← Star Fleet Math

Erdős Problem #267

Erdős Problem

www.erdosproblems.com/267

The problem

Let F1=F2=1F_1=F_2=1 and Fn+1=Fn+Fn−1F_{n+1}=F_n+F_{n-1} be the Fibonacci sequence. Let n1<n2<⋯n_1<n_2<\cdots be an infinite sequence with nk+1/nk≥c>1n_{k+1}/n_k \geq c>1. Must∑k1Fnk\sum_k\frac{1}{F_{n_k}}be 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 F1=F2=1F_1=F_2=1, Fn+1=Fn+Fn−1F_{n+1}=F_n+F_{n-1}, and let n1<n2<⋯n_1<n_2<\cdots be any infinite index sequence with a uniform ratio gap nk+1/nk≥cn_{k+1}/n_k\ge c for some c>1c>1. Erdős asked whether

∑k1Fnk \sum_k \frac{1}{F_{n_k}}

must be irrational.

For fast-growing gaps this is classical territory: when c≥2c\ge 2 the series is covered by known irrationality criteria for lacunary series (Badea, 1993). The genuinely open range was 1<c<21<c<2, 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 c=2c=2, the tail ∑j>k1/Fnj\sum_{j>k}1/F_{n_j} can be comparable to 1/Fnk1/F_{n_k}, so the classical "the fractional part cannot be that small" argument does not close.
  • Working modulo one denominator loses the structure. Individual FnF_n 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 1/Fn=5∑j(−1)njφ−(2j+1)n1/F_n=\sqrt5\sum_j(-1)^{nj}\varphi^{-(2j+1)n} converts the series into a Lambert-type series in φ−1\varphi^{-1}, 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 Z[φ]\mathbb Z[\varphi]. 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 00 and 11 — 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 c>1c>1.

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 c>1c>1 with c≤nk+1/nkc\le n_{k+1}/n_k for all kk. 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.

↓ Download full solution raw