For n,m∈ N, say that m has a medium divisor relative to n if there exists d∈ N such that n 0, there exists N∈ N such that for all n≥ N, all y≥ Y(epsilon,n), and all x∈ N, L(n,x,y)≤ epsilon y. Is there such a function Y for which, for every epsilon>0, lim_n→∞fracY(epsilon,n)n=0?

Erdős Problem #450 — Sparse Intervals with Medium Divisors

About 48 years old · traced to 1978

For $n,m\in\mathbb N$ , say that $m$ has a medium divisor relative to $n$ if there exists $d\in\mathbb N$ such that $n<d<2n$ and $d\mid m$ . For $n,x,y\in\mathbb N$ , let

L(n,x,y)=\bigl|\{m\in\mathbb N:x<m<x+y\text{ and }m\text{ has a medium divisor relative to }n\}\bigr|.

Say that a function $Y:\mathbb R\to\mathbb N\to\mathbb N$ is a sufficient scale if, for every $\epsilon>0$ , there exists $N\in\mathbb N$ such that for all $n\ge N$ , all $y\ge Y(\epsilon,n)$ , and all $x\in\mathbb N$ ,

L(n,x,y)\le \epsilon y.

Is there such a function $Y$ for which, for every $\epsilon>0$ ,

\lim_{n\to\infty}\frac{Y(\epsilon,n)}{n}=0?

References

Primary source

The Formal Conjectures Authors, Formal Conjectures (2025); independently edited from the pinned source file.

Additional references

Pinned Formal Conjectures source, Apache-2.0.

divisors number-theory

InErdős Problems

Posted 2 months ago · edited 3 weeks ago

Progress summary

Refreshed 1 month ago

Claimed progress

The problem is still open: known estimates give partial bounds and show that one natural interpretation can fail, but no complete answer is known.

The problem asks for the shortest interval length guaranteeing a prescribed sparsity of integers with a divisor between $n$ and $2n$ . The discussion notes that the intended quantifier on $x$ is unclear, and no complete resolution is recorded.

Known results

Ford, 2008: $H(x,y,2y)\asymp x/((\log y)^\delta(\log\log y)^{3/2})$ , where $\delta=0.086071\ldots$ .
Cambie: under the interpretation requiring the condition for every $x$ , if $\epsilon(\log n)^\delta(\log\log n)^{3/2}\to\infty$ , no such $y$ exists.
Cambie: if $\epsilon\ll 1/n$ , then $y(\epsilon,n)\sim 2n$ .
For fixed $\delta\in(0,1)$ and sufficiently large $n$ , every $2(1+\delta)n$ consecutive integers contain many integers divisible by an element of $(n,2n)$ ; this is not a full asymptotic formula.

Current status (as of March 2026): Ford’s estimates and Cambie’s observations give substantial partial information, but the quantifier issue and the full determination of $y(\epsilon,n)$ remain open.

Sources

No solutions have been posted yet.