A seemingly simple combinatorial object that must have an easy generating function 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.
 A: The number of unmatchable pairs can be computed with the formula:
$$F(n) = \sum_{k=0}^{2n-1} (-1)^k \sum_{0<m_1<\dots<m_k<2n} \left( \sum_{0\le s_1<\dots<s_k<n=s_{k+1}\atop s_i<m_i/2} g(s_1)\prod_{i=1}^k g(s_i+s_{i+1}-m_i)\right)^2,$$ 
where 
$$g(t) = \begin{cases} 
0, &\text{if}\ t<0;\\
1, &\text{if}\ t=0;\\
2^{t-1}, &\text{if}\ t\geq 1.
\end{cases}
$$
UPDATE #2 (02/14/16). The expression being squared in the formula for $F(n)$ equals the coefficient of $x_1^{m_1}x_2^{m_2-m_1}\cdots x_k^{m_k-m_{k-1}}x_{k+1}^{2n-m_k}$ in
$$G(x_1^2)\cdot \prod_{i=1}^k \frac{x_i x_{i+1}}{1-x_ix_{i+1}}\cdot G(x_{i+1}^2),$$
where $G(x)$ is the generating function for $g(s)$:
$$G(x) = \sum_{s=0}^\infty g(s)\cdot x^s = \frac{1-x}{1-2x}.$$
Without squaring this would lead to almost trivial summation of these coefficients, but summation with squaring remains a challenge I do not yet know how to address. Parseval's identity may be relevant here somehow. 
P.S. Midpoints have also a nice geometrical interpretation in terms of certain lattice paths, which I can explain later if there is interest. see update below.
P.S. #2 Here is my PARI/GP code, which implements the above formula for $F(n)$:
{ g(t) = if(t<0, return(0)); if(t==0,1,2^(t-1)); }
{ f(n,m) = my(r=0); if(#m==0,return(g(n))); forvec(s=vector(#m,i,[0,(m[i]-1)\2]), r += prod(i=0,#m, g(if(i<#m,s[i+1],n)-if(i>0,m[i]-s[i],0));); ,2); r; }
{ F(n) = my(r=0); for(k=0,2*n-1, forvec(m=vector(k,i,[1,2*n-1]), r+=(-1)^k*f(n,m)^2; ,2); ); r; }

UPDATE #1 (adjusted per suggestion of მამუკაჯიბლაძე). 
I will illustrate the path interpretation on the example of $a=(1,3,2,3,4)$ with the sum $n=13$. 
I define the "modified" middle points (obtained from the original ones by multiplying by 2 and subtracting 1) as
$$m_i = 2(a_1+\dots+a_{i-1})+a_i.$$
In our example, we have $m=(1,5,10,15,22)$. 
Let us represent these entities as a path:
$$(0,0) \to (m_1,-a_1) \to (m_2,a_2) \to (m_3,-a_3) \to \dots \to (2n,0).$$
It can be seen that this path consists of alternating diagonal -45° and +45° steps, each of which crosses the $x$-axis but does not visit it, except for the start and end points. In other words, the intermediate vertices in this path alternatingly lie below and above the $x$-axis and their $x$-coordinates correspond to the modified middle points of $a$. Clearly, vectors $a$ and such paths are in one-to-one correspondence.
Here is the path for our example.

