OBJECTIVE AND SOURCE REVIEW
Erdős #257; independent attempt, 2026-09-23. Is $S_A=\sum_{n\in A}(2^n-1)^{-1}$ irrational for every infinite $A\subset\mathbb N$? I read the complete
257.tex, whose periodic special case imports a Lambert-series theorem. Here is a separate elementary criterion covering irregular sparse sets, with an explicit obstruction for rational sums.
WORK: A NECESSARY GAP INEQUALITY
Enumerate $A=\{a_1<a_2<\cdots\}$. If $S_A=p/q$ is rational, let $Q_j=\prod_{i\le j}(2^{a_i}-1)$. The positive remainder after $j$ terms is a nonzero rational with denominator dividing $qQ_j$, whence \[
S_A-\sum_{i\le j}\frac1{2^{a_i}-1}\ge\frac1{qQ_j}
>\frac{2^{-\sum_{i\le j}a_i}}q.
\] On the other hand, using $1/(2^n-1)\le2^{1-n}$ and distinctness of the remaining exponents, \[
S_A-\sum_{i\le j}\frac1{2^{a_i}-1}
\le\sum_{n\ge a_{j+1}}2^{1-n}=2^{2-a_{j+1}}.
\] Consequently rationality forces, for every $j$, \[
a_{j+1}-\sum_{i\le j}a_i<2+\log_2q.
\] In particular, an unbounded limsup of the expression on the left proves irrationality. For example $a_j=3^j$ satisfies the criterion, giving an aperiodic family handled independently of the periodic theorem. More generally $a_{j+1}\ge(2+\delta)a_j$ eventually suffices: the earlier terms have sum at most $(1+1/(1+\delta))a_j+O(1)$, leaving a difference tending to infinity.
ADVERSARIAL VERIFICATION AND REMAINING GAP
The denominator used need not be reduced; this makes the rational lower bound weaker but valid. Infinite positivity ensures the remainder is nonzero. The geometric tail bound uses the fact that $A$ is a set, so exponents do not repeat. For dense or moderately sparse sets the gap expression can be negative, and the criterion says nothing. This is a standard denominator-tail mechanism, not a claim of new literature progress. It does not decide arbitrary aperiodic $A$.
SOURCES CHECKED
Complete local source:
gpt_pro_5.4/257.tex. Official indexed statement checked at
https://www.erdosproblems.com/257, 2026-09-23. The proof above does not depend on the external periodic-sequence theorem.
FINAL
UNRESOLVED. A self-contained lacunary criterion is established; the general infinite-subset question remains.
COMPLETION ESTIMATE
Conservative subjective progress, not a probability.
COMPLETION: 12%