In A binomial identity from counting permutations with cycles of length two. we have analyzed a certain hypergeometric sum by three different methods, firstly using a combinatorial, secondly two different analytical approaches. Here we want to generalize the results. To be specific we consider the following sum: \begin{equation} S_d(n) := n! \sum\limits_{k=0}^{\lfloor n/d \rfloor} \binom{-\frac{1}{d} + k}{k} \binom{n}{d \cdot k} \end{equation} where $d=2,3,4,5,\cdots$. For given value of $d$ we used Sister Celine's algorithm with $I=d$ and $J=1$ (see the other question for more information on the algorithm). As a result we found the following recursion relations for our sum in question: \begin{eqnarray} \tiny S_d(n) - (2 n-1) S_d(n-1)=0 & \quad \mbox{for $d=2$} \\ \tiny (n-2) S_d(n) - (3 n^2-7 n+3) S_d(n-1) + (n-1)^2 (3n-5) S_d(n-2) -2 (n-2)^2 S_d(n-3)=0 & \quad \mbox{for $d=3$} \\ \tiny (n-2)(n-3) S_d(n) - (n-3)(4 n^2-9 n+4) S_d(n-1) + (n-1)^2(28-27 n+6 n^2) S_d(n-2)-(n-2)^2(n-1)^2(4 n-11) S_d(n-3)=0 & \quad \mbox{for $d=4$} \end{eqnarray} In general we have: \begin{equation} \sum\limits_{i=0}^d (-1)^{d-i} \cdot \left\{ \begin{array}{rrr} (1+(-1)^{d-1}) & \mbox{$i=d$}\\ 1 & \mbox{$i\ne d$} \end{array}\right\} \cdot \left\{ \begin{array}{rrr} \frac{(n-i-2)!}{(n-d)!} & \mbox{$i\le d-2$}\\ 1 & \mbox{$i\ge d-1$} \end{array}\right\} \cdot \left\{ \begin{array}{rrr} [\frac{(n-1)!}{(n-i)!}]^2 & \mbox{$i\ge 1$}\\ 1 & \mbox{$i=0$} \end{array}\right\}\cdot {\mathcal A}_{i,d}(n) S_d(n-i) =0 \end{equation} where \begin{equation} {\mathcal A}_{i,d}(n) := \left\{ \begin{array}{rrr} 1 & \mbox{$i=0$}\\ d-(2 d+1) n+d \cdot n^2 & \mbox{$i=1$} \\ \frac{1}{i-1} \binom{d-2}{i-2}d (i d-1) - (2 d-1) \binom{d-1}{i-1} n + \binom{d}{i} n^2 & \mbox{$2 \le i \le d-2$} \\ -d(d-1)+1+d \cdot n & \mbox{$i=d-1$}\\ 1 & \mbox{$i=d$} \end{array} \right. \end{equation} Now, my question would be is it possible to find a closed form solution to those recurrences in the case $d > 2$? Another question asks about alternative ways of tackling this sum.
Closed form for a hypergeometric sum
1 Answers
Partial answer since I don't see any reasonable way to simplify the last result.
BC4 and GA refer to GouldBk.pdf
http://www.dsi.unifi.it/~resp/GouldBK.pdf
Which I am working my way through.
Note that the upper limit of the sum is automatically satisfied.
Rewriting for the Riordan transform
We use BC4:
$\mathcal{G}\left(\left(\begin{array}{c}
\frac{1}{d}+k\\
k
\end{array}\right)\right)=\frac{1}{\left(1-t\right)^{1+\frac{1}{d}}}$
i.e.: $ \left[t^{k}\right]\frac{1}{\left(1-t\right)^{1+\frac{1}{d}}}=\left(\begin{array}{c} \frac{1}{d}+k\\ k \end{array}\right)$
and rewrite:
$\frac{S_{d}\left(n\right)}{n!}={\displaystyle \sum_{k=0}^{\infty}\left(\begin{array}{c} n\\ d\cdot k \end{array}\right)\left[t^{k}\right]\left(\frac{1}{\left(1-t\right)^{1+\frac{1}{d}}}\right)}$
Using GA, z=1 :
Yields
$=\left[t^{n}\right]\frac{t^{0}}{\left(1-t\right)^{1}}\cdot\frac{1}{\left(1-\left(\frac{t^{d}}{\left(1-t\right)^{d}}\right)\right)^{1}\left(1-\left(\frac{t^{d}}{\left(1-t\right)^{d}}\right)\right)^{\frac{1}{d}}}$
Which gives the Ordinary Generating Function in $t^{n}$
$\frac{t^{0}}{\left(1-t\right)^{1}}\cdot\frac{1}{\left(1-\left(\frac{t^{d}}{\left(1-t\right)^{d}}\right)\right)^{1}\left(1-\left(\frac{t^{d}}{\left(1-t\right)^{d}}\right)\right)^{\frac{1}{d}}} $
I don't see much point in rearranging terms here.
One could just take the n th derivative or ask a symbolic algebra program to start expanding out the OGF series in the indexing variable $[t^{n}]$.