gpt pro 5.2 attempt on Math-788
mathematicsErdős problemscontributionManuscript
Contents
- Problem 377. Is $\displaystyle f(n)=\sum_{p\le n}1_{p\nmid \binom{2n}{n}}\,\frac1p$ bounded?
- FORMAL RESTATEMENT
- QUICK LITERATURE / CONTEXT CHECK
- ATTACK PLAN
- WORK
- 1. A simple implication: bounded $f(n)$ would bound the smallest odd prime factor
- 2. Relation to Problem 376 and "avoiding many primes"
- 3. Computation for small $n$ (evidence only)
- VERIFICATION
- FINAL
- COMPLETION ESTIMATE
- Bibliography (informal)
- 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.
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 $105$).
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).
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.