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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? $$\nu_3(f(n,k))>\nu_3(f(n,0)).$$

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  • $\begingroup$ In general, Kummer's theorem is a useful tool for these sorts of calculations: en.wikipedia.org/wiki/Kummer%27s_theorem $\endgroup$
    – Terry Tao
    Commented Jun 26, 2021 at 15:38
  • $\begingroup$ This is true, although in the present case it is a bit subtle to decide $2\nu_3(f(n,k))=k+s_3(n+k)+s_3(n-k)+s_3(2k)-s_3(2n+2k)>2s_3(n)-s_3(2n)$ unless one proceeds with a case-by-case analysis. Tedious? $\endgroup$
    – T. Amdeberhan
    Commented Jun 26, 2021 at 16:14

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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.

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