GPT 6 Astra Ultra attempt on Math-5473

Erdős Problem 169: reciprocal sums of progression-free sets

Statement and outcome

For $k\ge3$, let $f(k)$ be the supremum of $\sum_{n\in A}1/n$ over subsets of the positive integers with no nonconstant $k$-term arithmetic progression. Let $W(k)$ be the two-colour van der Waerden number. The problem asks for estimates and whether $f(k)/\log W(k)\to\infty$.

We prove the basic Ramsey lower bound, construct an explicit infinite admissible set, and establish a uniform vanishing reciprocal tail for $k=3$ using one explicitly external quantitative Roth theorem. The comparison as $k\to\infty$ remains unresolved in this attempt.

Finite truncations and the Ramsey comparison

Every nonnegative reciprocal sum is the supremum of its finite truncations. Consequently $f(k)=\sup_N F_k(N)$, where $F_k(N)$ maximizes the weighted sum on admissible subsets of $[1,N]$. Also $f(k)\le f(k+1)$, since a $(k+1)$-term progression contains a $k$-term one.

A colouring of $[1,W(k)-1]$ with neither colour containing a $k$-term progression partitions $H_{W(k)-1}$ into two admissible sums. Thus \[ f(k)\ge \frac12H_{W(k)-1}\ge\frac12\log W(k). \tag{1} \] The last step follows by comparing each $1/j$ with $\int_j^{j+1}dt/t$. Independently, $f(k)\ge H_{k-1}$. These elementary arguments also appeared in the earlier local attempt and have been checked.

A digit construction with a rigorous tail

Let $B$ consist of the nonnegative integers whose ternary digits all lie in $\{0,1\}$, and put $A=1+B$. Suppose $x+z=2y$ with $x,y,z\in B$. Modulo three, the allowed digits force the last digits of all three numbers to agree: unequal endpoint digits give residue one, which $2y$ cannot have. Subtract the common digit and divide by three. Repetition through all digits gives $x=y=z$. Hence $B$, and its translate $A$, are three-term-free.

For $m\ge1$, let $B_m$ restrict to the first $m$ ternary places. Then \[ \sum_{b\in B_m}\frac1{b+1} \le \sum_{a\in A}\frac1a \le \sum_{b\in B_m}\frac1{b+1} +3(2/3)^m. \tag{2} \] Indeed, numbers of $B$ whose highest nonzero place is $j\ge m$ number $2^j$ and are at least $3^j$. Summing their reciprocal bounds gives the geometric tail. This is an explicit, effectively evaluated lower construction for every $f(k)$, $k\ge3$, not a claim to the known numerical records.

Uniform reciprocal tails when $k=3$

We now use the following external input, not proved here: Bloom--Sisask, Theorem 1.1, gives absolute $c,C>0$ such that every three-term-free subset of an interval of length $X\ge3$ has size at most $CX/(\log X)^{1+c}$. Translation preserves the hypothesis.

For any three-term-free $A$ and integer $M\ge3$, partition its elements at least $M$ into the intervals $[2^jM,2^{j+1}M)$, $j\ge0$. Each contributes at most $C/(\log M+j\log2)^{1+c}$ to the reciprocal sum. Comparing this decreasing series with an integral gives \[ \sup_{A\text{ three-term-free}}\ \sum_{\substack{a\in A\\a\ge M}}\frac1a \le C\left((\log M)^{-1-c} +\frac{(\log M)^{-c}}{c\log2}\right). \tag{3} \] In particular $f(3)<\infty$, and the sufficiently-large-minimum question recorded on the problem page has an affirmative answer for $k=3$. Moreover $0\le f(3)-F_3(M-1)$ is bounded by the right side of $3$. This uniform statement supplies a genuine truncation error; convergence separately for each admissible set would not by itself give it.

Assessment and attribution

Source: https://www.erdosproblems.com/169. External theorem: T. F. Bloom and O. Sisask, Breaking the logarithmic barrier in Roth's theorem on arithmetic progressions, https://arxiv.org/abs/2007.03528. The old gpt_pro_5.2/169.tex correctly derived $1$, but treated even finiteness for $k=3$ as unavailable; $3$ records precisely what follows from the primary theorem. Neither this input nor the digit construction addresses general $k$ or improves the factor $1/2$ in $1$ uniformly. No novelty claim is made.

COMPLETION ESTIMATE

COMPLETION: 35%