are strictly lower than the lowest power.

Note 2:The Zeilberger algorithm (see book A==B for example) when applied to the left hand side for some fixed values of $n,j,j_1$ always returns a a two-point recursion relation-- with respect to the parameter $l$-- with some non-zero right hand side. This however does not give us a clue whether a closed form solution exist whereas our approach does give a closed form.

Now, the question is as the usual one. Can we prove identities of this kind using combinatorics?

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  • $\begingroup$ I don't know if your still interested but I might be able to solve it (or not). Notice that the summation limits can be removed by rewriting the middle term of the first expression as $\left(\begin{array}{c} l+j\\ s+1+j \end{array}\right)=\left(\begin{array}{c} l+j\\ \left(l-1\right)-s \end{array}\right)$ $\endgroup$ – rrogers Apr 27 '17 at 16:06
  • $\begingroup$ Are you interested in the generating function for these sequences? I am pretty sure I can do that. See my answer at math.stackexchange.com/questions/2194252 although I would have to work some more to include (-1)^x and have to make sure the summations are "natural". $\endgroup$ – rrogers Sep 7 '17 at 21:59

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