Erdős Problem #379 — Prime-power divisibility in binomial coefficients

At least 48 years old · documented by

Let S(n)S(n) be the largest integer ss such that, for every integer kk with 1≤k<n1\le k<n, there is a prime pp (which may depend on kk) for which psp^s divides the binomial coefficient (nk)\binom{n}{k}. Prove that

lim sup⁡n→∞S(n)=∞.\limsup_{n\to\infty}S(n)=\infty.
References

Progress summary

Refreshed
Claimed solved

The question has been answered yes: binomial coefficients can have arbitrarily deep prime-power divisibility across every nontrivial position.

Erdős Problem 379379 asks whether the divisibility depth S(n)S(n) is unbounded along some sequence of integers nn. The problem is attributed to Erdős; the source does not state when it was posed.

Affirmative solution (date not stated)

Cambie, Kovač, and Tao proved the assertion, and a short elementary complete proof was subsequently obtained. For r≥2r\ge2, choosing n=2ϕ(pr)n=2^{\phi(p^r)} with p>2r−1p>2^{r-1} gives S(n)≥rS(n)\ge r; hence lim sup⁡S(n)=∞\limsup S(n)=\infty. The argument shows every (nk)\binom{n}{k}, 1≤k<n1\le k<n, is divisible by either 2r2^r or prp^r. The discussion also records simpler constructions involving 32k3^{2^k}.

Current status (as of June 2026): The problem is resolved affirmatively; the complete proof establishes lim sup⁡S(n)=∞\limsup S(n)=\infty.

Sources

Solutions 0

No solutions have been posted yet.