Timeline for A conjectured formula for Apéry numbers

6 events

when toggle format	what		by	license	comment
Aug 22, 2014 at 13:44	vote	accept	Gerald Edgar
Aug 18, 2014 at 19:09	comment	added	Mark Wildon		Another proof : $\sum_{j} \binom{k}{j}^3 = \sum_r \binom{k}{r} \binom{k}{k-r} \binom{k}{r} = \sum_r \sum_t \binom{k}{r}\binom{k}{k-r}\binom{r}{t}\binom{k-r}{t} = \sum_r\sum_t \binom{k}{t}\binom{k-t}{r-t}\binom{k}{t}\binom{k-t}{k-r-t}= \sum_{t} \binom{k}{t}^2 \sum_r \binom{k-t}{r-t}\binom{k-t}{k-r-t} = \sum_{u} \binom{k}{u}^2 \sum_r \binom{u}{k-r}\binom{u}{r} = \sum_u \binom{k}{u}^2 \binom{2u}{k} = \sum_u \binom{k}{u} \binom{2u}{u} \binom{u}{k-u} = \sum_\ell \binom{k}{\ell}\binom{2(k-\ell)}{k-\ell} \binom{k-\ell}{\ell} = \sum_\ell \binom{k}{2\ell}\binom{2\ell}{\ell}\binom{2(k-\ell)}{k-\ell}.$
Aug 18, 2014 at 18:33	history	edited	Paata Ivanishvili	CC BY-SA 3.0	added 6 characters in body
Aug 18, 2014 at 18:13	comment	added	Timothy Chow		The identity involving $\sum_j {k \choose j}^3$ can also be proved by observing that both sides satisfy the recursion $(8k^2+32k+32)f(k+1) + (7k^2+35k+44)f(k+2) -(k^2+6k+9)f(k+3)$ and have the same initial values, where the recursion can be found by standard techniques (e.g., $\mathtt{listtorec}$ in Maple's $\mathtt{gfun}$ package).
Aug 18, 2014 at 17:59	history	edited	Paata Ivanishvili	CC BY-SA 3.0	added 295 characters in body
Aug 18, 2014 at 16:45	history	answered	Paata Ivanishvili	CC BY-SA 3.0