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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Claim. For a positive integer MM let RMR_M be the set of positive integers congruent to 00, 11 or 22 modulo MM, and put

A=⋃k≥0((R2q∩[4q29,4q2])∪(Rq∩[5q29,q2])),q=3k.A=\bigcup_{k\ge0}\left( \left(R_{2q}\cap\left[\frac{4q^2}{9},4q^2\right]\right) \cup \left(R_q\cap\left[\frac{5q^2}{9},q^2\right]\right) \right),\qquad q=3^k.

Every positive integer is m2+am^2+a with an integer m≥0m\ge0 and a∈Aa\in A, and, writing A(X)=∣A∩[1,X]∣A(X)=\lvert A\cap[1,X]\rvert for real XX,

lim inf⁡X→∞A(X)X=32,lim sup⁡X→∞A(X)X=92.\liminf_{X\to\infty}\frac{A(X)}{\sqrt X}=3\sqrt2, \qquad \limsup_{X\to\infty}\frac{A(X)}{\sqrt X}=\frac92.

The upper limit is witnessed along X=32kX=3^{2k}. Hence the smallest possible value of the limsup in the first question of Problem 33 is at most 9/29/2.

Covers. An upper bound of 9/29/2 for the smallest possible limsup, the first question of Problem 33. The value lies below the pending upper bounds of van Doorn, 2φ5/2≈6.662\varphi^{5/2}\approx6.66, and of the unreviewed write-up of 2026-09-07, 132/4≈4.59613\sqrt2/4\approx4.596. The claim does not determine the smallest limsup and asserts neither optimality of this set nor a bound A(X)≤(9/2)XA(X)\le(9/2)\sqrt X at every cutoff. The lower limit 323\sqrt2 is that of this one set and says nothing about the liminf question, which concerns every complement.

Standing. L3 is author-recorded at tier 0: a complete symbolic proof, including a uniform counting error at every real cutoff, without independent examination or a Lean proof. Catalog acceptance is pending.

Depends on. L3 (tier 0), which consumes no other native claim and no external theorem.