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Erdős problem 1051

Is it true that if a0<a1<a2<⋯a_0 < a_1 < a_2 < \cdots is a strictly increasing sequence of integers with lim inf⁡an1/2n>1\liminf a_n^{1/2^n} > 1, then the series ∑n=0∞1an⋅an+1\sum_{n=0}^\infty \frac{1}{a_n \cdot a_{n+1}} is irrational?

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FormalConjectures/ErdosProblems/

1051.lean

Retained formal statement1 of 2

Is it true that if a0<a1<a2<⋯a_0 < a_1 < a_2 < \cdots is a strictly increasing sequence of integers with lim inf⁡an1/2n>1\liminf a_n^{1/2^n} > 1, then the series ∑n=0∞1an⋅an+1\sum_{n=0}^\infty \frac{1}{a_n \cdot a_{n+1}} is irrational?

This was solved in the affirmative by Aletheia [Fe26]. This was extended by Barreto, Kang, Kim, Kovač, and Zhang [BKKKZ26], who essentially give a complete answer: if ϕ=1+52\phi=\frac{1+\sqrt{5}}{2} is the golden ratio and 1≤a1<a2<⋯1\leq a_1 < a_2 < \cdots is a monotonically increasing sequence of integers such that lim sup⁡an1/ϕn=∞\limsup a_n^{1/\phi^{n}}=\infty then ∑n=1∞1anan+1\sum_{n=1}^\infty \frac{1}{a_na_{n+1}} is irrational. Conversely, for any 1<C<∞1 < C < \infty there exists a sequence of integers 1≤a1<⋯1\leq a_1<\cdots such that lim⁡an1/ϕn=C\lim a_n^{1/\phi^{n}}=C where this infinite sum is a rational number.

(Further, more general, results are available in [BKKKZ26].)

This was formalized in Lean by Baretto.

FormalConjectures/ErdosProblems/1051.leanErdos1051.erdos_10511 lineExact file
True ↔ ∀ (a : ℕ → ℤ), StrictMono a → Erdos1051.GrowthCondition a → Irrational (Erdos1051.ErdosSeries a)
SolvedProof has a holelean4external proof

The proof uses `sorry`: part of the argument is written but not proved. Lean accepts the file; it does not accept the theorem.