Let $$a(n) = \sum\_{0 \leq k \leq n} {n \choose k}{{n+k} \choose k},$$ and define
$b(n) = \nu\_3 \bigl(a(n)\bigr)$, where $\nu\_3$ is the $3$-adic valuation. About twenty years ago or so, I discovered (empirically) the following conjectured expression for $b(n)$:


$$b(n) = \begin{cases}
 b\bigl(\lfloor n/3 \rfloor\bigr) + \bigl(\lfloor n/3 \rfloor \bmod 2\bigr), & \text{if
 $n \equiv 0,2$ (mod 3); } \\
 b\bigl(\lfloor n/9 \rfloor\bigr) + 1, & \text{if $n \equiv 1$ (mod 3).}
 \end{cases} \tag{$\*$}$$


But I have not been able to prove it. 


For some background, the problem has some similarity to the following theorem, a weaker version of which was originally suggested by N. Strauss:  

$$\text{If}\quad r(n) = \sum\_{0 \leq i < n} {{2i} \choose i},\quad\text{then}\quad \nu\_3 \bigl(r(n)\bigr) = \nu\_3\left ( n^2 {{2n} \choose n}\right),$$ which I proved by a kind of tedious argument, with Jean-Paul Allouche. Later, another more elegant proof was given by Don Zagier. See [here](http://people.mpim-bonn.mpg.de/zagier/files/amm/99/fulltext.pdf).


Can anybody prove $(\*)$?



3-adic valuation of a sum involving binomial coefficients
The $3$-adic evaluation you seek is compactly given by
$$\nu\_3(a\_{2n})=\nu\_3\left(\binom{2n}n\right) \qquad \text{and} \qquad
\nu\_3(a\_{2n+1})=\nu\_3\left(3(2n+1)\binom{2n}n\right),$$
which can be proved inductively using the well-known recurrence
$$na\_n=3(2n-1)a\_{n-1}-(n-1)a\_{n-2}$$
according to the parity of $n$. To address Cigler's request, here is an illustration for the case even: $(2n)a\_{2n}=3(4n-1)a\_{2n-1}-(2n-1)a\_{2n-2}$ and by induction assumption 
$$\nu\_3(3(4n-1)a\_{2n-1})=\nu\_3\left(9(4n-1)(2n-1)\binom{2n-2}{n-1}\right) \qquad\text{and}\qquad \nu\_3((2n-1)a\_{2n-2})=\nu\_3\left((2n-1)\binom{2n-2}{n-1}\right).$$ 
So, $\nu\_3(3(4n-1)a\_{2n-1})>\nu\_3((2n-1)a\_{2n-2})$ and 
$\nu\_3((2n)a\_{2n})=\nu\_3((2n-1)\binom{2n-2}{n-1})$; or,
\begin{align}
\nu\_3(a\_{2n})&=\nu\_3\left(\frac{(2n-1)}{2n}\binom{2n-2}{n-1}\right)
=\nu\_3\left(\frac14\binom{2n}n\right)=\nu\_3\left(\binom{2n}n\right)
\end{align}
as desired.


