This question is motivated by Richard Stanley's answer to this MO question.

Let $g(n)$ be the number of distinct monomials in the expansion of the determinant of an $n\times n$ generic "*skew-symmetric $+$ diagonal*" matrix.

For example, $g(3)=4$ since \begin{align*} \det\begin{pmatrix} x_{1,1}&x_{1,2}&x_{1,3} \\ -x_{1,2}&x_{2,2}&x_{2,3} \\ -x_{1,3}&-x_{2,3}&x_{3,3} \end{pmatrix} &=x_{1, 1}x_{2, 2}x_{3, 3}+x_{1, 1}x_{2, 3}^2+x_{1, 2}^2x_{3, 3} +x_{1, 3}^2x_{2, 2}. \end{align*} The sequence $g(n)$ seems to have found a match in OEIS with the generating function $$ \sum_{n\geq 0} g(n)\frac{x^n}{n!} = \frac{e^x}{1-\frac12x^2}.$$

QUESTION.Is it true and can you furnish a proof for $$g(n)=\sum_{k=0}^{\lfloor \frac{n}2\rfloor}\frac{n!}{(n-2k)!\,\,2^k}?$$

**POSTSCRIPT.** I'm convinced by Stanley's reply below, so let's correct the above as follows:
$$g(n)=\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}\frac{(2k)!}{k!}\cdot\prod_{i=1}^k\frac{4i-3}4\cdot\sum_{j=0}^{\lfloor n/2-k\rfloor}
\binom{n-2k}{2j}\frac{(2j)!}{4^j\,j!}.$$