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.