Math-788-A2 · revision r1 · on Math-788

gpt pro 5.2 attempt on Math-788

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Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Problem 377. Is $\displaystyle f(n)=\sum_{p\le n}1_{p\nmid \binom{2n}{n}}\,\frac1p$ bounded?
  2. FORMAL RESTATEMENT
  3. QUICK LITERATURE / CONTEXT CHECK
  4. ATTACK PLAN
  5. WORK
  6. 1. A simple implication: bounded $f(n)$ would bound the smallest odd prime factor
  7. 2. Relation to Problem 376 and "avoiding many primes"
  8. 3. Computation for small $n$ (evidence only)
  9. VERIFICATION
  10. FINAL
  11. COMPLETION ESTIMATE
  12. Bibliography (informal)
  13. References

Problem 377. Is $\displaystyle f(n)=\sum_{p\le n}1_{p\nmid \binom{2n}{n}}\,\frac1p$ bounded?

FORMAL RESTATEMENT

Define for $n\ge 1$ \[ f(n)\;:=\;\sum_{\substack{p\le n\\ p\text{ prime}}} \mathbf{1}_{p\nmid \binom{2n}{n}}\,\frac{1}{p}. \] Is there an absolute constant $C$ such that $f(n)\le C$ for all $n$?

QUICK LITERATURE / CONTEXT CHECK

Erdős--Graham--Ruzsa--Straus $1975$ introduced $f(n)$ and proved:
  • $\displaystyle \lim_{x\to\infty}\frac1x\sum_{n\le x} f(n)=\sum_{k\ge 2}\frac{\log k}{2^k}=:\gamma_0$ and $\lim_{x\to\infty}\frac1x\sum_{n\le x} f(n)^2=\gamma_0^2$, hence for almost all $n$, $f(n)=\gamma_0+o(1)$.
  • For all sufficiently large $n$, $f(n)\le c\log\log n$ for some absolute constant $c<1$.
  • They remark explicitly that they cannot decide whether $f(n)$ is unbounded.
The Erdős problems database (Problem 377) records the same status.

Numerically, $\gamma_0\approx 0.5078339229$.

ATTACK PLAN

To prove boundedness one would need a mechanism showing that for every $n$ there are "enough" primes $p\le n$ that do divide $\binom{2n}{n}$, forcing the complement (weighted by $1/p$) to have bounded harmonic mass.

To disprove boundedness one would need to construct infinitely many $n$ for which many small primes simultaneously satisfy $p\nmid\binom{2n}{n}$, so that $\sum 1/p$ over those primes grows.

Kummer's theorem again translates $p\nmid\binom{2n}{n}$ into the digit condition "all base-$p$ digits of $n$ are $\le (p-1)/2$", making the problem a question about how often $n$ can have small digits in many bases at once.

WORK

1. A simple implication: bounded $f(n)$ would bound the smallest odd prime factor

Let $g(n)$ be the smallest odd prime divisor of $\binom{2n}{n}$ (if it exists). If $g(n)>y$ then no odd prime $p\le y$ divides $\binom{2n}{n}$, hence \[ f(n)\ge \sum_{3\le p\le y} \frac1p. \] Since $\sum_{p\le y}1/p\to\infty$ with $y$, it follows that \[ \sup_n f(n)<\infty \ \Longrightarrow\ \sup_n g(n)<\infty. \] EGRS $1975$ note that even the boundedness of $g(n)$ is unknown.

2. Relation to Problem 376 and "avoiding many primes"

If for some finite set of primes $\mathcal{P}$ we can find $n$ with $p\nmid\binom{2n}{n}$ for all $p\in\mathcal{P}$, then \[ f(n)\ge \sum_{p\in\mathcal{P}}\frac1p. \] Thus unboundedness of $f(n)$ would follow from the ability to simultaneously avoid an increasing collection of small primes.

However, even avoiding $\{3,5,7\}$ (Problem 376) is open; and avoiding any two odd primes is known to be possible infinitely often (EGRS), which only yields a bounded lower bound.

3. Computation for small $n$ (evidence only)

Using Kummer's digit criterion to test $p\nmid\binom{2n}{n}$, a brute-force search over $1\le n\le 20000$ gives:
  • the maximum value of $f(n)$ in this range is \[\max_{1\le n\le 20000} f(n) \approx 1.1792429058\quad\text{attained at }n=3250,\]
  • other large values occur at $n=3160$ ($\approx 1.1155$) and at $n=756$ ($\approx 1.0770$), both of which are also solutions to Problem 376 (coprime to $105$).
This does not decide boundedness but is consistent with the possibility that $f(n)$ has a modest global supremum.

VERIFICATION

The logical implications in §1--§2 are direct. The numerical values in §3 were computed exactly from the digit condition (not by evaluating $\binom{2n}{n}$).

FINAL

LABEL: UNRESOLVED.
SUB-LABEL: N/A.
The existence of an absolute constant $C$ with $f(n)\le C$ for all $n$ remains open. Known results show $f(n)$ is typically close to the constant $\gamma_0\approx 0.5078$ and always satisfies $f(n)\le c\log\log n$ for some $c<1$ for large $n$, but no boundedness/unboundedness proof is known.

COMPLETION ESTIMATE

A full solution would require either:
  • a construction of integers $n$ for which $p\nmid\binom{2n}{n}$ for enough small primes $p$ that $\sum 1/p$ grows without bound (proving unboundedness), or
  • a uniform argument forcing the set of avoided primes to have bounded harmonic weight for every $n$ (proving boundedness).
Both directions appear to hinge on deep control of simultaneous small-digit expansions (or carries) in many bases.

Bibliography (informal)

References

  • [EGRS75] P. Erdős, R. L. Graham, I. Z. Ruzsa, and E. G. Straus, On the prime factors of $\binom{2n}{n}$, Math. Comp. 29 $1975$, 83--92.
  • [EG76] P. Erdős and R. L. Graham, On products of factorials, Bull. Inst. Math. Acad. Sinica 4 $1976$, 337--355.
  • [DNSS14] A. Dujella, F. Najman, N. Saradha, and T. N. Shorey, Products of three factorials, Publ. Math. Debrecen 85 $2014$, 3--13.
  • [BlCr25] T. F. Bloom and E. Croot, Integers with small digits in multiple bases, arXiv:2509.02835 $2025$.
  • [Yu22] H. Yu, Fractal projections with an application in number theory, Proc. Lond. Math. Soc. 125 $2022$, 1060--1109.