Wu’s Adjacent-Index Fibonacci Divisibility Conjecture

Abstract

An index whose Fibonacci number is divisible by itself and both neighbours ends in zero, two or eight.

Definition 1.1 (The A221018 assertion).

$$claim=(\forall k\in \mathbb{N},(2\le k)\Rightarrow ((k-1\mid \mathrm{Nat}.\mathrm{fib}\left(k\right))\Rightarrow ((k\mid \mathrm{Nat}.\mathrm{fib}\left(k\right))\Rightarrow ((k+1\mid \mathrm{Nat}.\mathrm{fib}\left(k\right))\Rightarrow ((k\mathrm{mod}10=0)\vee ((k\mathrm{mod}10=2)\vee (k\mathrm{mod}10=8)))))))$$

Formalization. D5/S3/Arith/Congruence/WuAdjacentFibonacciResidues.claim (✓ std3).

Citation. Chai Wah Wu; Alex Ratushnyak (2026). OEIS A221018, first FORMULA conjecture on adjacent-index Fibonacci divisibility. URL: https://oeis.org/A221018/internal.

Commentary.

OEIS A221018, revision 35, NAME: “Number k such that Fibonacci(k) is divisible by k, k + 1 and k - 1.” The first FORMULA line reads: “a(n) % 10 == 0, 2, or 8 (conjectured). - Chai Wah Wu, Aug 06 2026”. The web entry has no page number. The generator starts at two, so k ranges over natural numbers at least two. Nat.fib has F(0)=0, F(1)=1 and F(n+2)=F(n)+F(n+1). The three divisibilities are separate implication hypotheses. Subtraction is natural subtraction, which agrees with integer subtraction here; % denotes natural-number remainder. No converse is asserted.

Theorem 1.2 (The final-digit theorem).

$$\forall k\in \mathbb{N},(2\le k)\Rightarrow ((k-1\mid \mathrm{Nat}.\mathrm{fib}\left(k\right))\Rightarrow ((k\mid \mathrm{Nat}.\mathrm{fib}\left(k\right))\Rightarrow ((k+1\mid \mathrm{Nat}.\mathrm{fib}\left(k\right))\Rightarrow ((k\mathrm{mod}10=0)\vee ((k\mathrm{mod}10=2)\vee (k\mathrm{mod}10=8))))))$$

Proof. Machine-checked in Lean as D5/S3/Arith/Congruence/WuAdjacentFibonacciResidues.result (✓ std3). ∎

Resolves. Problems/wu221018-fibdiv (proved) by D5/S3/Arith/Congruence/WuAdjacentFibonacciResidues.result.

Source. Repository-derived.

Acknowledgement. Chai Wah Wu; Alex Ratushnyak (2026). OEIS A221018, first FORMULA conjecture on adjacent-index Fibonacci divisibility. URL: https://oeis.org/A221018/internal.

Commentary.

One of k and k+1 is even, so their divisibility into Nat.fib(k) makes that Fibonacci number even. The equivalence between Fibonacci divisibility and index divisibility, applied at indices three and four, forces 3 and 4 to divide k, hence 12 divides k. If k has remainder one or four modulo five, one neighbour is divisible by five. The same equivalence at index five then forces five to divide k, a contradiction. Evenness and these two exclusions leave final digits zero, two and eight.

References

• Truth anchor: D5/S3/Arith/Congruence/WuAdjacentFibonacciResidues.claim
• Truth anchor: D5/S3/Arith/Congruence/WuAdjacentFibonacciResidues.result
• Dependency: D5/S1/Recurrence/GoldenFibDivisibility