TITLE: Closed formula for $(-1)$-Baxter sequences
QUESTION [5 upvotes]: The number of the so-called Baxter permutations of length $n$ is computed by
$$a_n=\frac1{\binom{n+1}1\binom{n+1}2}\sum_{k=0}^{n-1}\binom{n+1}k\binom{n+1}{k+1}\binom{n+1}{k+2}.$$
There has also been a $q$-analogue of this sequence
$$a_n(q)=\frac1{\binom{n+1}1_q\binom{n+1}2_q}\sum_{k=0}^{n-1}q^{3\binom{k+1}2}
\binom{n+1}k_q\binom{n+1}{k+1}_q\binom{n+1}{k+2}_q.$$
Hence $a_n(1)=a_n$.
Now, if we focus on the special values at $q=-1$ then it appears that the resulting sequence does seem to exhibit small prime factors, suggesting that one might be able to find a closed formula. Notice that such is not true when $q\neq-1$. Here are the first few values of $a_n(-1)$ for $n\geq1$: $1, 0, 0, -2, -4, 0, 0, 18, 42, 0, 0, -240, -600, 0, 0, 3850, 10010, 0, 0, -68796,\dots$.
So, I like to ask:

QUESTION. Is there a nice formula for $a_n(-1)$? Of course, one may restrict to the non-zero values.

REPLY [5 votes]: 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 $n\rightarrow 4n+1$:
$$\frac1{2n+1}\sum_{k=0}^{2n}(-1)^k\binom{2n+1}k\binom{2n+1}{k+1}\binom{2n}k
=\frac{(-1)^n}{2n+1}\binom{3n+1}n\binom{2n+1}n.$$