Erdős Problem #257

Erdős Problems #256--#258: Round 2 continuation

Erdős Problem #257

1) ROUND-2 OBJECTIVE

I am using the fail-safe strict-progress option. The full problem remains open, so I cannot honestly claim path (A), (B), or (C) for the original statement. The round-2 goal is therefore:

The new target is:

If $A\subseteq \mathbb{N}$ is infinite and eventually periodic, then \[ \sum_{n\in A}\frac{1}{2^n-1} \] is irrational.

This is the most promising direction because Round 1 already identified the base-$2$ carry problem as the main obstruction for arbitrary $A$, whereas eventually periodic sets allow a direct Lambert-series theorem to be applied.

2) Round-1 FOUNDATION USED

I use the following Round-1 results.

1. The identity \[ \sum_{n\in A}\frac{1}{2^n-1}=\sum_{m\ge 1}\frac{f_A(m)}{2^m}, \qquad f_A(m)=\#\{d\in A:d\mid m\}. \]
2. The observation that cofinite sets are already handled as a trivial special case.
3. The explicit gap from Round 1: direct base-$2$ periodicity arguments become hard because of carrying in the divisor-counting expansion.

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is a theorem of Luca and Tachiya:

If $q$ is an integer with $|q|>1$ and $(u_n)_{n\ge 1}$ is an eventually periodic sequence of rational numbers which is not eventually zero, then \[ \sum_{n=1}^{\infty}\frac{u_n}{q^n-1} \] is irrational.

This bypasses the divisor-function expansion entirely. For an eventually periodic set $A$, the indicator sequence $u_n=\mathbf{1}_A(n)$ lies exactly in the range of this theorem.

4) ATTACK PLAN (ROUND-2)

Round 1 got stuck because arbitrary $A$ leads to irregular coefficients $f_A(m)$ and difficult carrying. The new plan is to avoid that representation altogether:

1. encode $A$ by its indicator sequence $u_n=\mathbf{1}_A(n)$;
2. observe that "$A$ eventually periodic" is equivalent to "$(u_n)$ eventually periodic";
3. apply the Luca--Tachiya irrationality theorem with $q=2$.

This removes the exact obstacle isolated in Round 1.

5) WORK (ROUND-2)

Recall that a set $A\subseteq \mathbb{N}$ is eventually periodic if there exist integers $N\ge 1$ and $p\ge 1$ such that \[ n\in A \iff n+p\in A\qquad \text{for all }n\ge N. \] Equivalently, its indicator sequence \[ u_n:=\mathbf{1}_A(n)\in\{0,1\} \] is eventually periodic.

Assume now that $A$ is infinite and eventually periodic. Then $(u_n)$ is an eventually periodic sequence of rational numbers. Because $A$ is infinite, the sequence $(u_n)$ is not eventually zero: if $u_n=0$ for all sufficiently large $n$, then $A$ would be finite.

Therefore the Luca--Tachiya theorem applies with $q=2$ and gives that \[ \sum_{n=1}^{\infty}\frac{u_n}{2^n-1} \] is irrational. But this is exactly \[ \sum_{n\in A}\frac{1}{2^n-1}. \] Hence we have proved the following.

Theorem. If $A\subseteq \mathbb{N}$ is infinite and eventually periodic, then \[ \sum_{n\in A}\frac{1}{2^n-1}\notin \mathbb{Q}. \]

This immediately yields several concrete corollaries.

1. If $A$ is a full residue class modulo $m$ (for example, all even integers, all odd integers, or $A=\{n:n\equiv r\pmod m\}$), then the sum is irrational.
2. If $A$ is a finite union of residue classes modulo $m$, then the sum is irrational.
3. If $A$ differs by a finite set from a periodic set, then the sum is irrational.

These all strictly extend the Round-1 cofinite case.

6) ADVERSARIAL VERIFICATION

1. Does infinity really imply "not eventually zero"? Yes. If an eventually periodic indicator sequence were eventually zero, then only finitely many integers would belong to $A$.
2. Are the theorem's hypotheses matched exactly? Yes: $q=2$ is an integer with $|q|>1$, and $u_n\in\{0,1\}\subset\mathbb{Q}$ is eventually periodic.
3. Could a finite initial segment break the argument? No. The external theorem is stated for eventually periodic sequences, so a preperiod is allowed.
4. What counterexamples remain possible? Any counterexample to the full problem must now have aperiodic indicator sequence. In particular, no eventually periodic infinite set can be a counterexample.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED)

The full problem is still open, but Round 2 proves a genuinely new theorem beyond Round 1:

For every infinite eventually periodic set $A\subseteq\mathbb{N}$, \[ \sum_{n\in A}\frac{1}{2^n-1} \] is irrational.

This covers, in one shot, all arithmetic progressions, all finite unions of residue classes, and all finite perturbations of periodic sets. Consequently, any counterexample to the full problem would have to come from a genuinely aperiodic infinite set.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 50%

9) REFERENCES

1. Florian Luca and Yohei Tachiya, Irrationality of Lambert series associated with a periodic sequence, International Journal of Number Theory 10 $2014$, 623--636.
2. Yohei Tachiya, Irrationality of Lambert series associated with periodic sequence (conference abstract based on joint work with Florian Luca), 2013. The abstract states the theorem for eventually periodic rational sequences.
3. T. F. Bloom (ed.), Erdős Problem #257, erdosproblems.com/257, accessed 2026-03-07.

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%

Erdős Problem #257

1) FORMAL RESTATEMENT

Let $A\subseteq\mathbb{N}:=\{1,2,3,\dots\}$ be an infinite set. Define \[ S(A):=\sum_{n\in A}\frac{1}{2^n-1}. \] Since $0<\frac{1}{2^n-1}\le 2^{-(n-1)}$ for $n\ge 2$, the sum converges absolutely for every $A$.

The problem asks whether the following statement is true: \[ (\dagger)\qquad \forall\text{ infinite }A\subseteq\mathbb{N}:\ S(A)\notin\mathbb{Q}. \]

2) QUICK LITERATURE/CONTEXT CHECK

(From the standard summary and cited papers.)

• For $A=\mathbb{N}$, one has \[S(\mathbb{N})=\sum_{n\ge1}\frac{1}{2^n-1}=\sum_{m\ge1}\frac{d(m)}{2^m},\] where $d(m)$ is the divisor-counting function; Erdős $1948$ proved this is irrational [Er48].
• Tao--Teräväinen $2025$ proved irrationality when $A$ is the set of primes, and more generally when $A$ is the set of prime powers [TaTe25].
• Erdős $1968$ proved $S(A)$ is irrational whenever $(a,b)=1$ for all distinct $a,b\in A$ and $\sum_{n\in A} \frac{1}{n}<\infty$ [Er68d].
• Erdős also speculated a far-reaching "bounded perturbation" generalisation $\sum_{n\in A}1/(2^n-t_n)$; this was disproved by Kovač--Tao $2024$, who give bounded $t_n$ for which such a sum can be rational [KoTa24].
• As of late 2025/early 2026 the general statement $(\dagger)$ remains listed as open on the Erdős Problems website [ErP257].

3) ATTACK PLAN

Two natural routes:

1. Lambert-series / base-2 expansion route. Rewrite $S(A)$ as a Lambert-type series $\sum f_A(m)/2^m$ where $f_A(m)$ counts divisors of $m$ lying in $A$; then try to show the base-$2$ expansion cannot be eventually periodic.
2. Construction route (for a counterexample). Try to choose an infinite $A$ whose induced coefficients $f_A(m)$ lead to an eventually periodic base-$2$ expansion after carrying.

I did not manage to complete either route for arbitrary infinite $A$.

4) WORK

Rewriting as a Lambert-type series.

For every $n\ge1$, \[ \frac{1}{2^n-1}=\frac{2^{-n}}{1-2^{-n}}=\sum_{k=1}^{\infty}2^{-kn} \] (geometric series). Hence, by Tonelli's theorem (all terms are nonnegative), \[ S(A)=\sum_{n\in A}\sum_{k\ge1}2^{-kn} =\sum_{m\ge1}\left(\#\{(n,k):n\in A,\ k\ge1,\ kn=m\}\right)\,2^{-m}. \] If we define \[ f_A(m):=\#\{d\in A: d\mid m\}, \] then the map $(n,k)\mapsto m=kn$ identifies the above count with $f_A(m)$, giving the identity \begin{equation}\label{wsref-0-65-71-3a-6c-61-6d-62-65-72-74} S(A)=\sum_{m\ge1}\frac{f_A(m)}{2^m}. \end{equation} In particular, when $A=\mathbb{N}$ we have $f_A(m)=d(m)$.

Tiny numerical experiments (illustrative only).

Below are high-precision numerical values (computed from long partial sums):

• $S(\mathbb{N})\approx 1.60669515241529176378\dots$ (the Erdős--Borwein constant).
• For $A=2\mathbb{N}$ (even indices), $S(A)=\sum_{k\ge1}\frac{1}{2^{2k}-1}\approx 0.42109768603342377729\dots$.
• For $A=\{2^k:k\ge0\}$, $S(A)=\sum_{k\ge0}\frac{1}{2^{2^k}-1}\approx 1.40393682788217832057\dots$.

These computations do not distinguish rational/irrational and are included only as sanity checks.

A simple fully proved special case (conditional on $S(\mathbb{N})$).

If $A$ is cofinite (i.e. $\mathbb{N}\setminus A$ is finite), then \[ S(A)=S(\mathbb{N})-\sum_{n\in\mathbb{N}\setminus A}\frac{1}{2^n-1}. \] The second term is a finite sum of rationals and hence rational. Therefore:

If $S(\mathbb{N})$ is irrational (known by Erdős [Er48]), then $S(A)$ is irrational for every cofinite $A$.

This does not address sparse $A$.

5) VERIFICATION

• Identity (eq:lambert) is correct: each $2^{-m}$ coefficient counts divisors $d\in A$ of $m$.
• The problem remains that $f_A(m)$ can be irregular and may be large, and carrying in base $2$ complicates direct periodicity arguments.

6) FINAL

UNRESOLVED

(i) Strongest fully proved partial result obtained here. The Lambert-series identity (eq:lambert) holds for every $A\subseteq\mathbb{N}$ (proved above). Also, if $A$ is cofinite then $S(A)$ is irrational assuming Erdős' theorem that $S(\mathbb{N})$ is irrational [Er48].

(ii) First gap. A proof of $(\dagger)$ for arbitrary infinite $A$ would need a method to rule out eventual periodicity (in base $2$) of the sum (eq:lambert) after carries, using only that $A$ is infinite. I did not find such an argument.

(iii) Top 3 next moves.

1. Try to extend Erdős' 1968 method [Er68d] to drop the pairwise-coprime hypothesis, perhaps by controlling overlaps in divisor structure of $A$.
2. Seek a structural dichotomy for infinite $A$ (e.g. "contains a dense multiplicative substructure" vs "very sparse") and prove irrationality in each regime.
3. Attempt an explicit counterexample search in the style of Kovač--Tao [KoTa24], but with the rigid choice $t_n\equiv1$; i.e. try to design $A$ so that the base-$2$ expansion of (eq:lambert) becomes eventually periodic.

(iv) Minimal counterexample structure. A counterexample would be an infinite $A$ such that $S(A)\in\mathbb{Q}$. Any such $A$ must be infinite (by hypothesis) and cannot be cofinite (since $S(\mathbb{N})$ is irrational). It would likely need to be extremely sparse and carefully engineered so that the induced coefficients $f_A(m)$ in (eq:lambert) produce eventual periodicity after carrying.

7) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 35%.