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Problem 878

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claims/: The 1 claim page of Problem 878, one per claimant's result; the problem's standing derives from them.


Statement. If n=∏1≤i≤tpikin=\prod_{1\leq i\leq t} p_i^{k_i} is the factorisation of nn into distinct primes then let

f(n)=∑piℓi,f(n)=\sum p_i^{\ell_i},

where ℓi\ell_i is chosen such that n∈[piℓi,piℓi+1)n\in [p_i^{\ell_i},p_i^{\ell_i+1}). Furthermore, let

F(n)=max⁡∑iaiF(n)=\max \sum_{i} a_i

where the maximum is taken over all distinct a1,…,ak≤na_1,\ldots,a_k\leq n such that (ai,aj)=1(a_i,a_j)=1 for i≠ji\neq j and all prime factors of each aia_i are prime factors of nn.

Is it true that, for almost all nn,

f(n)=o(nlog⁡log⁡n)f(n)=o(n\log\log n)

and

F(n)≫nlog⁡log⁡n?F(n) \gg n\log\log n?

Is it true that

max⁡n≤xf(n)∼xlog⁡xlog⁡log⁡x?\max_{n\leq x}f(n)\sim \frac{x\log x}{\log\log x}?

Is it true that (for all xx, or perhaps just for all large xx)

max⁡n≤xf(n)=max⁡n≤xF(n)?\max_{n\leq x}f(n)=\max_{n\leq x}F(n)?

Find an asymptotic formula for the number of n<xn<x such that f(n)=F(n)f(n)=F(n). Find an asymptotic formula for

H(x)=∑n<xf(n)n.H(x)=\sum_{n<x}\frac{f(n)}{n}.

Is it true that

H(x)≪xlog⁡log⁡log⁡log⁡x?H(x) \ll x\log\log\log\log x?

Formulation. In F(n)F(n) the aia_i are distinct, pairwise coprime integers with 2≤ai≤n2\le a_i\le n whose prime factors all divide nn, and their number is not restricted. The site's wording does not exclude ai=1a_i=1, which has no prime factors and is coprime to every integer. Allowing it would give F(n)≥f(n)+1F(n)\ge f(n)+1 for every nn, so f(n)=F(n)f(n)=F(n) would never hold, the two maxima would never agree, and the count asked for in the fourth question would be zero. Erdős's paper excludes it: he notes that f(n)=F(n)f(n)=F(n) when nn is a prime power, and his Theorem 1 gives integers nkn_k with ω(nk)=k\omega(n_k)=k and F(nk)=nkF(n_k)=n_k. His definition sets no number of summands. Requiring exactly ω(n)\omega(n) summands, each at least 22, would make each aia_i a power of a single prime of nn, so F=fF=f and the first question would have answer no. Under this reading the two maxima differ at x=210x=210, as Kevin Barreto noted in the site's comments and the site's remark records: max⁡n≤210f(n)=f(210)=383\max_{n\le210}f(n)=f(210)=383, while F(210)≥128+189+125=442F(210)\ge128+189+125=442. This refutes only the form of the third question for all xx; it says nothing about all large xx, so it settles no part. Erdős's own question (6) asks first whether m(x)=M(x)m(x)=M(x) for infinitely many xx, where mm and MM are the two maxima, and only suggests the site's two stronger forms.

Status. Open.

Source. erdosproblems.com/878, accessed 2026-09-04. Cite as: T. F. Bloom, Erdős Problem #878, https://www.erdosproblems.com/878.

References.

  • [Er84e] Erdős, P., On two unconventional number theoretic functions and on some related problems. (1984), 113-121.

Formalization. No statement in formal-conjectures. Kenta Kitamura's Lean development claiming proofs of the first two questions is recorded on Kitamura 2026.

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