This should probably be not that hard, but I would like to see a nifty way of proving it.
Consider the double-indexed sequence given by $$f(n,k)=\binom{2n + 2k}{n + k}\binom{n + k}{n - k}3^k.$$
QUESTION. For $1\leq k\leq n$, does this hold true for the $3$-adic valuations? $$\nu_3(f(n,k))>\nu_3(f(n,0)).$$
Notice that $$ 3^{-k-1}f(n,k+1)/3^{-k}f(n,k)=\frac{{2n+2k+2\choose n+k+1}}{{2n+2k\choose n+k}}\frac{{n+k+1\choose n-k-1}}{{n+k \choose n-k}}. $$ The first factor here equals $$ \frac{(2n+2k+2)!}{(2n+2k)!}\frac{(n+k)!^2}{(n+k+1)!^2}=\frac{2(2n+2k+1)}{n+k+1}. $$ The second is $$ \frac{(n+k+1)!}{(n+k)!}\frac{(n-k)!}{(n-k-1)!}\frac{(2k)!}{(2k+2)!}=(n+k+1)(n-k)\frac{(2k)!}{(2k+2)!}. $$ For instance, $$ \frac{3^{-k-1}(2k+2)!f(n,k+1)}{3^{-k}(2k)!f(n,k)}=2(n-k)(2n+2k+1) $$ is an integer. Multiplying several fractions of this type together, we see that for any $k\leq n$ we have $$ \frac{3^{-k}(2k)!f(n,k)}{3^{-0}0!f(n,0)}\in\mathbb Z. $$ Hence $$ \nu_3(3^{-k}(2k)!f(n,k))\geq \nu_3(f(n,0)). $$ Next, for $k>0$ $$ \nu_3((2k)!)=\left[\frac{2k}{3}\right]+\left[\frac{2k}{9}\right]+\ldots< $$ $$ <\frac{2k}{3}+\frac{2k}{9}+\ldots=k, $$ which means that $$ \nu_3(f(n,k))-\nu_3(f(n,0))\geq k-\nu_3((2k)!)>0, $$ as needed.
