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Source. Liu--Sawhney, On further questions regarding unit fractions, arXiv:2404.07113v1, Lemma 2.3, pp. 7--8; see the source digest.

Statement. Take $N$ sufficiently large and $2\leq t\leq N^{1/4}$, and let $Y$ consist of the positive integers having at least one prime-power divisor $q>N/t$. Then

$$|Y\cap[1,N]|\leq\frac{2N\log t}{\log N}.$$

Proof. The union bound over prime powers gives

$$|Y\cap[1,N]|\leq N\sum_{N/t<q\leq N}\frac1q.$$

Put $u=N/t\geq N^{3/4}$. There are $O(N^{1/2}+N^{1/3}\log N)=O(N^{1/2})$ proper prime powers $p^a\leq N$ with $a\geq2$: count squares first and then use $p\leq N^{1/3}$ and $a\leq\log_2N$ for the remaining powers. Each of the powers exceeding $u$ contributes at most $1/u$, so their contribution after multiplication by $N$ is $O(N^{3/4})$. For the primes, Theorem 2.1 therefore gives, uniformly in $t$,

$$|Y\cap[1,N]| \leq N\log\!\left(\frac{\log N}{\log(N/t)}\right) +O\!\left(\frac{N}{(\log N)^2}\right).$$

Writing $a=\log t/\log N$, we have $\log2/\log N\leq a\leq1/4$. Hence

$$\frac1{1-a}\leq1+\frac43a.$$

The difference between $\log(1+3a/2)$ and $\log(1+4a/3)$ is bounded below by an absolute positive constant times $a$ on this range. Since $a\geq\log2/\log N$, this absorbs the preceding $O((\log N)^{-2})$ error for sufficiently large $N$. Thus

$$|Y\cap[1,N]| \leq N\log(1+3a/2) \leq\frac32Na\leq2Na,$$

as required. This expands the source's error absorption and its dismissal of proper prime powers.

Source notation corrections. The first sum in the printed proof has lower limit $t\leq q$, whereas the set in the statement requires $N/t<q$. That line also writes $|Y|$ although $Y$ itself is infinite; the quantity being bounded throughout is $|Y\cap[1,N]|$. The proof above uses the limits and finite intersection specified by the statement.

Dependencies. Theorem 2.1 and elementary divisor counting.

Bears on. #298 and #299, through smooth-denominator reductions in the quantitative reciprocal-sum criterion.