we-solve (beta)

Math-5538-A2 · revision r1 · on Math-5538

gpt pro 5.2 attempt on Math-5538

No claim recorded · LLM-generated · gpt pro 5.2 ·

Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Problem #263
  2. 1. Formal Restatement
  3. 2. Quick Literature / Context Check
  4. 3. Attack Plan
  5. 4. Work (provable partial results)
  6. Remark.
  7. 5. Verification / Sanity Checks
  8. 6. FINAL
  9. Why stuck / what remains open
  10. Strongest partial result proved here
  11. Next steps (concrete)
  12. 7. Completion Estimate
  13. References

Problem #263

1. Formal Restatement

Call a strictly increasing sequence of positive integers $(a_n)_{n\ge 1}$ a Type-2 irrationality sequence if for every integer sequence $(b_n)_{n\ge 1}$ with \[ \frac{b_n}{a_n}\to 1\qquad (n\to\infty), \] the series $\sum_{n\ge 1} 1/b_n$ is irrational (assuming the sum converges, which it does if $b_n\to\infty$ sufficiently fast).

Questions:

  1. Is $a_n=2^{2^n}$ a Type-2 irrationality sequence?
  2. Must every Type-2 irrationality sequence satisfy $a_n^{1/n}\to\infty$?

2. Quick Literature / Context Check

A "folklore" sufficient condition states that $\sum 1/a_n$ is irrational if $\lim a_n^{1/2^n}=\infty$. Recent work of Kovač and Tao studies several "types" of irrationality sequences; in particular they show a broad class of sequences are not Type-2 when $a_{n+1}/a_n^2\to 0$ (under convergence of $\sum 1/a_n$), while the folklore condition yields Type-2 examples under the much stronger growth $\liminf a_{n+1}/a_n^{2+\varepsilon}>0$. The borderline case $a_{n+1}\asymp a_n^2$ includes $a_n=2^{2^n}$ (where $a_{n+1}=a_n^2$) and remains open.

3. Attack Plan

Two natural directions:
  1. Try to prove Type-2 for $2^{2^n}$. One would need a digit/carry or Diophantine approximation argument showing that any perturbation $b_n=2^{2^n}(1+o(1))$ forces the sum $\sum 1/b_n$ to avoid all rationals.
  2. Try to disprove Type-2 for $2^{2^n}$. Construct a specific integer sequence $b_n$ with $b_n/2^{2^n}\to 1$ such that $\sum 1/b_n\in\mathbb{Q}$. This resembles building an "Egyptian-fraction-like" expansion with very large denominators constrained to lie in thin relative intervals.

4. Work (provable partial results)

We record the known partial implications as clean lemmas (with proofs only where elementary).
Lemma (A sufficient growth condition implying Type-2): Assume there exists $\varepsilon>0$ and $c>0$ such that for all sufficiently large $n$, \[ a_{n+1} \ge c\, a_n^{2+\varepsilon}. \] Then $(a_n)$ is Type-2.
Proof[Proof sketch]: Under the hypothesis, $a_n$ grows faster than doubly-exponential in the sense that $a_n^{1/2^n}\to\infty$. If $b_n/a_n\to 1$ then also $b_n^{1/2^n}\to\infty$. The folklore irrationality criterion (see the cited sources) then gives that $\sum 1/b_n$ is irrational. Hence $(a_n)$ is Type-2. ∎
Lemma (A broad obstruction (Kovač--Tao)): Suppose $(a_n)$ is strictly increasing, $\sum 1/a_n$ converges, and \[ \frac{a_{n+1}}{a_n^2}\to 0. \] Then $(a_n)$ is not Type-2.
Remark.
The sequence $a_n=2^{2^n}$ satisfies $a_{n+1}/a_n^2=1$, so it lies exactly on the boundary not covered by Lemma Lemma (A broad obstruction (Kovač--Tao)) and does not satisfy the sufficient condition in Lemma Lemma (A sufficient growth condition implying Type-2).

5. Verification / Sanity Checks

  • For $a_n=2^{2^n}$, indeed $a_{n+1}=2^{2^{n+1}}=(2^{2^n})^2=a_n^2$, so $a_{n+1}/a_n^2\equiv 1$.
  • The "folklore" criterion is consistent with classical irrationality proofs for rapidly growing Egyptian-fraction denominators.

6. FINAL

UNRESOLVED

Why stuck / what remains open

Existing theorems cover either much slower growth (showing not Type-2) or much faster growth (showing Type-2). The sequence $2^{2^n}$ is precisely the "critical" growth rate $a_{n+1}\asymp a_n^2$ where current methods (per the cited references) do not decide Type-2.

The second question (whether Type-2 forces $a_n^{1/n}\to\infty$) also remains open; current results only give partial constraints.

Strongest partial result proved here

Lemma Lemma (A sufficient growth condition implying Type-2): a super-quadratic recurrence $a_{n+1}\ge c a_n^{2+\varepsilon}$ guarantees Type-2.

Next steps (concrete)

  1. Attempt a "critical" carry/approximation analysis for $a_{n+1}=a_n^2$ (or close): show that any rational approximation would force impossible congruence/size constraints on $b_n$.
  2. Try to explicitly construct $b_n=2^{2^n}+o(2^{2^n})$ so that partial sums hit a fixed rational within an error smaller than the remaining tail.
  3. Search for intermediate criteria between $a_{n+1}/a_n^2\to 0$ and $a_{n+1}\ge a_n^{2+\varepsilon}$ that can decide the borderline.

7. Completion Estimate

35%. We can cleanly state the known sufficient/obstruction regimes and locate $2^{2^n}$ at the critical boundary, but the main question is still open.

References

  • [Bloom263] T. F. Bloom, Erdős Problem #263, https://www.erdosproblems.com/263 (accessed 2026-01-17).
  • [BoLo90] P. Borwein and L. Loring, Some questions of Erdős and Graham on numbers of the form $\sum g_n/2^{g_n}$, Amer. Math. Monthly 97 $1990$, 22--26.
  • [TUZ20] S. Tengely, M. Ulas, and J. Zygadło, On a Diophantine equation of Erdős and Graham, Revista Colombiana de Matemáticas 54 $2020$, 445--459.
  • [KoTa24] V. Kovač and T. Tao, The Ahmes series and irrationality sequences, arXiv:2406.17593 (version v4, 2025-07-14).