Let $a(n) = \sum_{0 \leq k \leq n} {n \choose k}{{n+k} \choose k}$ and define
$b(n) = \nu_3 (a(n))$, 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(\lfloor n/3 \rfloor) + (\lfloor n/3 \rfloor \bmod 2), & \text{if
        $n \equiv 0,2$ (mod 3); } \\
        b(\lfloor n/9 \rfloor) + 1, & \text{if $n \equiv 1$ (mod 3).}
        \end{cases}$

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:  If $r(n) = \sum_{0 \leq i < n} {{2i} \choose i},$ then $ \nu_3 (r(n)) = \nu_3 ( n^2 {{2n} \choose n})$, 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][1].


  [1]: http://people.mpim-bonn.mpg.de/zagier/files/amm/99/fulltext.pdf