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Source. Liu--Sawhney, On further questions regarding unit fractions, arXiv:2404.07113v1, Lemma 2.2, p. 7; see the source digest.

Statement as printed. For sufficiently large $N$,

$$\bigl|\{n\in[1,N]:\Omega(n)>5\log\log N\}\bigr| \ll N(\log N)^{-3}.$$

The paper defines $\Omega(n)=\sum_i a_i$ when $n=\prod_i p_i^{a_i}$ with distinct primes $p_i$. Thus prime factors are counted with multiplicity, not by the distinct-prime function $\omega$.

Status of the printed statement. The counting bound is false for $\Omega$ with multiplicity, as the counterargument below shows. The separate reciprocal-mass deduction at the end supplies the weaker input needed for Theorem 1.1. Neither correction is attributed to an author erratum or to the uninspected published version.

Printed argument. Set $k=\lceil5\log\log N\rceil$ and $S=\sum_{q\leq N}1/q$, where $q$ ranges over prime powers. The source uses the fact that the number of multiples of $d$ up to $N$ is at most $N/d$, and then asserts

$$\bigl|\{n\in[1,N]:\Omega(n)\geq5\log\log N\}\bigr| \leq\frac{NS^k}{k!} \leq N\left(\frac{eS}{k}\right)^k. \tag{1}$$

The remainder of that argument is explicit. By Theorem 2.1, the prime terms contribute $\log\log N+O(1)$. The other powers contribute a bounded amount because

$$\sum_p\sum_{a\geq2}\frac1{p^a} =\sum_p\frac1{p(p-1)}<\infty.$$

Consequently $S=\log\log N+O(1)$, while $k!\geq(k/e)^k$. The logarithm of the final factor in (1) is

$$k\log\frac{eS}{k} =5(1-\log5)\log\log N+O(1).$$

Since $5(\log5-1)>3$, (1) would imply the stated bound.

The first inequality in (1), however, is not justified by the supplied multiple-counting argument. An integer with $\Omega(n)\geq k$ has at least $k$ distinct prime-power divisors, but those divisors need not be coprime. For selected divisors $q_1,\ldots,q_k$, their simultaneous divisibility counts multiples of $\operatorname{lcm}(q_1,\ldots,q_k)$, not multiples of $q_1\cdots q_k$. For example, $2$ and $4$ both divide $4$, while their product does not. Thus the factorial-moment expression in (1) does not follow in the way it would for distinct prime divisors.

Counterargument to the printed count

Write $L=\log N$, $\ell=\log\log N$, and set

$$a=\lceil(21/5)\ell\rceil,\qquad T=\lfloor N/2^a\rfloor,\qquad \kappa=(21/5)\log2<3.$$

Then $T\asymp N/L^\kappa$ and $\log\log T=\ell+O(\ell/L)$. The external Hardy–Ramanujan normal-order theorem gives $\Omega(m)>0.9\log\log T>0.8\ell$ for $(1-o(1))T$ integers $m\le T$. For each such $m$, $n=2^am\le N$ and complete additivity gives $\Omega(n)=a+\Omega(m)>5\ell$. This map is injective, so

$$\#\{n\le N:\Omega(n)>5\ell\}\gg N/L^\kappa.$$

Its ratio to $N/L^3$ tends to infinity because $\kappa=2.911218\ldots<3$, contradicting the printed bound. The argument does not require $m$ to be odd.

External input. Hardy–Ramanujan, The normal number of prime factors of a number n, Quarterly Journal of Mathematics 48 (1917), 76–92, Theorems B and C. In the collected-paper reproduction, Theorem B is on printed p. 336/PDF p. 11, and Theorem C and its normal-order consequence are on printed p. 340/PDF p. 15. The latter extends the concentration statement to prime factors counted with multiplicity. This external theorem is stated and cited, not reproved.

Sufficient reciprocal-mass deduction

For sufficiently large $N$ and $\beta=5\log(3/2)-3/2=0.527325\ldots>0$,

$$\sum_{\substack{n\le N\\\Omega(n)>5\log\log N}}\frac1n \ll(\log N)^{-\beta}=o(1).$$

This is an elementary deduction included in the compilation, using Theorem 2.1. It replaces the use of the false counting statement in the main proof; it neither proves that statement nor gives the source's stronger claimed $O((\log N)^{-2})$ reciprocal loss.

Proof. Put $z=3/2$. Every $n\le N$ occurs with its exact weight in the positive expansion of the finite Euler product, so

$$\sum_{n\le N}\frac{z^{\Omega(n)}}n \le\prod_{p\le N}(1-z/p)^{-1}.$$

Every local geometric series converges since $z/p\le3/4<1$. Uniformly for primes $p\ge2$, $-\log(1-z/p)=z/p+O(p^{-2})$. Theorem 2.1 and convergence of $\sum_p p^{-2}$ therefore give

$$\prod_{p\le N}(1-z/p)^{-1} =\exp(z\log\log N+O(1))\ll L^z.$$

For $\Omega(n)>5\ell$, we have $z^{\Omega(n)}>L^{5\log z}$. Dividing the weighted bound by this factor gives $O(L^{z-5\log z})=O(L^{-\beta})$, as claimed. It bounds the loss on any subset of $[1,N]$, including the localized set in Theorem 1.1.

Verification scope. Independent source-based reviews checked the counterargument and the reciprocal deduction separately; see the preliminary review and source checks. The false literal v1 count remains recorded; it is not used as a proved input.

Bears on. #298 and #299, through the paper's quantitative reciprocal-sum criterion. This source-level limitation does not alter their independent resolution by Bloom's theorem.