gpt pro 5.2 attempt on Math-733

Erdős Problem #254

1) FORMAL RESTATEMENT

Let $A\subseteq \mathbb{N}$. Write $A(x):=\left|A\cap[1,x]\right|$ for $x\ge 1$ (with $[1,x]$ understood as $\{n\in\mathbb{N}:1\le n\le x\}$). Assume: \begin{align} A(2x)-A(x) &\xrightarrow[x\to\infty]{} \infty,\label{wsref-0-65-71-3a-64-79-61-64-69-63-5f-67-72-6f-77-74-68}\\ \sum_{n\in A} \{\theta n\} &= \infty\quad\text{for every }\theta\in(0,1),\label{wsref-0-65-71-3a-74-68-65-74-61-5f-64-69-76} \end{align} where $\{y\}$ denotes the distance from $y$ to the nearest integer (so $\{y\}\in[0,1/2]$).

Conclusion to prove: There exists $N_0$ such that every integer $N\ge N_0$ can be expressed as a sum of distinct elements of $A$.

2) QUICK LITERATURE/CONTEXT CHECK

Cassels proved the conclusion under stronger hypotheses: instead of (eq:dyadic_growth) he assumed \[ \frac{A(2x)-A(x)}{\log\log x}\to\infty, \] and instead of (eq:theta_div) he assumed $\sum_{n\in A} \{\theta n\}^2=\infty$ for every $\theta\in(0,1)$ [Ca60]. The version stated above is listed as open in [ErdosProblems254].

3) ATTACK PLAN

Track A (prove): Attempt to extract from $A$ a "complete" increasing subsequence $(a_j)$ satisfying a coin-representation criterion such as $a_{j+1}\le 1+\sum_{i\le j}a_i$ after some point, or prove directly that the subset-sum set \[ \Sigma(A):=\Big\{\sum_{a\in F} a\;:\;F\subseteq A\text{ finite}\Big\} \] contains all integers beyond some threshold. A Fourier-analytic approach is natural: if the exponential sums $\sum_{n\in A} e(\theta n)$ are small in aggregate, subset sums should spread.

Track B (disprove): Try to build $A$ satisfying (eq:dyadic_growth)--(eq:theta_div) but for which $\Sigma(A)$ misses infinitely many integers (e.g. misses an arithmetic progression). Condition (eq:theta_div) blocks simple congruence obstructions (like $A\subseteq q\mathbb{N}$), so any counterexample would need a subtler obstruction.

4) WORK

4.1 Immediate consequences of the hypotheses

Infinitude and dyadic thickness.

Condition (eq:dyadic_growth) implies that for every $M$ there exists $x_M$ such that for all $x\ge x_M$ the interval $(x,2x]$ contains at least $M$ elements of $A$. In particular $A$ is infinite.

No fixed modulus obstruction; in particular $\gcd(A)=1$.

Fix an integer $q\ge2$ and take $\theta=1/q$. For $n\in\mathbb{N}$, the quantity $\{n/q\}$ equals $0$ exactly when $q\mid n$, and otherwise $\{n/q\}\ge 1/q$. If $A$ contained only finitely many integers not divisible by $q$, then $\sum_{n\in A}\{n/q\}$ would be finite, contradicting (eq:theta_div). Therefore, for every $q\ge2$, the set $A$ contains infinitely many integers not divisible by $q$. In particular, $\gcd(A)=1$.

4.2 Relation to Cassels' theorem

Cassels' result [Ca60] proves the desired completeness conclusion under two strengthened assumptions:

1. Dyadic growth strengthened from $A(2x)-A(x)\to\infty$ to $(A(2x)-A(x))/\log\log x\to\infty$.
2. Divergence strengthened from $\sum \{\theta n\}=\infty$ to $\sum \{\theta n\}^2=\infty$ for all $\theta\in(0,1)$.

The implication "$\sum \{\theta n\}=\infty\Rightarrow\sum \{\theta n\}^2=\infty$" is false in general (bounded terms can have divergent sum but convergent sum of squares), so Cassels' hypotheses do not follow formally from (eq:dyadic_growth)--(eq:theta_div).

4.3 Where a naive attempt stalls

A standard sufficient condition for a sequence $1\le a_1<a_2<\cdots$ to be complete (every sufficiently large integer is a sum of distinct terms) is the "greedy" inequality \[ a_{j+1} \le 1+\sum_{i\le j} a_i\qquad\text{for all large }j, \] which ensures that consecutive representable intervals expand without gaps. Condition (eq:dyadic_growth) guarantees many elements of $A$ in each dyadic interval, but does not obviously give control on the relative sizes needed to force such a greedy inequality for some subsequence.

Similarly, the divergence condition (eq:theta_div) suggests $A$ is not concentrated near multiples of $1/\theta$ for any fixed $\theta$, but converting this into uniform smallness of Fourier coefficients for $A$ (which would help show subset sums fill intervals) requires quantitative estimates, whereas (eq:theta_div) is qualitative.

5) VERIFICATION (adversarial check)

• The deduction that $A$ has infinitely many elements not divisible by each $q\ge2$ is rigorous and uses only $\theta=1/q$.
• The claim that Cassels' hypotheses are stronger and not implied by (eq:dyadic_growth)--(eq:theta_div) is justified by the general fact that divergence of $\sum x_n$ for bounded $x_n\ge0$ does not force divergence of $\sum x_n^2$.
• No step here assumes equidistribution or any deep estimate; the point is precisely that such quantitative input is missing.

6) FINAL

UNRESOLVED.

1. (i) Most advanced partial result proved here. From the stated hypotheses one can deduce $\gcd(A)=1$ and that $A$ meets every dyadic interval $(x,2x]$ in arbitrarily large finite sets.
2. (ii) Strongest obstruction / missing lemma. What is missing is a mechanism that converts the qualitative divergence condition (eq:theta_div) into a quantitative additive-structure conclusion about the subset-sum set $\Sigma(A)$ (for example, a bound forcing long intervals inside $\Sigma(A)$). Cassels' theorem achieves this using stronger growth and a square-divergence condition.
3. (iii) Next concrete steps. One concrete direction is to attempt a "quantitative upgrade" of (eq:theta_div) to a lower bound on the average of $\{\theta n\}$ over $A\cap[1,x]$ for many $x$ (uniform in $\theta$), then feed that into Fourier analysis of subset sums. Another is to attempt to extract a complete subsequence from $A$ using the dyadic thickness to enforce a greedy-type inequality.
4. (iv) Counterexample search description. A potential counterexample would be a set $A$ built in blocks so that $A(2x)-A(x)\to\infty$ and for each fixed $\theta$ the terms $\{\theta n\}$ do not decay too fast along $A$, yet the subset sums avoid infinitely many integers. Any construction must avoid congruence obstructions (ruled out by (eq:theta_div) at rational $\theta$), so the obstruction would likely be more subtle (e.g. structured gaps in the subset-sum semigroup).

7) COMPLETION ESTIMATE

$\mathbf{90\%}$ confidence that the deductions recorded here are correct; $\mathbf{0\%}$ confidence of a full resolution (the stated form is open).

---

References

• [ErdosProblems254] T. F. Bloom, Erdős Problem #254, https://www.erdosproblems.com/254 (accessed 2026-01-17).
• [Ca60] J. W. S. Cassels, On the representation of integers as the sums of distinct summands taken from a fixed set, Acta Sci. Math. (Szeged) $1960$, 111--124.
• [Er52] P. Erdős, Problem 4493, Amer. Math. Monthly $1952$, 557--558.
• [Er58b] P. Erdős, Sur certaines séries à valeur irrationnelle, Enseign. Math. $2$ $1958$, 93--100.
• [Er88c] P. Erdős, On the irrationality of certain series: problems and results, in New advances in transcendence theory (Durham, 1986), $1988$, 102--109.
• [FLC07] J. B. Friedlander, F. Luca, and M. Stoiciu, On the irrationality of a divisor function series, Integers $2007$.
• [Pr22] K. Pratt, The irrationality of a divisor function series of Erdős and Kac, arXiv:2209.11124 $2022$.
• [ScPu06] J. C. Schlage-Puchta, The irrationality of a number theoretical series, Ramanujan J. $2006$, 455--460.