Closed form for $\sum_{l=m}^{\lfloor n/2\rfloor} (-1)^l \binom{2(n-l)}{n}\binom{n}{l}\binom{l}{m}$ I have a sum in terms of the coefficients 
$$a_{m,n} = \sum_{l=m}^{\lfloor n/2\rfloor} (-1)^l\binom{2(n-l)}{n}\binom{n}{l}\binom{l}{m}.$$
Given the answer to my previous question, I suspect a closed form might exist for this rather similar sum. I've tried applying a method similar to the one used in the answer to that problem, but the upper bound of $\lfloor n/2\rfloor$ here is proving to be troublesome. Does anybody have a suggestion or an answer?
 A: I have found the solution; as it turns out, the method in the linked question does apply to this sum as well. We can rewrite the sum as
$$\binom{n}{m}\sum_{l=m}^{\lfloor n/2 \rfloor} (-1)^l \binom{n-m}{l-m}\binom{2(n-l)}{n}$$
Using the integral representation of the binomial coefficients, this becomes 
$$\frac{1}{2\pi i}(-1)^m\binom{n}{m}\oint \frac{dz}{z^{n+1}} \sum_{l=m}^{\lfloor n/2 \rfloor} \binom{n-m}{l-m} (1+z)^{2(n-l)}$$
The integral vanishes for $2l>n$, so we can extend the summation to $l=n$. Then we recognize a binomial expansion of $(1-(1+z)^{-2})^{n-m}$, and so after some cancellation we have
$$\frac{1}{2\pi i}(-1)^m\binom{n}{m}\oint \frac{(z+2)^{n-m}\,dz}{z^{m+1}}$$
The remaining integral is almost the integral representation of $\binom{n-m}{m}$, except the residue will have an extra factor of $2^{n-2m}$. Thus, the closed form is
$$(-1)^m\binom{n}{m}\binom{n-m}{m}2^{n-2m}$$
A: Suppose we seek to find a closed form of
$$a_{n,m} = \sum_{l=m}^{\lfloor n/2\rfloor} 
(-1)^l {2n-2l\choose n} {n\choose l} {l\choose m}.$$
We have
$${n\choose l} {l\choose m}
= \frac{n!}{(n-l)! m! (l-m)!}
= {n\choose m} {n-m\choose l-m}$$
and obtain for the sum
$${n\choose m} \sum_{l=m}^{\lfloor n/2\rfloor} 
(-1)^l {2n-2l\choose n} {n-m\choose l-m}.$$
Introduce 
$${2n-2l\choose n}  = {2n-2l\choose n-2l} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{n-2l+1}} (1+z)^{2n-2l} \; dz$$
Note that this vanishes  when $2l\gt n$ so we may raise  $l$ to end at
$n$, getting for the inner sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{n+1}} (1+z)^{2n} 
\sum_{l=m}^n (-1)^l {n-m\choose l-m}
\frac{z^{2l}}{(1+z)^{2l}}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{n+1}} (1+z)^{2n} \frac{(-1)^m z^{2m}}{(1+z)^{2m}}
\sum_{l=0}^{n-m} (-1)^l {n-m\choose l}
\frac{z^{2l}}{(1+z)^{2l}}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{(-1)^m}{z^{n-2m+1}} (1+z)^{2n-2m}
\left(1-\frac{z^2}{(1+z)^2}\right)^{n-m}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{(-1)^m}{z^{n-2m+1}} (1+2z)^{n-m}
\; dz.$$
We thus have the closed form
$${n\choose m} (-1)^m 2^{n-2m} {n-m\choose n-2m}
= {n\choose m} (-1)^m 2^{n-2m} {n-m\choose m}.$$
This is zero when $2m\gt n$ just like the sum that we started with.
Same as in the link by the OP we see having reached the result that we
did not make  use of the differential in the  integral which means the
above also works using formal power series only.
