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Claim. Vjekoslav Kovač, An improved bound in Erdős problem #1054, a note dated 17 May 2026 in its current version, proves (Theorem 1) that there is a $c>0$ such that for all sufficiently small $\delta>0$ and all $X\ge1$

$$\#\{N\le X: f(N)\le\delta N\}\ll\exp\bigl(-e^{(1/\delta)^c}\bigr)\,X .$$

In particular the proportion of $N\le X$ with $f(N)\le\delta N$ is $O_M(\delta^M)$ for every $M>0$, and the upper density of $\{N: f(N)\le\delta N\}$ is $O(\exp(-e^{(1/\delta)^c}))$. The first version, posted on 11 May 2026, proved the $O_M(\delta^M)$ bound; the revision of 17 May 2026 gives the doubly exponential bound, after a suggestion of the site's curator in the thread. The note builds on the argument Terence Tao posted in the site's comments, which gives upper density $\ll\delta^2$; its abstract also acknowledges Liam Price's AI-generated proof. The posting states that no AI was used.

Since the set where $f(N)\le\delta N$ has upper density below $1$ for small $\delta$, $f(N)=o(N)$ fails, and it fails along every set of density one.

Submission note. Posted to the site's forum by Vjekoslav Kovač on 11 May 2026:

Since there have been many comments scattered around (thanks Thomas for moving them all here), I wrote up a short self-contained proof that the fraction of the numbers $n\in[1,x]$ satisfying $f(n)\leq \delta n$ is $\ll_M \delta^M$ for every $M>0$ (uniformly in $x>0$). [EDIT: After Thomas's comment, this is now improved to quite rapid decay $\ll \exp(-e^{(1/\delta)^c})$ for some small $c>0$.] It is based on Terry's proof, so all credit goes to him. This is just a minor modification and even a simplification - no decay in $k$ estimate is shown first. (No AI was used. Note that Terry's comment is from Nov 2025.)

Posted to the site's forum by Vjekoslav Kovač on 17 May 2026:

I updated the short note to give the quantitatively stronger result (motivated by Thomas's comment and promised above): The fraction of the numbers $n\in[1,x]$ satisfying $f(n)\leq \delta n$ is $\ll \exp(-e^{(1/\delta)^c})$ for some small constant $c>0$ and for all sufficiently small $\delta>0$, uniformly in $x>0$.

I also now rather prefer to talk about the uniform fraction bound

$$> \sup_{x>0}\frac{\#\{n\in[1,x] \,:\, f(n)\leq \delta n\}}{x} \ll > \exp(-e^{(1/\delta)^c})$$

rather than the asymptotic upper density

$$> \limsup_{x\to\infty}\frac{\#\{n\in[1,x] \,:\, f(n)\leq \delta n\}}{x} \ll > \exp(-e^{(1/\delta)^c}).$$

Before I thought that the upper density is the

right way of quantifying the failure of Erdos's conjecture, but I'm no longer sure. The former is a stronger property (and it is safer to formulate the result that way), as the density could theoretically be $0$. However, I don't know how to prove/disprove this for every small $\delta>0$, and I don't even have an opinion on whether the density should really be $0$ or not for every sufficiently small $\delta>0$.

Experiments based on the table of values of f are also a bit indecisive: For $\delta=1/2$ and for $x=1000, 5000, 10000, 20000$ the fractions are: $0.0830, 0.0670, 0.0602, 0.05625$ and it is unclear if this goes to $0$ or stabilizes at a positive number. For $\delta=1/10$ and for $x=1000, 5000, 10000, 20000$ the decay seems more convincing, as the fractions are now $0.0020, 0.0004, 0.0002, 0.0001$, but it might only be the case that these numbers stabilize much later (necessarily to a much smaller quantity).

For large $\delta$ (probably already $\delta>1$) it should be easy to see that the upper density is in fact strictly positive. Perhaps this is to be added if the whole thing starts converging to some paper. Speaking of large ratios $f(n)/n$, the $\limsup$ part of the original question is also something to think about.

EDIT: A brave conjecture would be that the distributions of the ratios $f(n)/n$ for $n\in[1,x]$ converge to some limit distribution as $x\to\infty$. An even braver conjecture is that the support of this limit distribution is the whole half-line $[0,\infty)$. Its approximation with $x=20000$ then looks like this (drawn in Wolfram Mathematica 13). Both of the above claims (positive density of $f(n)\leq\delta n$ for every $\delta>0$ and $\limsup f(n)/n=\infty$) would be consequences of this conjecture.

Covers. Parts (i) and (ii) of Problem 1054 as its Formulation reads them: $f(n)=o(n)$ is false, and it is false for almost all $n$. Part (iii), the limsup, is not addressed.

Standing. Claimed. The note is posted on the author's web page and in the site's thread, with no refereed publication or review recorded; the site labels the problem OPEN, so its commentary on Tao's bound is not acceptance. The theorem is restated with a proof as Theorem 1.1 of [[problems/divisors/E1054/claims/2026_10_03_chae_fraiture_hou_kovac_kudeba_shakov_vidal|the collaboration paper]]. Nothing here is independently reviewed by this project.