Erdős was right?

The claim, in plain words

Add up the reciprocals of a set of whole numbers. For the squares the sum converges; for the primes it diverges, though slowly (about 3.17 for the primes up to 100 million, by our count). “Erdős asked whether every set A of positive integers for which Σ 1/a diverges contains arithmetic progressions of every finite length”, citing a 1974 problem list. As the paper recounts, the question extends a density problem: Erdős and Turán conjectured in 1936 that sets without three-term progressions have density tending to zero, Roth proved that case, and Szemerédi proved in 1975 that sets of positive upper density contain progressions of every length.

Write r_k(N) for the size of the largest subset of {1, …, N} with no k-term progression. The paper’s link is a counting argument: a progression-free set has at most r_k(2^m) members between 2^m and 2^m+1, so if the densities r_k(2^m)/2^m have a finite sum, every progression-free set has a finite reciprocal sum. For three terms, Bloom and Sisask crossed that line in 2020, and Kelley and Meka went further. For five or more terms, the paper notes, Leng, Sah and Sawhney’s bound “does not give the dyadic summability”; as far as we know, Green and Tao’s four-term bound does not either.

Result 159 is a single OpenAI preprint, Quasipolynomial Bounds for Arithmetic Progressions (198 pages, dated September 23, 2026). Theorem 1.1: for each fixed k ≥ 3 there are positive constants C_k, c_k, ε_k with r_k(N) ≤ C_k N exp(−c_k(log N)^ε_k) for every N ≥ 2; equivalently, a set of density α in {1, …, N} contains a k-term progression once log N ≥ A_k(2 + log 1/α)^A_k, a quasipolynomial threshold. Corollary 1.2: every set with divergent reciprocal sum contains progressions of every finite length. The paper also recovers Green and Tao’s theorem on dense subsets of the primes, and shows that even a divergent sum of (log a)^B/a forces progressions of every length.

What it doesn’t claim

No constant is given: “the exponent is not optimized”. By our own inference ε_k ≤ 1/2, since Behrend’s construction gives r_k(N) ≥ r₃(N) ≥ N exp(−C√log N). The paper claims no improvement for three terms (“no improvement of the three-term exponent is claimed”), where Kelley and Meka’s and later bounds already have explicit exponents, and it says nothing about how the constants grow with k. It calls the statement “Erdős’s conjecture”; it does not use the name “Erdős–Turán conjecture” for it, and mentions no prize.

Where it stands

Posted as a preprint

OpenAI, 198 pages, dated September 23, 2026

Peer reviewed

Not yet

Checked in Lean

Erdős’s statement itself, by a weaker route; the paper’s main bound is not formalized. We have not rebuilt it

Last checked October 19, 2026. The preprint has not been revised since it was added to the repository in its initial commit on October 6, 2026. OpenAI published a Lean formalization of Erdős’s statement for arbitrary sets of natural numbers (comparator), in 4,774 files and 1,122,983 lines (excluding Mathlib); we found no unproved steps, and we have not rebuilt it. It reaches the statement through its own, weaker density bound, r_k(N) ≤ C N exp(−c(log log N)^1+η), which still suffices for the doubling-block sum; the paper’s Theorem 1.1 is not formalized. OpenAI also released an abridged summary of the model’s reasoning for this result. The paper is not among those OSVerify reviewed, and we have seen no expert review.

How the proof goes

1. A density increment. If a set of density α has no k-term progression, it cannot look random: higher-order Fourier analysis, built on Leng, Sah and Sawhney’s quasipolynomial inverse theorem, finds a structured region where it is denser by a fixed factor. Repeating, after about log(1/α) rounds the density would exceed one. The difficulty is the cost: every round shrinks and complicates the region.
2. Triangular polynomial cells. A number u lies in a cell when a ladder of polynomial expressions all land within narrow widths of integers, each rung using the actual integers found on the rungs below it. Each rung has a precision budget, and the key estimate (Theorem 2.1) makes the extra precision at a rung depend, besides the density and the dimensions, only on the rungs above it, never on its own budget or the rungs below, so the budgets stay polynomial over all rounds.
3. One round. Prepare the cell by rank cuts; descend along random affine paths to sampled boxes; apply an increment that adds a number of new coordinates independent of the box’s dimension (adapting Kelley and Meka’s sifting and using Schoen and Sisask’s almost-periodicity); return through every layer keeping every old equation exact; extract a new cell. After about log(1/α) rounds the target density would exceed one, which gives the threshold.

The numbers in the video

• Our sums: the reciprocals of the primes up to 10², 10⁴, 10⁶, 10⁸ add to about 1.80, 2.48, 2.89, 3.17; those of the squares tend to π²/6 ≈ 1.64.
• Our checks: 199 + 210j is prime for j = 0, …, 9; no four of the first 2,500 squares are evenly spaced.
• Our exact search: r₃(74) = 22, r₃(64) = 20, r₄(48) = 25; in the 22-element set we found in 1 to 74, the nine numbers 1, 4, …, 25 contain five members.
• Our budget toy uses a made-up loss rule: charged only from higher layers, the total grows like a power of the number of rounds; charged by itself, its number of digits doubles every round.
• The Lean counts: 4,774 files and 1,122,983 lines, our count of the import closure.

Read the source

• Quasipolynomial Bounds for Arithmetic Progressions (PDF), 198 pages.
• The Lean scope note.
• The collection’s catalog, where this is result 159.

The paper

OpenAI groups related papers into one result; this one has a single paper.

• Quasipolynomial Bounds for Arithmetic Progressions

Transcript

Spot a mistake?

This episode simplifies on purpose, but it shouldn’t be wrong. If something is, open an issue on GitHub or leave a comment on YouTube. Corrections are noted here with a date.