A combinatorial sum

Question

• Any idea to evaluate the sum $$ \sum_{j=m}^{k}\frac{\binom{m}{2m - j\,\,}}{\binom{k}{j}} \quad\mbox{with}\quad m \leq k < 2m - 1. $$
• I have found the sum for $k=2m-1$. In fact, it is verified that $$ \sum_{j = m}^{2m - 1}\ \frac{\binom{m}{2m-j\,\,}} {\binom{2m - 1\,\,}{j}} = 2m\left(H_{2m} - H_{m}\right), $$ where $H_{j}$ is the $j$-th harmonic number.

Answer 1

For a short time, I hoped to be able to express the result in terms of hypergeometric functions but I failed.

If we let $k=m+n$, the problem reduces to $$S_n=\sum_{j=m}^{m+n} \frac{\Gamma (j+1)\,\,\Gamma (m+n+1-j)}{\Gamma (m+n+1)} \,\binom{m}{2 m-j} $$ which can write $$S_n=\frac{\Gamma(m+1)}{n!\ \Gamma(m+n+1)} P_{n}(m)$$ where $P_n$ is a polynomial of degree $2n$ (it does not seem to be factorable for any $n$).

The very first are $$\left( \begin{array}{cc} n & P_n \\ 0 & 1 \\ 1 & m^2+m+1 \\ 2 & m^4+2 m^3+m^2+4 \\ 3 & m^6+3 m^5-2 m^4-9 m^3+13 m^2+18 m+36 \\ 4 & m^8+4 m^7-10 m^6-44 m^5+53 m^4+184 m^3+100 m^2+576 \end{array} \right)$$

I have not been able to find any pattern for the coefficients [except that the constant term is $(n!)^2$ and that the coefficient of $m^{2n}$ is $1$ (!!) ].

The only thing I observed is that, if $n$ is even, the term is $m$ is systematically missing.

Without any proof of it, I do not think that a closed form could exist for a general $n$.

Answer 2

Here maxima's Zeilberger package (take a peek at Petkovsek, Wilf, Zeilberger "A = B" for the gory details) says your sum isn't Gosper summable. That means there is no simple closed form for it.