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The story of Erdős problem #1026

Author not recordedWhat's newPublished Dec 9, 2025Original︎ ↗Archived copy

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The story of Erdős problem #102630 passages · 1 marked

This gave a somewhat complicated but predictable conjecture for the values of the sequence . On posting this, Boris found a clean formulation of the conjecture, namely that

For k≥1 and −k≤a≤k, the constant satisfies c(k^2 + 2a + 1) = k/(k^2 + a).Unassessed

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Proposition 2 For any , one has3 passages · 1 marked

This idea of using packing to prove Erdős-Szekeres type results goes back to a 1959 paper of Seidenberg , although it was a discrete rectangle-packing argument that was not phrased in such an elegantly geometric form. It is possible that Aristotle was “aware” of the Seidenberg argument via its training data, as it had incorporated a version of this argument in its proof.

The packing approach to proving Erdős–Szekeres-type results originates in Seidenberg's 1959 paper.Unassessed

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Here is an illustration of the above argument using the AlphaEvolve-provided example2 passages

Theorem 3 For any , one has1 passage

Theorem 4 For any and with , one has3 passages

Pick . Then we have a grid1 passage

By the packing condition, we have0 passages

From (2) we have1 passage

Inserting this bound and rearranging, we conclude that0 passages

Taking the supremum over we conclude that5 passages

We can pack by unit squares. We can similarly pack2 passages

Applying Theorem 3 , we conclude that0 passages

The right-hand side is16 passages

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Not found in the text 9

As quoted

given a sequence of {k^2+1} distinct real numbers, one can find a subsequence of length {k+1} which is either increasing or decreasing; and that one cannot improve the {k^2+1} to {k^2}

Every sequence of k^2+1 distinct reals has a monotone subsequence of length k+1, and k^2+1 is optimal.Unassessed

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As quoted

a result of Hanani that every sequence of length {k(k+3)/2} can be decomposed into the union of {k} monotone sequences

Every sequence of length k(k+3)/2 decomposes into a union of k monotone sequences.Unassessed

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As quoted

a similar construction to the one above, in which Alice divides the coins into {k^2} pairs that are almost even, in such a way that the longest monotone sequence is of length {k}, gives the upper bound {c(k^2) \leq 1/k}

The constant c satisfies the upper bound c(k^2) ≤ 1/k.Unassessed

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As quoted

the Hanani result mentioned above gives the lower bound {c(n) \geq (\frac{1}{\sqrt{2}}-o(1))/\sqrt{n}}

The constant c satisfies c(n) ≥ (1/√2 − o(1))/√n.Unassessed

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As quoted

this was enough data for Stijn to conjecture that {c(k^2)=1/k}, which would also imply that {\sqrt{n} c(n) \rightarrow 1} as {n \rightarrow \infty}

The coin-game constant satisfies c(k^2) = 1/k, and √n·c(n) → 1 as n → ∞.Unassessed

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As quoted

If {k \geq 1} and {-k \leq a \leq k}, then {c(k^2+2a+1) \leq \frac{k}{k^2+a}}

For k≥1 and −k≤a≤k, c(k^2+2a+1) ≤ k/(k^2+a).Unassessed

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As quoted

For any {n}, one has

For all n, the coin-game constant satisfies c(n) ≥ 1/f(n), where f is the square-packing constant.Unassessed

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As quoted

For any {k \geq 1}, one has

For all k≥1, the axis-parallel square-packing constant satisfies f(k^2+1) ≤ k.Unassessed

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As quoted

For any {k \geq 1} and {c \in {\bf Z}} with {k^2+2c+1 \geq 1}, one has

For k≥1 and integer c with k^2+2c+1≥1, f(k^2+2c+1) ≤ k + c/k.Unassessed

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Kind: unknown
Authors: Not recorded
Publisher: What's new
Published: Dec 9, 2025
DOI: None
Link: https://terrytao.wordpress.com/2025/12/08/the-story-of-erdos-problem-126/
Archived copy: https://web.archive.org/web/20261011082250/https://terrytao.wordpress.com/2025/12/08/the-story-of-erdos-problem-126/
Stored text: 32,135 characters, retrieved Sep 13, 2026 · 133 sections, 420 passages
Claims extracted: Not recorded
Versions: One known
Copies: None recorded
Corrections: None recorded
Facts checked: Nov 9, 2026

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Nov 9, 2026DividedSegmenter · Divided into 552 parts by its own structure
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