We show for $n\geq 2$ the validity of $F(n)=1$ with the help of generating functions. It is somewhat more convenient to consider $F(n+2)$ instead of $F(n)$ and we claim:
The following is valid for $n\geq 0$:
\begin{align*}
\color{blue}{\sum_{k\geq 1}\left(\binom{n+1}{2k-2}-(n+2)\binom{n}{2k-1}-\binom{n}{2k}\right)(2k-1)!!=1}\tag{1}
\end{align*}
The sum (1) is finite since $\binom{p}{q}=0$ for $p,q$ non-negative integers with $q>p$. We use the coefficient of operator $[z^q]$ to denote the coefficient of $z^q$ in a series. We consider each of the parts at the left-hand side of (1) separately and put them together at the end.
First part: We obtain
\begin{align*}
\color{blue}{\sum_{k=1}^{\left\lfloor\frac{n+3}{2}\right\rfloor}}&\color{blue}{\binom{n+1}{2k-2}(2k-1)!!}\\
&=\sum_{k=1}^{\left\lfloor\frac{n+3}{2}\right\rfloor}\frac{(n+1)!}{(2k-2)!(n+3-2k)!}\,\frac{(2k )!}{(2k)!!}\\
&=\sum_{k=1}^{\left\lfloor\frac{n+3}{2}\right\rfloor}\frac{(n+1)!}{(n+3-2k)!}\,\frac{2k(2k-1)}{2^kk!}\\
&=\sum_{k=1}^{\left\lfloor\frac{n+3}{2}\right\rfloor}\frac{(n+1)!}{(n+3-2k)!}\,\frac{2k-1}{2^{k-1}(k-1)!}\\
&=\sum_{k=0}^{\left\lfloor\frac{n+1}{2}\right\rfloor}\frac{(n+1)!}{(n+1-2k)!}\,\frac{2k+1}{2^kk!}\tag{2}\\
&=\sum_{k=1}^{\left\lfloor\frac{n+1}{2}\right\rfloor}\frac{(n+1)!}{(n+1-2k)!}\,\frac{1}{2^{k-1}(k-1)!}
+\sum_{k=0}^{\left\lfloor\frac{n+1}{2}\right\rfloor}\frac{(n+1)!}{(n+1-2k)!}\,\frac{1}{2^{k}k!}\tag{3}\\
&=\sum_{k=0}^{\left\lfloor\frac{n+1}{2}\right\rfloor}\frac{(n+1)!}{(n-1-2k)!}\,\frac{1}{2^{k}k!}
+\sum_{k=0}^{\left\lfloor\frac{n+1}{2}\right\rfloor}\frac{(n+1)!}{(n+1-2k)!}\,\frac{1}{2^{k}k!}\tag{4}\\
&\,\,\color{blue}{=(n+1)!\left([z^{n-1}]+[z^{n+1}]\right)\exp\left(z+\frac{z^2}{2}\right)}\tag{5}
\end{align*}
Comment:
In (2) we shift the index $k$ by one and start with $k=0$.
In (3) we split the sums according to $2k+1$ and cancel at the left-hand sum. Since the term with $k=0$ is zero at the left-hand sum, we start with $k=1$.
In (4) we shift the index $k$ of the left-hand sum by one.
In (5) we observe we have the coefficients $[z^{n-1}]$ and $[z^{n+1}]$ of $\exp\left(z\right)\exp\left(\frac{z^2}{2}\right)=\exp\left(z+\frac{z^2}{2}\right)$. The derivation for it is similar to that of the third part and is shown there.
Second part: We obtain
\begin{align*}
\color{blue}{\sum_{k=1}^{\left\lfloor\frac{n+1}{2}\right\rfloor}}&\color{blue}{\binom{n}{2k-1}(2k-1)!!}\\
&=\sum_{k=1}^{\left\lfloor\frac{n+1}{2}\right\rfloor}\frac{n!}{(2k-1)!(n+1-2k)!}\,\frac{(2k )!}{(2k)!!}\\
&=\sum_{k=1}^{\left\lfloor\frac{n+1}{2}\right\rfloor}\frac{n!}{(n+1-2k)!}\,\frac{2k}{2^kk!}\\
&=\sum_{k=1}^{\left\lfloor\frac{n+1}{2}\right\rfloor}\frac{n!}{(n+1-2k)!}\,\frac{1}{2^{k-1}(k-1)!}\\
&=\sum_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\frac{n!}{(n-1-2k)!}\,\frac{1}{2^kk!}\\
&\,\,\color{blue}{=n![z^{n-1}]\exp\left(z+\frac{z^2}{2}\right)}\tag{6}
\end{align*}
Comment:
- In (6) we observe we have a coefficient of $[z^{n-1}]$ of $\exp\left(z+\frac{z^2}{2}\right)$. The derivation for it is similar to that of the third part and is shown there.
Third part: The third part is very similar to the parts before, but it has additionally a nice twist which provides us with the constant $1$. We obtain
\begin{align*}
\color{blue}{\sum_{k=1}^{\left\lfloor\frac{n}{2}\right\rfloor}}&\color{blue}{\binom{n}{2k}(2k-1)!!}\\
&=\sum_{k=1}^{\left\lfloor\frac{n}{2}\right\rfloor}\frac{n!}{(2k)!(n-2k)!}\,\frac{(2k )!}{(2k)!!}\\
&=\sum_{k=1}^{\left\lfloor\frac{n}{2}\right\rfloor}\frac{n!}{(n-2k)!}\,\frac{1}{2^kk!}\tag{7}\\
&\,\,\color{blue}{=n![z^{n}]\left(\exp\left(z+\frac{z^2}{2}\right)-\exp(z)\right)}\tag{8}
\end{align*}
Comment:
- In (8) we have the funny intermediate result:
\begin{align*}
n![z^{n}]\exp(z)=n![z^n]\sum_{j=0}^\infty \frac{z^j}{j!}\color{blue}{=1}
\end{align*}
and the further procedure is clear: We expect that the sum of the coefficients of $\exp\left(z+\frac{z^2}{2}\right)$ cancel away leaving the constant $n![z^{n}]\exp(z)=1$. But at first we have to show the validity of (8).
Derivation of (8): We get
\begin{align*}
\color{blue}{\exp}&\color{blue}{\left(z+\frac{z^2}{2}\right)-\exp(z)}\\
&=\exp(z)\left(\exp\left(\frac{z^2}{2}\right)-1\right)\\
&=\left(\sum_{j=0}^\infty\frac{z^j}{j!}\right)\left(\sum_{k=1}^\infty\frac{z^{2k}}{2^kk!}\right)\\
&=\sum_{n=0}^\infty\left(\sum_{{j+2k=n}\atop{j\geq 0, k\geq 1}}\frac{1}{j!}\,\frac{1}{2^kk!}\right)z^n\\
&\,\,\color{blue}{=\sum_{n=0}^\infty\left(\sum_{k=1}^{\left\lfloor\frac{n}{2}\right\rfloor}\frac{n!}{(n-2k)!}\,\frac{1}{2^kk!}\right)\frac{z^n}{n!}}
\end{align*}
and the claim (8) follows.
Now it's (nearly) time to harvest. Let's collect the representation of the parts (5),(6) and (8) we have so far:
Final step: We obtain from (1), (5), (6) and (8)
\begin{align*}
\color{blue}{\sum_{k\geq 1}}&\color{blue}{\left(\binom{n+1}{2k-2}-(n+2)\binom{n}{2k-1}-\binom{n}{2k}\right)(2k-1)!!}\\
&=(n+1)!\left([z^{n-1}]+[z^{n+1}]\right)\exp\left(z+\frac{z^2}{2}\right)\\
&\qquad-(n+2)n![z^{n-1}]\exp\left(z+\frac{z^2}{2}\right)\\
&\qquad-n![z^{n}]\left(\exp\left(z+\frac{z^2}{2}\right)-\exp(z)\right)\\
&=\left((n+1)![z^{n+1}]-n![z^n]-n![z^{n-1}]\right)\exp\left(z+\frac{z^2}{2}\right)\color{blue}{+1}\tag{9}
\end{align*}
We want to show the coefficients in (9) cancel away. In order to do so, we consider
\begin{align*}
A(z)&=\sum_{n=0}^\infty a_nz^n=\exp\left(z+\frac{z^2}{2}\right)\\
\\
\frac{d}{dz}A(z)&=\sum_{n=1}^\infty na_nz^{n-1}\tag{10}\\
\frac{d}{dz}\exp\left(z+\frac{z^2}{2}\right)&=(1+z)\exp\left(z+\frac{z^2}{2}\right)\tag{11}\\
\end{align*}
Coefficient comparison of (10) and (11) gives:
\begin{align*}
(n+1)a_{n+1}&=\color{blue}{(n+1)[z^{n+1}]A(z)}\\
&=[z^n](1+z)\exp\left(z+\frac{z^2}{2}\right)\\
&\,\,\color{blue}{=\left([z^n]+[z^{n-1}]\right)A(z)}
\end{align*}
which shows that in (9)
\begin{align*}
\left((n+1)![z^{n+1}]-n![z^n]-n![z^{n-1}]\right)\exp\left(z+\frac{z^2}{2}\right)=0
\end{align*}
and the claim (1) follows.
