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Here is the answer I hinted at in a comment, in real detail. Took me a while, but I had no idea how tiresome such arguments are to expose... Yes, it is true: see Corollary 6 (b) below. The proof reli …

Let $n$ be a nonnegative integer, and let $B$ be the $n \times n$-matrix (over the rational numbers) whose $\left(i, j\right)$-th entry is $\dbinom{n+1}{2j-i}$ for all $i, j \in \left\{ 1, 2, \ldots, …

Let $l$ and $m$ be two integers such that $l\geq m\geq0$. You want me to prove the identity \begin{align} & \left( 1-\left( 2m+1\right) \left( m+1\right) \right) \dbinom{l+1} {m}+\sum_{k=m+1}^{l …

This is correct. Let me prove a more general fact: Theorem 1. Let $i$, $j$ and $n$ be three nonnegative integers such that $i\leq n$ and $j\leq n$. Let $P\in\mathbb{Q}\left[ X\right] $ be a …

No, there are no others. In fact, define a function $q : \mathbb N\to\left\lbrace 1,-1\right\rbrace$ by $q\left(i\right) = \left(-1\right)^i p\left(i\right)$ for every $i\in\mathbb N$. Then, $\sum\li …

Wadim, isn't that 95% of the proof? First let me correct your first displayed equation (thanks to fherzig for pointing this out): It is not sufficient for the proof, but $$ \sum_{i=0}^{n-1}\binom{4n}{ …