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Printed statement

Let $N$ be sufficiently large. Suppose $A\subseteq[N^{1-1/\log\log N},N]\cap\mathbb N$ and $1\le y\le z\le(\log N)^{1/500}$ satisfy:

1. $R(A)\ge2/y+(\log N)^{-1/200}$.
2. Each $n\in A$ has positive integer divisors $d_1,d_2$ with $y\le d_1$ and $4d_1\le d_2\le z$.
3. Every prime power dividing any $n\in A$ is at most $N^{1-6/\log\log N}$.
4. $\frac{99}{100}\log\log N\le\omega(n)\le2\log\log N$ for every $n\in A$.

Then $R(S)=1/d$ for some $S\subseteq A$ and integer $d\in[y,z]$.

Source. Bloom, arXiv:2112.03726v2, Proposition 1, p. 4; printed proof on pp. 18–19.

Source discrepancy and the variant proved here

The printed proof sets $\ell=\log\log N$, $M=N^{1-1/\ell}$, $K=MN^{-2/\ell}$ and $\eta=1/[2(\log N)^{1/100}]$. However, Proposition 2 requires

$$q\le\frac{c\eta MK^2}{N^2(\log N)^2} =\frac{cN^{1-7/\ell}}{2(\log N)^{2+1/100}}.$$

The printed assumption $q\le N^{1-6/\ell}$ does not imply this. Consequently this page does not assert a complete proof of the printed constant $6$.

For sufficiently large integer $N$, the existing Bloom–Mehta formalization instead proves a variant with $8$ in place of $6$ and the additional hypothesis $4y+4\le z$: technical_prop, lines 1528–1537. The following rewritten proof establishes precisely that stronger-hypothesis variant, following the paper's argument with these explicitly sourced parameters. It suffices for Theorem 2 and Theorem 3. The additional smoothness exclusion costs the same asymptotic amount in both applications.

Rewritten proof of the formalization variant

Use the notation $A_q$, $\mathcal Q_A$, $R(A;q)$ from Lemma 6, and put

$$L=\log N,\quad \ell=\log\log N,\quad M=N^{1-1/\ell},\quad K=N^{1-3/\ell},\quad \eta=\tfrac12 L^{-1/100},\quad \rho=L^{-1/200}.$$

All constants below are uniform for $1\le y$ and $4y+4\le z\le L^{1/500}$. For every integer $1\le u\le z$ and sufficiently large $N$,

$$\frac2u>2\rho,\qquad \frac2u-\frac1M\ge\frac2{u+1}+\rho, \qquad N^{1-8/\ell}\le ML^{-1/100}.$$

For the middle inequality, the gap $2/[u(u+1)]$ is at least a positive constant times $L^{-2/500}$, whereas $\rho=L^{-1/200}$ has a strictly larger decay exponent. Thus Lemma 7 can successively tune the mass just below $2/u$.

For consecutive integers $d_i=\lceil y\rceil+i$ up to $\lfloor z/4\rfloor$, construct nested nonempty sets $A_i\subseteq A$ with

$$R(A_i)\in[2/d_i-1/M,2/d_i),\qquad R(A_i;q)\ge L^{-1/100}\quad(q\in\mathcal Q_{A_i}).$$

The first application of Lemma 7 is allowed by $2/d_0\le2/y$; the inequalities above allow every subsequent one. Nonemptiness follows from $2/d_i-1/M>0$.

Choose the first $j$ for which $A_j$ contains a multiple of $d_j$. Such a $j$ exists: otherwise an element of the final set would, by nesting, be divisible by none of the integers in $[y,z/4]\cap\mathbb N$, contradicting condition 2. By minimality, no integer in $[y,d_j)\cap\mathbb N$ divides an element of $A_j$. The inclusive endpoint $\lfloor z/4\rfloor$ is intentional: condition 2 allows $d_1=z/4$; the printed proof's $\lceil z/4\rceil-1$ omits that case when $z/4$ is integral.

We may apply Proposition 3 to $A_j$: the regularity is inherited, $M\ge N^{1/2}$, and $R(A_j)\ge2/z-1/M\ge L^{-1/101}$. If its second alternative holds, we apply Proposition 2 with $k=d_j$ and the above $M,K,\eta$. Here $k$ divides the least common multiple because $A_j$ contains a multiple of it, and the mass and interval hypotheses are exactly those already arranged. The size requirements $M\ge N^{3/4}$, $N^{3/4}\le K\le M/2$, $k\le cM$ and $0<\eta<1$ hold for large $N$. Finally, both smoothness bounds hold:

$$\frac{N^{1-8/\ell}}{M/k} =kN^{-7/\ell}\longrightarrow0, \qquad \frac{N^{1-8/\ell}}{\eta MK^2/[N^2L^2]} =2L^{2+1/100}N^{-1/\ell}\longrightarrow0,$$

uniformly for $k\le z$. Thus they are below the fixed constant $c$. Proposition 2 gives a subset of mass $1/d_j$.

If instead the first alternative of Proposition 3 is needed, obtain $B\subseteq A_j$ with

$$R(B)\ge\tfrac13R(A_j)\ge\frac2{3d_j}-\frac1M, \qquad\sum_{q\in\mathcal Q_B}\frac1q\le\tfrac23\ell.$$

For large $N$, the lower bound is at least $1/(2d_j)+\rho=2/(4d_j)+\rho$, because the difference $1/(6d_j)-1/M$ dominates $\rho$. Apply Lemma 7 successively at the integers $e_i=4d_j+i$ through $\lfloor z\rfloor$ to obtain nested nonempty $B_i\subseteq B$ with

$$R(B_i)\in[2/e_i-1/M,2/e_i),\qquad R(B_i;q)\ge L^{-1/100}\quad(q\in\mathcal Q_{B_i}).$$

Every $n\in A_j$ has a pair $d_1,d_2$ from condition 2. The preceding minimality shows $d_1\ge d_j$, hence $4d_j\le d_2\le z$. Therefore some $B_s$ contains a multiple of $e_s$, by the same nesting argument. Since $\mathcal Q_{B_s}\subseteq\mathcal Q_B$, its prime-power reciprocal mass is at most $2\ell/3$. The final clause of Proposition 3 now guarantees its second alternative for $B_s$. All the checks for Proposition 2 just made remain valid with $k=e_s\le z$. It supplies $S\subseteq B_s\subseteq A$ with $R(S)=1/e_s$, as required.

Dependencies and verification scope

Lemma 7, Proposition 2, and Proposition 3. No proof of the printed constant-$6$ criterion is supplied here. The constant-$8$ variant is supported by the existing Lean source; that project was inspected, not built in this compilation.

Bears on

• Problem 298
• Problem 299