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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 $(1)$?

Your comments are welcome!

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    $\begingroup$ Could you link to your paper (using the AMS site version, not the paywall JSTOR version)? $\endgroup$ – YCor Oct 15 at 19:18
  • $\begingroup$ I have added the link. Thank you for the suggestion. $\endgroup$ – Zhi-Wei Sun Oct 15 at 19:25
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    $\begingroup$ @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. $\endgroup$ – Jack L. Oct 15 at 20:06
  • $\begingroup$ The 'any' in "any prime $p \le k$" is 'every', not 'some', right? $\endgroup$ – LSpice Nov 5 at 19:55
  • $\begingroup$ Yes, "any" means "every" or "each". $\endgroup$ – Zhi-Wei Sun Nov 5 at 22:35
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Thanks for this interesting question. From your Proposition 1.2 it would seem that the non-universal positive integers contains the set $$ \{k\in\mathbb{Z}_{\geq 0}: \exists p\leq k\mbox{ such that } \prod_{\substack{i=0\\k_i>0}}^r (p-k_i) <p-1\} $$ 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 $0\leq n_i<p$. 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 $$A_{p,m}=\{n\in\mathbb{Z}: p < n = \sum_{i=0}^m n_ip^i < x, \prod_{\substack{i=0\\n_i>0}}^m(p-n_i)<p-1 \}.$$ 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 $$B_{p,x}=\{n\in\mathbb{Z}: p < n = \sum_{i=0}^m n_ip^i < x, \prod_{\substack{i=0\\n_i>0}}^m(p-n_i)<p-1 \}.$$ Then it is less clear what $$ \lim_{x \rightarrow \infty}\frac{|\cup_{p<x}B_{p,x}|}{x} $$ is.

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