Closed form for $\sum_{k=n}^{2n-1} k(-1)^k \sum_{i=k}^{2n-1} {2n \choose i+1} {i \choose n}$?

I've found this sum:

$$S(n) = \sum_{k=n}^{2n-1} k(-1)^k \sum_{i=k}^{2n-1} {2n \choose i+1} {i \choose n}$$

The inner sum is elliptical iirc, but perhaps the double sum has a nice expression. We can change the order of summation and the expression becomes a tiny bit better to manipulate, but I went straight to try some induction on it. So here is the sequence $S(n)$ as $n$ goes from $1$ to $10$:

$$-1,5,-61,614,-5762, 52093, -460099, 3999356,-34363396, 292676603,...$$

I tried some polynomial interpolations but they didn't stabilize on the values I tried (something like $n=6$). Then I did suppose $S(n) = (-1)^n a^n p(n)$ for $p(n)$ a polynomial and $a=2$ or $a=4$, it also didn't stabilize. I have good reasons to believe

$$S(n) = a \frac{(-1)^n}{n+d} { 4n+b \choose n+c}$$

for some constants $a,b,c$ and $d$ but I'm not sure how to find those constants. Any ideas on how to solve this problem?

EDIT: I made a Python program that verifies those constants and I got nothing for each on the interval $[0,19]$, perhaps $S(n) = \mathcal{o}((\frac{256}{27})^n\frac1{n \sqrt n})$, indeed $\frac1n{4n \choose n}$ begin dominating a bit after $n=40$

EDIT2: changing order of summation: $$n \le k \le i \le 2n-1 \implies S(n) = \sum_{i=n}^{2n-1} \sum_{k=n}^i {2n \choose i+1} {i \choose n}k(-1)^k = \\ = \sum_{i=n}^{2n-1} {2n \choose i+1} {i \choose n} \sum_{k=n}^i (-1)^k k = \frac14 \sum_{i=n}^{2n-1} {2n \choose i+1} {i \choose n}[(-1)^i(2i+1) + (-1)^n(2n-1)] $$

EDIT3: apparently we have an hypergeometric function:

$$_2F_1(1-n,n+1;n+2;-1) = - \frac{(n+1)!}{2(2n+1)!}f^{(n)}(0), f(z) = (1-z)^{-n}(1+z)^{2n+1}$$

EDIT4: By "closed form" I mean a finite expression of few terms which involves only adition, multiplication, potentiation, rational roots, factorials, combinations, floors functions of integers related to $n$. Not involving any explicit algorithm $\mathcal O(n)$ in the final result nor any function not expressed above.