<s>[INST] How to calculate the sum of general type $\sum_{k=0}^n {n\choose k} {n\choose k+a} {2 k- n + a \choose r }$? 
QUESTION. How to calculate the sum of such general type?
$$\sum_{k=0}^n {n\choose k} {n\choose k+a} {2 k - n + a \choose r }. $$

Some particular examples
$$\sum_{k=0}^n {n\choose k} {n\choose k+a}  = {2n\choose n+a},$$
$$\sum_{k=0}^n {n\choose k} {n\choose k+a} \binom{2 k - n + a}1 = 0, $$
$$\sum_{k=0}^n {n\choose k} {n\choose k+a} {2 k - n + a \choose 2}  = -\frac{(a+ n)(a-n)}{2 (2n-1)} {2n\choose n+a}, $$
$$\sum_{k=0}^n {n\choose k} {n\choose k+a} {2 k - n + a \choose 3}  = \frac{(a+ n)(a-n)}{2 (2n-1)} {2n\choose n+a}. $$
I am struggling to calculate at least
$$\sum_{k=0}^n {n\choose k} {n\choose k+a} {2 k - n + a \choose 4}   $$
Can someone  help me with this one?
 [/INST] Alternatively, one can get an explicit expression for a fixed $r$ via generating functions by noticing that the given sum equals
$$[y^{n+a}z^r]\ (y+1+z)^n (1+y(1+z))^n (1+z)^{-n}$$
and rewriting it as
$$[y^{n+a}z^r]\ \big((1+y)^2 + \frac{z^2}{1+z}y\big)^n = [y^{n+a}z^r]\ \sum_{i=0}^{\lfloor r/2\rfloor} \binom{n}{i} (1+y)^{2(n-i)} y^i\left(\frac{z^2}{1+z}\right)^i,$$
which we evaluate as
$$[y^{n+a}]\ \sum_{i=0}^{\lfloor r/2\rfloor} \binom{n}i\binom{-i}{r-2i} (1+y)^{2(n-i)}y^i = (-1)^r \sum_{i=0}^{\lfloor r/2\rfloor}  \binom{n}i \binom{r-i-1}{r-2i} \binom{2(n-i)}{n+a-i}.$$
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