Closed form for $\sum\limits_i (-1)^i \binom{m}{i} \sum\limits_{j \ge k+1} \binom{2k+1-m}{j-i}.$ I would like a closed form for the sum $S(m,k) = \sum\limits_{i=0}^m (-1)^i \binom{m}{i} \sum\limits_{j = k+1}^{2k+1} \binom{2k+1-m}{j-i}.$
For $m=0$ we always get $4^k,$ so I assume $k>0.$ For $1 \le m \le 2k+1,$ the sums are $(-2, 0, 2)$ for $k=1,$ $(6, 0, -2, 0, 6)$ for $k=2,$ $(20, 0, -4, 0, 4, 0, -20)$ for $k=3,$ and $(70, 0, -10, 0,-6, 0, -10, 0, 70)$ for $k=3.$
The symmetry for odd $k$ and antisymmetry for even $k$ plus the fact the sum is $0$ for even $k>0$ means there's a lot of patterns. Due to all these patterns, I'm hoping there's a closed form.
Update: Let $r = j-i,$ then the coefficient of $\binom{2k+1-m}{r}$ in the sum is $\sum\limits_{i \ge k+1-r} (-1)^i \binom{m}{i}$ since $i$ gets counted whenever $r+i = j \ge k+1 \Rightarrow i \ge k+1-r.$ This sum is $(1-1)^m = 0$ for $r \ge k+1,$ otherwise we get $\sum\limits_{i=0}^{k-r} (-1)^i \binom{m}{i}.$ Thus, $S(m,k) = \sum\limits_r \binom{2k+1-m}{r} \sum\limits_{i=0}^{k-r} (-1)^i \binom{m}{i} = \sum\limits_r \binom{2k+1-m}{k-r} \sum\limits_{i=0}^r (-1)^i \binom{m}{i}$ by reindexing. We can prove the inner sum is $(-1)^r \binom{m-1}{r}$ by induction, hence $$S(m,k) = \sum\limits_i \binom{2k+1-m}{k-i} \binom{m-1}{i} (-1)^i $$
after relabeling $r \to i$ for convenience. Without the term $(-1)^i,$ this would be $\binom{2k}{k}$ by Vandermonde's identity.
The closed form is $S(m,k) = \begin{cases} 0, m \text{ even} \\ (-1)^{n+1} \frac{(2n)!}{n!} \frac{(2k-2n)!}{k!(k-n)!}, n = (m-1)/2, \, m \text{ odd}\end{cases}.$
There is a sign error somewhere in the previous work. By going back to the double sum right before induction and pairing opposite terms, we can prove the $m$ even case. Plugging in $m = 1, 3, 5$ was enough to spot the $k$ terms and alternating sign, plugging in $m = 7$ gave me enough data on the coefficients to conjecture they're $\frac{(2n)!}{n!}.$ I have confirmed this works for $m = 9, 11$ and other small $n$ via Desmos, so all that remains is a proof.
Another interesting thing, you can rewrite the last expression as $(-1)^{n+1}\frac{\binom{k}{n}\binom{2k}{k}}{\binom{2k}{2n}}.$

If there is a closed form, it would not be very nice since if
$$ A_i(m,k)=\sum\limits_{j = k+1}^{2k+1} \binom{2k+1-m}{j-i}$$
using the Gaussian hypergeometric function
$$ A_i(m,k)=\binom{2 k-m+1}{k+1-i} \, _2F_1(1,-i-k+m;k+2-i;-1)-$$ $$\binom{2
   k-m+1}{2 k+2-i} \, _2F_1(1,m+1-i;2 k+3-i;-1)$$
For
$$S(m,k)=\sum_{i=0}^m (-1)^i\, \binom{m}{i}\,A_i(m,k)$$ Mathematica provides an implicit closed form which more than nasty including two recurrence equations.
But, using your last result (I think that there is a sign error)
$$S(m,k) = \sum\limits_i  (-1)^{i+1}\,\binom{2k+1-m}{k-i} \binom{m-1}{i}$$ this becomes quite simple
$$S(m,k) =-\binom{2 k-m+1}{k} \,\, _2F_1(-k,1-m;k-m+2;-1)$$
