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There is an absolute constant $C$ such that the following holds for all sufficiently large integers $N$. Put $L=\log N$, $\ell=\log\log N$, and suppose

$$N^{.9999}\le S\le K\le M\le N/10, \qquad \frac N{L^{10}}\le K\le10^{-7}\frac NL, \tag{1}$$ $$S\le\min\left\{\frac{M^2}{CN}, \frac{K^3}{CN^2\ell^6}\right\}. \tag{2}$$

Let $A$ be all integers in $[M,N]$ whose prime-power divisors are at most $S$ and which satisfy $\widetilde\Omega(n)\le5\ell$, $\Omega(n)\le10\ell$. Here $\Omega$ counts prime factors with multiplicity and $\widetilde\Omega$ is the maximum exponent. Let

$$\mathcal Q_A=\{q:q\text{ is a prime power dividing some }n\in A\}, \qquad Q=\operatorname{lcm}(A)=\operatorname{lcm}(\mathcal Q_A).$$

Choose probabilities $1/\ell\le p_n\le1/2$ for $n\in A$. If $1\le x\le Q$ is an integer and $\sum_{n\in A}p_n/n=x/Q$, the independently sampled subset $B$ obeys

$$\mathbb P\left(\sum_{n\in B}\frac1n=x/Q\right)\ge\frac1{4Q}. \tag{3}$$

Source and precise scope. This reconstructs the method of Liu–Sawhney, arXiv:2404.07113v1, Proposition 3.2, pp. 9–12, in a sufficient range for Theorem 1.2. The printed period has a counterexample. The proof here uses the actual period and the stronger sixth power in (2), supplies the carrier step, and makes the residue and cyclic-counting conventions explicit. It does not prove the statement as printed or its whole fifth-power parameter range. These are compilation corrections, not claims about the uninspected published version.

The density input is Lemma 3.3, and the local Fourier inputs are Fact 2.5 and Lemma 3.1. The external probability input is Azuma–Hoeffding. The prime number theorem $\pi(X)\sim X/\log X$ and the prime-power product estimate at Theorem 2.1 are external. One bounded-scale choice also uses external Bertrand's postulate: for every integer $m\ge1$ there is a prime in $(m,2m]$.

Bears on. Problem 297.

Proof

Write $R(B)=\sum_{n\in B}1/n$, $e(t)=\exp(2\pi it)$, and $A_d=\{n\in A:d\mid n\}$ for every positive integer $d$. For a finite set $D$ of positive integers, write $[D]$ for its least common multiple, with $[\varnothing]=1$.

Density, orthogonality, and the major arc

Lemma 3.3 applies because $S\le K<N/2$, and gives $|A|\ge .89N$. Consequently $Q\ge\max A\ge.89N>M\ge K$. Also, by the external prime-power product estimate,

$$Q\le\prod_{q\le S}q\le e^{5S} \tag{4}$$

for large $N$. The product ranges over all prime powers.

Every $n\in A$ divides $Q$, so finite cyclic orthogonality gives

$$\begin{aligned} \mathbb P(R(B)-x/Q\in\mathbb Z) &=\frac1Q\sum_{-Q/2<h\le Q/2} \operatorname{Re}\left(e(-hx/Q) \prod_{n\in A}(1-p_n+p_ne(h/n))\right). \end{aligned} \tag{5}$$

All frequencies are integers in the stated half-open interval. The major arc $|h|\le M/2$ lies inside it. Since $M\ge N^{.9999}$, $|A|\ge.89N\ge N^{.95}$, and $[1/\ell,1/2]\subseteq[L^{-2},1-L^{-2}]$ eventually, Lemma 3.1 gives a contribution at least $3/(4Q)$.

To recover equality from the integer-congruence event in (5), reveal the Bernoulli indicators one at a time. The centered increments are bounded by $1/n$, so their squared bounds sum to at most $N/M^2$. Azuma–Hoeffding gives

$$\mathbb P(|R(B)-x/Q|\ge1) \le2\exp(-M^2/(2N))\le e^{-6S}<\frac1{4Q} \tag{6}$$

for large $N$, using (2), $C\ge14$, and (4). Every nonzero integer difference is in this tail. It remains to show that the normalized minor-arc absolute contribution in (5) is at most $1/(4Q)$.

Residues and minor-arc decay

For each $n$, let $h_n$ be the unique representative of $h\pmod n$ in $(-n/2,n/2]$. Put

$$t=\frac{100N^2L\ell^2}{K^2},\qquad I_h=(h-K/2,h+K/2),$$ $$D_h=\{q\in\mathcal Q_A: |\{n\in A_q:|h_n|\ge K/2\}|<t\}, \quad T_q=\{n\in A_q:|h_n|<K/2\}. \tag{7}$$

Thus $T_q$ is a set, and $|A_q\setminus T_q|<t$ for $q\in D_h$. Equality at $K/2$ belongs to the bad set. These definitions repair the missing absolute value on p. 10 and the set/cardinality mismatch in the p. 11 display.

Let $W(h)=\prod_{n\in A}|1-p_n+p_ne(h/n)|$. Each factor is in $[0,1]$, and every $n$ has exactly $\Omega(n)\le10\ell$ prime-power divisors. Hence

$$W(h)^{10\ell}\le \prod_{q\in\mathcal Q_A}\prod_{n\in A_q}|1-p_n+p_ne(h/n)|.$$

For $|h_n|\ge K/2$, Fact 2.5 and $p_n\le1/2$ imply

$$|1-p_n+p_ne(h/n)| \le\exp(-p_nK^2/N^2) \le\exp(-K^2/(N^2\ell)).$$

Every $q\notin D_h$ has at least $t$ such factors. Taking the $10\ell$-th root proves

$$W(h)\le N^{-10|\mathcal Q_A\setminus D_h|}. \tag{8}$$

Averaging and the stronger parameter condition

Fix $q\in D_h$. For any collection of candidate primes $p'$,

$$\sum_{p'}|A_{qp'}\setminus T_q| \le\sum_{n\in A_q\setminus T_q}\Omega(n)<10\ell t. \tag{9}$$

Use the cutoff

$$Z=10^6t\ell^4/L=10^8N^2\ell^6/K^2.$$

Uniformly under (1),

$$10^{22}L^2\ell^6\le Z\le10^8L^{20}\ell^6, \qquad \log Z\le21\ell$$

eventually. PNT therefore gives

$$\pi(Z)\ge\frac{Z}{2\log Z} \ge\frac{10^6}{42}\frac{t\ell^3}{L} >20000\frac{t\ell^3}{L}.$$

After excluding the one prime underlying $q$, (9) supplies $p'\le Z$ with $(p',q)=1$ and

$$|A_{qp'}\setminus T_q|\le\frac{L}{1000\ell^2}. \tag{10}$$

By (2), choosing $C\ge10^8$ ensures

$$qp'\le SZ\le10^8SN^2\ell^6/K^2\le K. \tag{11}$$

This explains the sixth power in (2). To infer (10) from (9) needs order $t\ell^3/L$ prime candidates. The source's preliminary $t\ell^2/L$ cutoff, and its later $10^6t\ell^3/L$ cutoff on p. 11, do not give this many: the latter has logarithm of order $\ell$ and only order $t\ell^2/L$ primes. The fourth-power cutoff above supplies the required additional factor.

A divisor between $2K$ and $100K$

Bertrand's postulate implies that, for every real $v\ge1$, a prime lies in $(v,2v]$. For $1\le v<2$ use 2. Otherwise apply the integer statement to $\lfloor v\rfloor$: the resulting prime exceeds $v$ and is at most $2v$.

Put $a=K/(qp')\ge1$. We construct $r$, a product of one or two distinct primes at most $S$, coprime to $qp'$, such that

$$2K\le qp'r\le100K. \tag{12}$$

If $S\ge100a$, three disjoint dyadic intervals starting at $2a$ provide three distinct primes in $(2a,16a]$. At most two divide $qp'$, so one is a legal $r$ and is at most $S$.

If $S<100a$, PNT supplies a prime $r_1\in[S/2,S]$ avoiding those two divisors. Then $qp'r_1<100K$. If it is at least $2K$, use $r=r_1$. Otherwise set $a_2=K/(qp'r_1)>1/2$. Four disjoint dyadic intervals starting at $2a_2>1$ provide four primes in $(2a_2,32a_2]$. At most three divide $qp'r_1$, so choose a remaining prime $r_2$. It satisfies

$$r_2\le32a_2\le64K/S<S$$

for large $N$, since $S\ge N^{.9999}$ and $K\le N$. Now $r=r_1r_2$ gives (12), with upper bound $32K$. This verifies the bounded real endpoints as well as the distinctness of the prime factors.

Common primes and nonempty admissible fibers

Let $\mathcal P$ be the primes in $[20L,40L]$. PNT gives $|\mathcal P|\ge10L/\ell$ eventually. Define

$$\mathcal P_q=\{p\in\mathcal P:(p,qp'r)=1, \ A_{qp'rp}\subseteq T_q\}.$$

By (10), at most $L/(1000\ell^2)$ bad denominators lie in $A_{qp'r}$. Each excludes at most $10\ell$ primes from $\mathcal P_q$, and at most four additional primes divide $qp'r$. Thus

$$|\mathcal P_q|\ge|\mathcal P|-L/(100\ell)-4 \ge .9|\mathcal P|. \tag{13}$$

For $p\in\mathcal P_q$, put $b=qp'rp$. The assertion $A_b\subseteq T_q$ is useful only after showing $A_b$ nonempty. By (1) and (12),

$$40KL\le b\le4000KL\le N/2000, \qquad 2000\le N/b\le L^9/40.$$

There are at most five distinct prime divisors of $b$. The prime exponent in $q$ is at most $5\ell$, since $q$ divides an actual member of $A$. All other factors $p'$, the primes in $r$, and $p$ are distinct and avoid it. Hence $\widetilde\Omega(b)\le5\ell$ and $\Omega(b)\le5\ell+4$. Every prime-power divisor of $b$ is at most $S$: this holds for $q$ by definition, for the factors in $r$ by construction, and for $p'\le Z$ and $p\le40L$ because both upper bounds are smaller than $S\ge N^{.9999}$ eventually.

The carrier lemma therefore supplies $n\in A_b\cap[N/2,N]$. This justifies the implicit multiple-selection step on source p. 12, including both exponent restrictions.

Since $n\in T_q$, $h-h_n$ is a multiple of $n$ in $I_h$. The interval has length $K$, whereas $qp'r\ge2K$, so it contains at most one multiple of $qp'r$. All choices $p\in\mathcal P_q$ yield the same integer $x_q\in I_h$, and every such $p$ divides $x_q$.

One common multiple and the cyclic count

For $q_1,q_2\in D_h$, (13) gives $|\mathcal P_{q_1}\cap\mathcal P_{q_2}|\ge.8|\mathcal P| \ge8L/\ell$. The product of these distinct primes is at least $(20L)^{8L/\ell}>N$. It divides $x_{q_1}-x_{q_2}$, whose absolute value is less than $K\le N$. Therefore all $x_q$ are equal, and $[D_h]$ divides their common value in $I_h$. If $D_h$ is empty, use the integer $h\in I_h$ and $[D_h]=1$ instead.

For a minor-arc frequency $|h|>M/2$, we cannot have $D_h=\mathcal Q_A$. Otherwise a multiple of $Q$ would lie within distance $K/2$ of $h\in(-Q/2,Q/2]$. Since $Q>K$, the only possible multiple is zero, which $|h|>M/2\ge K/2$ excludes.

Fix $D\subseteq\mathcal Q_A$ and count residues cyclically modulo $Q$. Since $[D]\mid Q$, wrapping preserves divisibility by $[D]$. There are $Q/[D]$ multiples in the cycle. Each has at most $K+1$ integer-frequency residues at distance less than $K/2$, so

$$\#\{h:D_h=D\}\le(K+1)Q/[D].$$

If $s=|\mathcal Q_A\setminus D|$, then $Q/[D]\le\prod_{q\notin D}q\le N^s$ and $K+1\le N$. There are at most $N^s$ missing sets of size $s$. Using (8), the normalized total absolute minor-arc contribution is at most

$$\frac1Q\sum_{s\ge1}N^{s+1}N^sN^{-10s} \le\frac2{QN}\le\frac1{4Q}$$

for sufficiently large $N$. Adding the major arc proves that (5) is at least $1/(2Q)$. Subtracting (6) proves (3).

Limits of the statement

All thresholds are sufficiently-large thresholds, with no explicit finite $N_0$ certified here. The actual-period, sixth-power form is enough for the source's counting choices. Other applications of the printed Proposition 3.2, including the later denominator theorems, require their own parameter and target checks.