<s>[INST] Power of $2$ dividing a specialized Mittag-Leffler polynomial While studying the so-called Mittag-Leffler Polynomials, denoted $M_n(x)$, I was looking into the sequence $\frac1{n!}M_n(n)$ which takes the following form
$$a_n:=\sum_{k=1}^n\binom{n-1}{k-1}\binom{n}k2^k.$$

QUESTION 1. Let $\nu_2(m)$ denote the $2$-adic valuation of $m\in\mathbb{N}$. Is this true?
$$\nu_2(a_n)=\begin{cases} \,\,\,\,\,\,1 \qquad \,\,\,\,\,\text{if $n$ is odd} \\
3\nu_2(n) \qquad \text{if $n$ is even}.
\end{cases}
$$
QUESTION 2. Is it true that $a_n$ is never divisible by $5$, for $n\geq1$?

Postscript. I've recently solved QUESTION 2, so only QUESTION 1 remains.
 [/INST] I will confine myself to Question 1 since you mentioned that you know how to do Question 2.  Also the case when $n$ is odd is easy, and let us restrict to $n$ being divisible exactly by $2^r$ with $r\ge 1$, and we need to show that the exact power of $2$ dividing $a_n$ is $3r$.  Thus in what follows we may discard any terms that are divisible by a power of $2$ larger than $3r$.  For example, we can restrict attention to the terms with $k\le 3r$ in the sum.  The case $r=1$ can be easily checked by hand, and we assume below that $r\ge 2$, and $k \le 3r$.
Consider the $k$-th term in the sum defining $a_n$:
$$ 
2^{k} \binom{n}{k} \binom{n-1}{k-1} = 2^k \frac{n}{k} \prod_{j=1}^{k-1} \Big(1 -\frac{n}{j}\Big)^2. 
$$
Let us now expand the product above, discarding any terms that are divisible by $2^{3r+1}$.  Note that $n/j$ is a $2$-adic integer, since the power of $2$ dividing $j$ is at most $\lfloor \log_2(k-1)\rfloor \le \lfloor \log_2 (3r-1)\rfloor \le r$.
We first observe that when expanding the product out terms that ``have $n^4$" in them (so using $3$ factors of $n/j$ in the product) can be omitted, and therefore terms that have higher powers of $n$ may also be omitted.  Indeed a term using $3$ factors of $n/j$ in the product is divisible by a power of $2$
$$
\ge k + r -v_2(k) + 3r - 3\lfloor \log_2(k-1)\rfloor. 
$$
A small calculation shows that $k-v_2(k)-3\lfloor \log_2(k-1) \rfloor \ge -1$ for all $k\ge 1$, and therefore the above is $\ge 3r+r-1 \ge 3r+1$, as desired.
Now consider terms with $n^3$ in them (so using $2$ terms of the form $n/j$ from the product).  We claim that there is exactly one such term that is relevant, and this happens only in the case $k=4$, and the two terms from the product are (both) $n/2$. Indeed the power of $2$ in term using two factors $n/j$ is
$$ 
\ge k + r-v_2(k) + 2 r -2 \lfloor \log_2 (k-1)\rfloor,
$$
and we can check that $k-v_2(k) -2 \lfloor \log_2(k-1)\rfloor \ge 1$ for $k\ge 1$, except in the case $k=4$. In the case $k=4$ we can quickly check that the only relevant term with two factors $n/j$ must be $(n/2)(n/2)$.
Putting all these observations together, we find that $a_n$ equals (after omitting any terms divisible by $2^{3r+1}$)
\begin{align*}
&\sum_{k\le 3r} 2^k \frac{n}{k} \Big( 1 - 2 \sum_{j=1}^{k-1} \frac{n}{j}\Big) + 2^4 \frac{n}{4} \frac n2 \frac n2
\\
&= n \sum_{k\le 3r} \frac{2^k}{k} -n^2 \sum_{k\le 3r} \frac{2^k}{k} 
\sum_{j=1}^{k-1}\Big( \frac 1j + \frac{1}{k-j}\Big) +n^3\\
&=  n \sum_{k\le 3r} \frac{2^k}{k} -n^2 \sum_{k\le 3r} \sum_{j=1}^{k-1} \frac{2^k}{j(k-j)} + n^3.
\end{align*}
Now here it is convenient to extend the sums over $k$ to infinity, noting that the extra terms are divisible by $2^{3r+1}$.  (Thus one checks that for $k>3r$ one has $r+k -v_2(k) \ge 3r+1$, and that $2r + k -2 \lfloor \log_2 (k-1)\rfloor \ge 3r+1$.)
Now comes the magical bit.  In the $2$-adics one has
$$ 
\sum_{k=1}^{\infty} \frac{2^k}{k} =0,
$$
and therefore also
$$ 
\sum_{k=1}^{\infty} 2^k \sum_{j=1}^{k-1} \frac{1}{j(k-j)} = \Big( \sum_{k=1}^{\infty} \frac{2^k}{k} \Big)^2 =0. \tag{*} 
$$
We are then left with $a_n$ being $n^3$ up to multiples of $2^{3r+1}$, which is what we want.
For the evaluation of $\sum_{k=1}^{\infty} 2^k/k$, note that if $|x|_2 <1$ then the series
$$ 
\sum_{k=1}^{\infty} \frac{x^k}{k}
$$
converges in the $2$-adics, and this clearly "looks like" $-\log(1-x)$.  In our case we'd get $-\log (1-2) = \log (-1) =0$ (upon "using" $\log (-1) + \log (-1) = \log 1 = 0$).  This last bit can all be made precise; a lovely write up of what is involved can be found in Keith Conrad's notes (see Example 8.10 there).
</s>