Are $F(2)$ and $F(4)$ are the only integer values of $F(x)=\sum_{n=1}^{\infty}(-1)^{n-1}{(2n+1)^x-1\over n\cdot n!}$ for $x\in\mathbb{N}$? Given 
$$F(x)=\sum_{n=1}^{\infty}(-1)^{n-1}{(2n+1)^x-1\over n\cdot n!}\tag1$$
The $F(2)$ and $F(4)$ yield 4 and 8 respectively.
As shown below by expanding out (1)
$$4={3^2-1\over1\cdot 1!}-{5^2-1\over 2\cdot 2!}+{7^2-1\over 3\cdot3!}-\cdots\tag2$$
$$8={3^4-1\over1\cdot 1!}-{5^4-1\over 2\cdot 2!}+{7^4-1\over 3\cdot3!}-\cdots\tag3$$

Are $F(2)$ and $F(4)$ the only integer outcomes of $F(n)$ for $n\in\mathbb{N}$?

 A: We may exploit the following fact:
$$ \forall k\in\mathbb{N},\qquad\sum_{n\geq 1}\frac{(-1)^{n-1}\binom{n}{k}}{n!}=\frac{(-1)^{k+1}}{k! e}\tag{1} $$
that comes from evaluating $\frac{d^k}{dx^k}\left(x^k e^{-x}\right)$ at $x=1$. For every value of $m\in\mathbb{N}$, the polynomial
$$ p_m(n) = \frac{(2n+1)^m-1}{n} \tag{2} $$
can be represented with respect to the binomial base, and the value of
$$ \sum_{n\geq 1}\frac{(-1)^{n-1} p_m(n)}{n!} \tag{3}$$
can be computed through $(1)$, so it is always a linear combination of $1$ and $\frac{1}{e}$ with rational coefficients. In order that such value is actually an integer, we need a great conspiracy between the coefficients of the expansion of $p_m(n)$ in the binomial base and the reciprocal of factorials.
So, at least, it isn't unlikely that your conjecture holds, but an actual proof probably requires an accurate study of the arithmetic behaviour of Stirling numbers of the second kind.
We also have:
$$\sum_{n\geq 1}\frac{(-1)^{n-1}n^a}{n!}=\frac{\eta(a)}{e}\tag{4}$$
where $\eta(a)$ is an arithmetic function (related with Bell numbers) whose absolute value grows quite fast. So it is highly unlikely that for large values of $m$ the contribute to $(3)$ given by the leading term of $p_m(n)$ get canceled out.
