TITLE: A seemingly simple combinatorial object that must have an easy generating function
QUESTION [14 upvotes]: One more question related to my earlier "Special" meanders.
I am trying to isolate simplest problems related to it. Here is one.
For a composition (i. e. a tuple of natural numbers) $\boldsymbol a=(a_1,...,a_k)$, define the set of its midpoints by
$$
\operatorname{mid}(\boldsymbol a):=\{a_1+...+a_{i-1}+\frac{a_i+1}2\mid1\leqslant i\leqslant k\}.
$$
For example, $\operatorname{mid}(4,1,3)=\{\frac52,5,7\}$.
Call compositions $\boldsymbol a$ and $\boldsymbol b$ unmatchable if $\operatorname{mid}(\boldsymbol a)\cap\operatorname{mid}(\boldsymbol b)=\varnothing$.
I need any explicit information (formula, generating function, ...) for the numbers
$F(n):=$ number of unmatchable pairs $\langle\boldsymbol a,\boldsymbol b\rangle$ with $\sum a_i=\sum b_j=n$.
The sequence starts $0,2,6,24,78,284,960,3402,11710,41020,...$
My attempts so far have led to increasingly absurdly complicated approaches (like inverting infinite matrices with power series coefficients) and gave nothing in the end.
I believe a specialist can either give an answer immediately or relate it to some known difficult problem.

REPLY [5 votes]: I've now got an explicit formula for the generating function; after several simplifications it is still quite messy. Still decided to show it (in a separate answer, adding this to the old one would make it too long I think), maybe somebody can use it for a better answer. The proof is equally messy and uninspiring, it consists in making the functional equation from the recursion which I described in my previous answer.
So let $f(q)=\sum F(n)q^n$. Then,
$$
f(q)=\frac AB,
$$
where
$$
A=\sum_{n=0}^\infty
(-1)^{n-1}\prod_{k=1}^n
\frac{(1-2q^k)^2}{(1-q^k)(1-4q^{k+1})}
\frac{1+q-4q^{n+1}}{1-2q^n}\left(\frac{q^2}{2(1-q)}\right)^n
$$
and
$$
B=\sum_{n=0}^\infty
(-1)^n\prod_{k=1}^n
\frac{(1-2q^k)^2}{(1-q^k)(1-4q^{k+1})}
\left(\frac{1-q}{1-2q^n}+2\frac{1-2q}{(1-2q^n)^2}\right)\left(\frac{q^2}{2(1-q)}\right)^n.
$$
Or, if you prefer,
\begin{multline*}
A=1-3 q
+\frac{(1-2 q) \left(1-4 q^2+q\right) }{(1-q) \left(1-4 q^2\right)}\frac{q^2}{2 (1-q)}
-\frac{(1-2 q)^2 \left(1-2 q^2\right) \left(1-4
   q^3+q\right) }{(1-q) \left(1-4 q^2\right) \left(1-q^2\right) \left(1-4 q^3\right)}\left(\frac{q^2}{2 (1-q)}\right)^2
+\frac{(1-2 q)^2 \left(1-2
   q^2\right)^2 \left(1-2 q^3\right) \left(1-4 q^4+q\right) }{(1-q) \left(1-4 q^2\right) \left(1-q^2\right) \left(1-4
   q^3\right) \left(1-q^3\right) \left(1-4 q^4\right)}\left(\frac{q^2}{2 (1-q)}\right)^3\mp...
\end{multline*}
and
\begin{multline*}
B=1-3 q-\frac{\left(3-7 q+2 q^2\right)}{(1-q) \left(1-4 q^2\right)}\frac{q^2}{2 (1-q)}
+\frac{(1-2 q)^2 \left(3-5 q-2 q^2+2 q^3\right)
   }{(1-q) \left(1-4 q^2\right) \left(1-q^2\right) \left(1-4 q^3\right)}\left(\frac{q^2}{2 (1-q)}\right)^2
-\frac{(1-2 q)^2 \left(1-2 q^2\right)^2 \left(3-5
   q-2 q^3+2 q^4\right) }{(1-q) \left(1-4 q^2\right) \left(1-q^2\right) \left(1-4 q^3\right) \left(1-q^3\right) \left(1-4
   q^4\right)}\left(\frac{q^2}{2 (1-q)}\right)^3\pm...
\end{multline*}
(Later)
Let me add one more version:
$$
f(q)=\frac{(1-4 q) \, _2\phi _1\left(\begin{smallmatrix}2,\ 2 q\\4 q\end{smallmatrix};q,\frac{-q^2}{2 (1-q)}\right)+q \, _2\phi _1\left(\begin{smallmatrix}2,\ 2 q\\4 q^2\end{smallmatrix};q,\frac{-q^2}{2 (1-q)}\right)}{2 (1-2 q) \,
   _2\phi _1\left(\begin{smallmatrix}2,\ 2\\4 q^2\end{smallmatrix};q,\frac{-q^2}{2 (1-q)}\right)-(1-q) \, _2\phi _1\left(\begin{smallmatrix}2,\ 2 q\\4 q^2\end{smallmatrix};q,\frac{-q^2}{2 (1-q)}\right)}
$$
where $_2\phi _1$ is the basic hypergeometric series