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Statement

There is an absolute constant C>0C>0 such that, for all sufficiently large integers NN, every A⊆{1,…,N}A\subseteq\{1,\ldots,N\} satisfying

R(A)≥Clog⁡log⁡log⁡Nlog⁡log⁡Nlog⁡NR(A)\ge C\frac{\log\log\log N}{\log\log N}\log N

contains S⊆AS\subseteq A with R(S)=1R(S)=1.

Source. Bloom, arXiv:2112.03726v2, Theorem 3, p. 2; proof pp. 6–7. The sufficiently-large-NN quantifier is explicit in the existing Lean statement reproduced in Appendix B, p. 20. It also avoids undefined or negative iterated logarithms for small NN.

Rewritten proof

Write L=log⁡NL=\log N, ℓ=log⁡log⁡N\ell=\log\log N, and ε=log⁡ℓ/ℓ\varepsilon=\log\ell/\ell. All bounds below have absolute constants. Discard n<Nεn<N^\varepsilon. Since ∑n<Nε1/n≤εL+O(1)\sum_{n<N^\varepsilon}1/n\le\varepsilon L+O(1), the remaining A′A' has R(A′)≥(C/2)εLR(A')\ge(C/2)\varepsilon L once CC and NN are large.

Let XX consist of integers with no prime divisor in [5,L1/1200][5,L^{1/1200}]. For Nε/2≤x≤NN^\varepsilon/2\le x\le N, Lemma 1 gives

∣X∩[x,2x]∣≪x/ℓ.|X\cap[x,2x]|\ll x/\ell.

Its parameter condition is valid because L1/1200≤log⁡xL^{1/1200}\le\log x for large NN. Covering [Nε,N][N^\varepsilon,N] by O(L)O(L) dyadic intervals, each with reciprocal contribution O(1/ℓ)O(1/\ell), yields R(X∩[Nε,N])≪L/ℓR(X\cap[N^\varepsilon,N])\ll L/\ell.

Next let YY consist of integers failing

99100ℓ≤ω(n)<101100ℓ.\tfrac{99}{100}\ell\le\omega(n)<\tfrac{101}{100}\ell.

Uniformly for x≥Nε/2x\ge N^\varepsilon/2 and n∈[x,2x]∩[1,N]n\in[x,2x]\cap[1,N], we have log⁡log⁡n=ℓ+O(log⁡ℓ)\log\log n=\ell+O(\log\ell). Thus, for sufficiently large NN, an integer in YY differs from its local center log⁡log⁡n\log\log n by at least ℓ/200\ell/200. The external Turán estimate

∑2≤n≤t(ω(n)−log⁡log⁡n)2≪tlog⁡log⁡t\sum_{2\le n\le t}(\omega(n)-\log\log n)^2\ll t\log\log t

therefore implies ∣Y∩[x,2x]∣≪x/ℓ|Y\cap[x,2x]|\ll x/\ell. Another dyadic summation gives R(Y∩[Nε,N])≪L/ℓR(Y\cap[N^\varepsilon,N])\ll L/\ell. Since εL=(log⁡ℓ)L/ℓ\varepsilon L=(\log\ell)L/\ell, both deletions are absorbed, and the set B=A′∖(X∪Y)B=A'\setminus(X\cup Y) has R(B)≥(C/4)εLR(B)\ge(C/4)\varepsilon L.

Put θ=1−1/ℓ\theta=1-1/\ell and Ni=NθiN_i=N^{\theta^i}. Partition BB among the intervals (Ni+1,Ni](N_{i+1},N_i]; half-open intervals avoid double counting boundary integers. Nonempty pieces have Ni≥NεN_i\ge N^\varepsilon. As θi≤e−i/ℓ\theta^i\le e^{-i/\ell}, there are at most 2ℓlog⁡(1/ε)2\ell\log(1/\varepsilon) relevant indices for large NN. One piece BiB_i consequently has

R(Bi)≥CεL8ℓlog⁡(1/ε).R(B_i)\ge\frac{C\varepsilon L}{8\ell\log(1/\varepsilon)}.

Set T=⌊Ni⌋T=\lfloor N_i\rfloor, ℓT=log⁡log⁡T\ell_T=\log\log T. The piece is nonempty, so it contains an integer at least NεN^\varepsilon, whence T≥NεT\ge N^\varepsilon. Since εL≤log⁡T≤L\varepsilon L\le\log T\le L, we have ℓT=ℓ+O(log⁡ℓ)\ell_T=\ell+O(\log\ell), and the interval and regularity conditions needed at scale TT hold:

Bi⊆[T1−1/ℓT,T],99100ℓT≤ω(n)≤2ℓT.B_i\subseteq[T^{1-1/\ell_T},T],\qquad \tfrac{99}{100}\ell_T\le\omega(n)\le2\ell_T.

For the first inclusion, T≤NiT\le N_i and ℓT≤ℓ\ell_T\le\ell give T1−1/ℓT≤Ni1−1/ℓ=Ni+1T^{1-1/\ell_T}\le N_i^{1-1/\ell}=N_{i+1}; the exponents are nonnegative for large NN. Every integer in the piece is at most TT. Every n∈Bin\in B_i also has a prime divisor between 55 and L1/1200≤(log⁡T)1/500L^{1/1200}\le(\log T)^{1/500}.

Remove from BiB_i all integers divisible by a prime power q>T1−8/ℓTq>T^{1-8/\ell_T}, obtaining Bi′B_i'. Writing U=T1−8/ℓTU=T^{1-8/\ell_T}, the reciprocal mass removed is at most

∑U<q≤T∑n≤Tq∣n1n≤∑U<q≤T1+log⁡(T/q)q≪log⁡TℓT∑U<q≤T1q≪log⁡TℓT2≪Lℓ2.\begin{aligned} \sum_{U<q\le T}\sum_{\substack{n\le T\\q\mid n}}\frac1n &\le\sum_{U<q\le T}\frac{1+\log(T/q)}q\\ &\ll\frac{\log T}{\ell_T}\sum_{U<q\le T}\frac1q\\ &\ll\frac{\log T}{\ell_T^2} \ll\frac L{\ell^2}. \end{aligned}

The first inequality uses the harmonic bound ∑m≤T/q1/m≤1+log⁡(T/q)\sum_{m\le T/q}1/m\le1+\log(T/q), including when qq is near TT. The second follows from log⁡(T/q)≤8log⁡T/ℓT\log(T/q)\le8\log T/\ell_T and log⁡T/ℓT→∞\log T/\ell_T\to\infty. The third uses Mertens to obtain ∑U<q≤T1/q≪1/ℓT\sum_{U<q\le T}1/q\ll1/\ell_T.

Because ε/[ℓlog⁡(1/ε)]≍1/ℓ2\varepsilon/[\ell\log(1/\varepsilon)]\asymp1/\ell^2, choosing CC sufficiently large makes this loss at most half the mass lower bound for BiB_i. In particular

R(Bi′)≫Lℓ2≥(log⁡T)1/200R(B_i')\gg\frac L{\ell^2}\ge(\log T)^{1/200}

for large NN. All the conditions of the constant-88 version of Corollary 1 are now met at scale TT. It provides a subset of Bi′⊆AB_i'\subseteq A with reciprocal sum one.

Source details

Three source-level details are explicit in this rewrite. The smoothness constant 88, in place of the printed 66, is the variant of Proposition 1 used by the existing formal proof. The Turán estimate on p. 6 is claimed there down to x=exp⁡log⁡Nx=\exp\sqrt{\log N} while centering ω\omega at log⁡log⁡N\log\log N; that range is too broad. Only x≥Nε/2x\ge N^\varepsilon/2 is used here, where the local and global centers differ by O(log⁡log⁡log⁡N)O(\log\log\log N). The harmonic bound on p. 7 is written with log⁡(T/q)\log(T/q) alone; the necessary additive 11 is retained above and has no effect on the final estimate.

Dependencies and existing formalization

Lemma 1, the constant-88 proof on Corollary 1, and the external Mertens and Turán estimates. The accessible existing Lean 3 theorem is unit_fractions_upper_log_density. Appendix B reports its formal verification; no build was run here.

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