I calculated, using Mathematica, that for $4\leq k \leq 100$,
$$ \sum_{j=k}^{2k} \sum_{i=j+1-k}^j (-1)^j 2^{j-i} \binom{2k}{j} S(j,i) s(i,j+1-k) = 0,$$
where $s(i,j)$ and $S(i,j)$ are Stirling numbers of the first and second kinds, respectively.
Here the code:
F[k_] := Sum[(-1)^j 2^(j – i) Binomial[2 k, j] StirlingS2[j,
i] (StirlingS1[i, j + 1 – k]), {j, k, 2 k}, {i, j – k + 1, j}];Table[F[k], {k, 4, 100}]
How do I prove it holds for all $k \geq 4$ ?
Best Answer
It looks like my first response interprets the problem statement to use unsigned Stirling numbers of the first kind. We find for signed ones,
$$S_n = (-1)^{n+1} \sum_{j=n}^{2n} \sum_{k=j+1-n}^j (-1)^k 2^{j-k} {2n\choose j} {j\brace k} {k\brack j+1-n}.$$
With the usual EGFs we get
$$(-1)^{n+1} \sum_{j=n}^{2n} \sum_{k=j+1-n}^j (-1)^k 2^{j-k} {2n\choose j} j! [z^j] \frac{(\exp(z)-1)^k}{k!} \\ \times k! [w^k] \frac{1}{(j+1-n)!} \left(\log\frac{1}{1-w}\right)^{j+1-n}.$$
Now we have
$${2n\choose j} j! \frac{1}{(j+1-n)!} = \frac{(2n)!}{(2n-j)! \times (j+1-n)!} = \frac{(2n)!}{(n+1)!} {n+1\choose j+1-n}.$$
This yields for the sum
$$(-1)^{n+1} \frac{(2n)!}{(n+1)!} \sum_{j=n}^{2n} {n+1\choose j+1-n} 2^j \\ \times [z^j] \sum_{k=j+1-n}^j (-1)^k 2^{-k} (\exp(z)-1)^k [w^k] \left(\log\frac{1}{1-w}\right)^{j+1-n} \\ = (-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^n \sum_{j=0}^{n} {n+1\choose j+1} 2^j \\ \times [z^{n+j}] \sum_{k=j+1}^{j+n} (-1)^k 2^{-k} (\exp(z)-1)^k [w^k] \left(\log\frac{1}{1-w}\right)^{j+1}.$$
Observe that $(\exp(z)-1)^k = z^k + \cdots$ and hence we may extend the inner sum beyond $j+n$ due to the coefficient extractor $[z^{n+j}].$ We find
$$(-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^n \sum_{j=0}^{n} {n+1\choose j+1} 2^j \\ \times [z^{n+j}] \sum_{k\ge j+1} (-1)^k 2^{-k} (\exp(z)-1)^k [w^k] \left(\log\frac{1}{1-w}\right)^{j+1}.$$
Furthermore note that $\left(\log\frac{1}{1-w}\right)^{j+1} = w^{j+1} +\cdots$ so that the coefficient extractor $[w^k]$ covers the entire series, producing
$$(-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^n \sum_{j=0}^{n} {n+1\choose j+1} 2^j [z^{n+j}] \left(\log\frac{1}{1+(\exp(z)-1)/2}\right)^{j+1}.$$
Working with formal power series we are justified in writing
$$[z^{n+j}] \left(\log\frac{1}{1+(\exp(z)-1)/2}\right)^{j+1} = [z^{n-1}] \frac{1}{z^{j+1}} \left(\log\frac{1}{1+(\exp(z)-1)/2}\right)^{j+1}$$
because the logarithmic term starts at $(-1)^{j+1} z^{j+1}/2^{j+1}.$ To see this write
$$-\frac{\exp(z)-1}{2} + \frac{1}{2} \frac{(\exp(z)-1)^2}{2^2} - \frac{1}{3} \frac{(\exp(z)-1)^3}{2^3} \pm \cdots$$
We continue
$$(-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^{n-1} \\ \times [z^{n-1}] \sum_{j=0}^{n} {n+1\choose j+1} 2^{j+1} \frac{1}{z^{j+1}} \left(\log\frac{1}{1+(\exp(z)-1)/2}\right)^{j+1} \\ = (-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^{n-1} \\ \times [z^{n-1}] \sum_{j=1}^{n+1} {n+1\choose j} 2^{j} \frac{1}{z^{j}} \left(\log\frac{1}{1+(\exp(z)-1)/2}\right)^{j}.$$
The term for $j=0$ in the sum is one and hence only contributes to $n=1$ so that we may write
$$-[[n=1]] + (-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^{n-1} \\ \times [z^{n-1}] \sum_{j=0}^{n+1} {n+1\choose j} 2^{j} \frac{1}{z^{j}} \left(\log\frac{1}{1+(\exp(z)-1)/2}\right)^{j} \\ = -[[n=1]] + (-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^{n-1} \\ \times [z^{n-1}] \left(1+\frac{2}{z} \log\frac{1}{1+(\exp(z)-1)/2}\right)^{n+1}.$$
Finally observe that
$$\left(1+\frac{2}{z} \log\frac{1}{1+(\exp(z)-1)/2}\right)^{n+1} \\ = \left(1+\frac{2}{z} \left( -\frac{\exp(z)-1}{2} + \frac{1}{2} \frac{(\exp(z)-1)^2}{2^2} - \frac{1}{3} \frac{(\exp(z)-1)^3}{2^3} \pm \cdots \right)\right)^{n+1} \\ = \left( -\frac{1}{4} z - \cdots \right)^{n+1}$$
and furthermore
$$[z^{n-1}] \left((-1)^{n+1} \frac{1}{4^{n+1}} z^{n+1} + \cdots \right) = 0$$
which is the claim.