Erdős Problem #267
Erdős ProblemThe problem
Let and be the Fibonacci sequence. Let be an infinite sequence with . Mustbe irrational?
(Erdős Problem #267 — irrationality — https://www.erdosproblems.com/267)
Result
For every infinite sequence n₁ < n₂ < ⋯ with a uniform ratio gap n_{k+1}/n_k ≥ c for some c > 1, the sum Σ 1/F_{n_k} of reciprocal Fibonacci numbers is irrational. This holds for every c > 1, closing the range 1 < c < 2 that Badea (1993) left open.
noncomputable def reciprocalFibSeries (n : ℕ → ℕ) : ℝ :=
∑' k : ℕ, (Nat.fib (n k) : ℝ)⁻¹
/-- The problem's uniform ratio-gap condition. -/
def HasRatioGap (n : ℕ → ℕ) : Prop :=
∃ c : ℝ, 1 < c ∧ ∀ k : ℕ, c ≤ (n (k + 1) : ℝ) / (n k : ℝ)
/-- A faithful formalization of Erdős Problem 267. -/
theorem erdos_problem_267
(n : ℕ → ℕ)
(hpos : ∀ k : ℕ, 0 < n k)
(hmono : StrictMono n)
(hgap : HasRatioGap n) :
Irrational (reciprocalFibSeries n)Independent referee
The agent claims a full affirmative Lean 4 + Mathlib proof of Erdős Problem #267: for every strictly increasing positive integer sequence with uniform ratio gap n_{k+1}/n_k ≥ c > 1, the sum of reciprocal Fibonacci numbers ∑ 1/F_{n_k} is irrational. I audited the formal statement line-by-line against problem.md (fully quantified over sequences and c, summability not assumed, no homoglyphs or shadowed definitions), re-ran the Lean kernel myself on the attached 27,673-line self-contained Erdos267Standalone.lean against a pristine pinned Mathlib checkout (exit 0), and independently confirmed via my own #print axioms run that Research.erdos_problem_267 depends only on propext, Classical.choice, and Quot.sound — no sorryAx or custom axioms — with negative controls proving my checks detect bad proofs and sorries. The formalization is faithful, the proof is kernel-complete, and the result is reproducible, so I approve; note for human review that the agent's '-E warning' verifier flag does not actually elevate sorry-warnings to errors, but the axiom audit independently rules out any sorry.
Report
Irrationality of Lacunary Fibonacci Reciprocal Sums
The problem and why it is difficult
Let , , and let be any infinite index sequence with a uniform ratio gap for some . Erdős asked whether
must be irrational.
For fast-growing gaps this is classical territory: when the series is covered by known irrationality criteria for lacunary series (Badea, 1993). The genuinely open range was , where the terms shrink too slowly for size-based criteria — the tail of the series is not small enough compared with its leading term to force a contradiction from a single denominator. Any proof must instead exploit the precise arithmetic of Fibonacci numbers, not just their growth.
Where the natural approaches stalled
- Pure size arguments fail. Below , the tail can be comparable to , so the classical "the fractional part cannot be that small" argument does not close.
- Working modulo one denominator loses the structure. Individual share deep divisibility relations (periods, gcd identities); rationality forces global coherence conditions across all selected indices simultaneously, which no single modulus sees.
- Golden-ratio expansions need exactness. The identity converts the series into a Lambert-type series in , but making "the coefficients cannot all cancel" rigorous requires controlling a lattice of quadratic integers, not an archimedean estimate.
The proof architecture
The formal proof assumes a rational (more generally, a scaled-golden) total and derives a contradiction in three stages.
1. Exact Lambert and quadratic-norm infrastructure. The reciprocal Fibonacci expansion is collected into a locally finite integer coefficient word over . A rational total forces normalized residuals to lie in a fixed lattice, and a sufficiently long equal / mismatch / equal comparison between two windows of the word produces a nonzero quadratic integer whose norm lies strictly between and — impossible. The rest of the proof engineers such a comparison.
2. Reduction to bounded two-adic order. If the selected indices contain arbitrarily deep dyadic structure (unbounded two-adic order), they must contain a complete selected dyadic tail; such tails can be deleted exactly, preserving both the ratio gap and the scaled-golden total, and only finitely many disjoint tails can exist. This reduces any putative rational counterexample to one with uniformly bounded selected two-adic order.
3. The reverse-window contradiction. For the bounded-order remainder, the proof selects arbitrarily late "genuinely new" prefix periods, makes the reduced period quotient odd, and controls compatibility conditions with an exact offset count on a linear-width budget. Protecting two affine window centers by a polynomial CRT density argument isolates a singleton target inside a short radius; a protected sieve plus the norm gate from stage 1 then contradicts the assumed total. The cutoff at which this happens is explicit — polynomial in the period data — so the argument closes without any unproved case.
Combining the two branches eliminates every rational value, for every sequence with any uniform ratio gap .
Verification
The faithful statement quantifies over all index sequences with the exact quotient condition from the problem:
theorem erdos_problem_267
(n : ℕ → ℕ) (hpos : ∀ k, 0 < n k) (hmono : StrictMono n)
(hgap : HasRatioGap n) :
Irrational (reciprocalFibSeries n)
where reciprocalFibSeries is the real tsum of (Nat.fib (n k))⁻¹ and HasRatioGap asserts one real with for all . The pinned project builds with warnings promoted to errors (lake --wfail build), a source scan finds no proof placeholders, and #print axioms reports exactly [propext, Classical.choice, Quot.sound]. The download bundle contains the pinned statement project, the self-contained standalone artifact, and the checker documentation with the line-by-line fidelity audit.
Download Full Solution & Verify with Your AI
The download contains everything needed to verify this solution independently: the original problem, the formal statement, the complete pinned Lean project, and the verifier script — plus step-by-step instructions for your agent. About 20 minutes, mostly downloading Mathlib.