# Numbers $k$ with $\{\binom nk:\ n\in\mathbb N\}$ dense in $\mathbb Z_p$ for any prime $p\le k$

Let $$k$$ be a positive integer and let $$p$$ be a prime. In my 2011 PAMS paper joint with my former student W. Zhang [Proc. Amer. Math. Soc. 139(2011), 1569-1577], we studied when $$S(k)=\left\{\binom nk:\ n=0,1,2,\ldots\right\}$$ is a dense subset of the ring $$\mathbb Z_p$$ of $$p$$-padic integers (i.e., $$S(k)$$ contains a complete system of residues modulo any powers of $$p$$).

We call a positive integer $$k$$ universal if $$S(k)$$ is dense in $$\mathbb Z_p$$ for any prime $$p\le k$$. I and W. Zhang showed that $$1,\ 2,\ 3,\ 4,\ 5,\ 9,\ 11,\ 17,\ 29\tag{1}$$ are universal. We believe that there are no other universal numbers.

Question. Can one show that there is no universal number not listed in $$(1)$$?

• Could you link to your paper (using the AMS site version, not the paywall JSTOR version)? – YCor Oct 15 at 19:18
• I have added the link. Thank you for the suggestion. – Zhi-Wei Sun Oct 15 at 19:25
• @Zhi-Wei Sun: Could you, perhaps, summarize what the obstacles were in obstructing your approach in Theorem 1.2 of the linked paper from resolving the problem. Was it a matter of computation (as probably hinted by your statement right after Theorem 1.3; perhaps a formulation of a “combinatorial” conjecture —arising from Theorem 1.2–might be equivalent (or otherwise) to your conjecture that there are no other universal natural besides those stated above. – Jack L. Oct 15 at 20:06
• The 'any' in "any prime $p \le k$" is 'every', not 'some', right? – LSpice Nov 5 at 19:55
• Yes, "any" means "every" or "each". – Zhi-Wei Sun Nov 5 at 22:35

Thanks for this interesting question. From your Proposition 1.2 it would seem that the non-universal positive integers contains the set $$\{k\in\mathbb{Z}_{\geq 0}: \exists p\leq k\mbox{ such that } \prod_{\substack{i=0\\k_i>0}}^r (p-k_i) where $$k=\sum_{i=0}^rk_ip^i$$ and $$p$$ is prime.
Let $$x=p^{m+1}$$. Let $$p$$ be prime and $$n=\sum_{i=0}^m n_ip^i$$ be the expansion of $$n$$ in base $$p$$ where $$0\leq n_i. Let $$\tau_k(n)=\sum_{\substack{(d_1,\dots,d_k)\in\mathbb{Z}^k\\1\leq d_1,\dots,d_k\leq n\\d_1\dots d_k=n}}1.$$ Let $$A_{p,m}=\{n\in\mathbb{Z}: p < n = \sum_{i=0}^m n_ip^i < x, \prod_{\substack{i=0\\n_i>0}}^m(p-n_i) Then $$\begin{eqnarray*}&&|A_{p,m}|\\ &=&\sum_{j=1}^{p-2}|\{n\in\mathbb{Z}: p < n = \sum_{i=0}^m n_ip^i < x, \prod_{\substack{i=0\\n_i>0}}^m(p-n_i)=j \}|\\ &=&\sum_{j=1}^{p-2}|\{n\in\mathbb{Z}: p \leq n = \sum_{i=0}^m n_ip^i < x, \prod_{\substack{i=0\\n_i>0}}^m(p-n_i)=j \}|\\ &=&\left(\sum_{j=1}^{p-2}|\{n\in\mathbb{Z}: 0 \leq n = \sum_{i=0}^m n_ip^i < x, \prod_{\substack{i=0\\n_i>0}}^m(p-n_i)=j \}|\right)-(p-1)\\ &=&\left(\sum_{j=1}^{p-2}\sum_{k=1}^{m+1}{m+1 \choose k}\tau_k(j)\right) + 1 - (p-1) \end{eqnarray*}$$ as $$\prod_{\substack{i=0\\n_i>0}}^m(p-n_i)=1$$ when $$n=0,p-1$$ and $$\prod_{\substack{i=0\\n_i>0}}^m(p-n_i)=p-n$$ when $$1< n < p-1$$ and furthermore, the number of ways to distribute $$m+1$$ digits $$n_i\in\{0,\dots,p-1\}$$ such that $$\prod_{\substack{i=0\\n_i>0}}^m(p-n_i)=j$$ where the fixed $$j$$ satisfies $$1 \leq j < p-1$$ and $$k$$ of the $$m+1$$ digits $$n_i$$ are in $$\{2,\dots,p-1\}$$, is $${m+1 \choose k}\tau_k(j)$$.
It is known that (https://terrytao.wordpress.com/2016/08/31/heuristic-computation-of-correlations-of-higher-order-divisor-functions/) $$\sum_{n \leq y}\tau_k(n) \sim \frac{\log^{k-1}y}{(k-1)!}y$$ and hence $$\begin{eqnarray*} |A_{p,m}| &=& \left(\sum_{k=1}^{m+1}{m+1 \choose k}\sum_{j=1}^{p-2}\tau_k(j)\right) - (p-2)\\ &\ll&\sum_{k=1}^{m+1}{m+1 \choose k}\frac{\log^{k-1}p}{(k-1)!}p\\ &\ll&2^{m+1}p\log^m(p) \end{eqnarray*}$$ as $$p \rightarrow \infty$$ for fixed $$m$$. Hence if $$x=p^{m+1}$$, $$|A_{p,m}|/x \rightarrow 0$$ as $$p\rightarrow\infty$$ for fixed $$m$$.
On the other hand, fix $$x$$ and let $$B_{p,x}=\{n\in\mathbb{Z}: p < n = \sum_{i=0}^m n_ip^i < x, \prod_{\substack{i=0\\n_i>0}}^m(p-n_i) Then it is less clear what $$\lim_{x \rightarrow \infty}\frac{|\cup_{p is.