I'm trying to prove the following identities: $$\forall t\geqslant2,\forall 0\leqslant i\leqslant\left\lfloor\frac{t}{2}\right\rfloor,S_{\text{even}}(i, t)=\sum_{k=0}^{\left\lfloor\frac{t}{2}\right\rfloor}(-1)^k\binom{t}{2k}(2i+2k-1)!!(2t-2i-2k-1)!!=(-1)^it!$$ $$\forall t\geqslant2,\forall 0\leqslant i\leqslant\left\lfloor\frac{t-1}{2}\right\rfloor,S_{\text{odd}}(i, t)=\sum_{k=0}^{\left\lfloor\frac{t-1}{2}\right\rfloor}(-1)^k\binom{t}{2k+1}(2i+2k+1)!!(2t-2i-2k-3)!!=(-1)^it!$$ Numerically, this holds for the first values of $i$ and $t$. What I've tried so far:
- Induction on $t$: it changes the summand too much for it to work in the general case. We do have $S_{\text{odd}}(0,t+1)=(t+1)S_{\text{even}}(0, t)$ though, so if we manage to show that increasing $i$ flips the sign, we could consider only the case $i=0$ and that would be great.
- "Induction" on $i$, or more exactly trying to prove that increasing or decreasing $i$ leads to flipping the sign of the result. Clearly, the summand of $S_{\text{even}}(i+1, t)$ and $S_{\text{odd}}(i, t)$ are very similar, but the binomial coefficient prevents us from getting an immediate relation. We can sum them using the summation formula for binomial coefficient, but I don't see an immediate way to justify that this sum is nil.
- Find an existing formula. I found two potentially relevant paper concerning this point, namely this one and this one. The first one contain sums that are kind of similar but that are not alternates, the second one contains alternate sums but with stirling numbers.
- Use combinatorics. Notably, the fact that this is an alternate sum makes me think of the onclusion-exclusion principle, but I didn't find a clever way to map this quantity to something that can be counted.
Is there a simple way to prove these identities? Or even better, are they already proven or are they a direct application of some more general identity that I've missed?