1

With the help of Mathematica I have discovered a following identity.

Let $T>1$ be an integer, $x$ be a real number and let q be a positive even integer and $l=0,1,\cdots,q/2$. The following identity holds: \begin{eqnarray} &&\sum\limits_{m=0}^\infty \frac{(2m+T+q-1)!}{(m!)^2 (m+l+\frac{T-3}{2}+1)!} (m+l)! x^{2m}= \frac{2(\frac{q}{2}-l)!l!}{\left(1-4 x^2\right)^{\frac{T}{2}+q}} \frac{(T+q-2)!}{\left(\frac{T+q-3}{2}\right)!} \cdot \\ &&\sum\limits_{p_1=0}^{\frac{q}{2}-l}\sum\limits_{p_2=0}^l \binom{\frac{T-1+q}{2}}{\frac{q}{2}-l-p_1} \binom{-\frac{T+q}{2}}{p_1} \binom{p_1+l}{l-p_2} \binom{-\frac{T+q}{2}-p_1}{p_2} \left(-4 x^2\right)^{p_1+p_2} \left(1-4 x^2\right)^{\frac{q}{2}-p_1-p_2} \end{eqnarray} If $q$ is an odd integer and $l=0,\cdots,(q-1)/2$ then we have: \begin{eqnarray} &&\sum\limits_{m=0}^\infty \frac{(2m+T+q-1)!}{(m!)^2 (m+l+\frac{T-3}{2}+1)!} (m+l)! x^{2m}= \frac{(\frac{q-1}{2}-l)!l!}{\left(1-4 x^2\right)^{\frac{T}{2}+q}} \frac{(T+q-1)!}{\left(\frac{T+q-2}{2}\right)!} \cdot \\ &&\sum\limits_{p_1=0}^{\frac{q-1}{2}-l}\sum\limits_{p_2=0}^l \binom{\frac{T-2+q}{2}}{\frac{q-1}{2}-l-p_1} \binom{-\frac{T+q+1}{2}}{p_1} \binom{p_1+l}{l-p_2} \binom{-\frac{T+q+1}{2}-p_1}{p_2} \left(-4 x^2\right)^{p_1+p_2} \left(1-4 x^2\right)^{\frac{q-1}{2}-p_1-p_2} \end{eqnarray}

Now, is there another way to prove that identity using pure combinatorial arguments?

Przemo
  • 11,331
  • I think the right way to look at this is that this formula is true at the level of formal power series in $x$ (and therefore true for real $x$ as well.) So you're trying to show that two different formula power series are identical; the Snake Oil method of Wilf is an obvious suggestion. – Semiclassical Aug 27 '14 at 15:20
  • The right hand side is of the form $\left(1/\sqrt{1-4x^2}\right)^N * f(x)$, for some polynomial $f$ and positive integer $N$. This indicates that for fixed $T,q,l$, the right side is a finite, though complicated, manipulation of the sequence $C(2n,n)$. Since the right side is algebraic and hence D-finite/P-recursive, in principle there does exist a (probably complex) combinatorial model via http://dl.acm.org/citation.cfm?id=2385116. For fixed $T,q,l$, the left side also looks like it satisfies a polynomial recursion, so there may be hope. But it looks really complicated! – Hugh Denoncourt Aug 29 '14 at 16:33

0 Answers0