Absolute values $1,4,42,600,10010,183456,\ldots$ of the non-zero terms not covered by Brian Hopkins answer seem to coincide with the sequence $$n\longmapsto {3n-2\choose n-1}{2n-1\choose n-1}\frac{1}{2n-1}$$ which is absent of the OEIS.
This corresponds to Fedor's case $n\rightarrow 4n+1$: $$\frac1{2n+1}\sum_{k=0}^{2n}(-1)^k\binom{2n+1}k\binom{2n+1}{k+1}\binom{2n}k =\frac{(-1)^n}{2n+1}\binom{3n+1}n\binom{2n+1}n.$$
