slug: fried-fibonacci-shift-non-fibonacci bibkey: fried2025proofs doi: null url: https://cs.uwaterloo.ca/journals/JIS/VOL28/Fried/fried15.pdf triage: theorem motivation_gids:

• D5/S3/Arith/FriedFibonacciShiftNonFibonacci.result

Fried’s Fibonacci Shift Conjecture

Problem

Fried, Journal of Integer Sequences 28 (2025), Article 25.4.3, Section 6, states:

Nevertheless, we were not able to show that the two sets { nF(n/2 + 3) − (n − 1)F(n/2 + 2) : n ≥ 1 is even }, { F((n+1)/2 + 2) : n ≥ 1 is odd } are disjoint, or, equivalently, that for every n ∈ N, the number F(n + 2) + 2nF(n + 1) is not a Fibonacci number. We conjecture that this is so.

Here F(1) = F(2) = 1. Both sets are indexed by n ≥ 1, so the N of the equivalent form starts at one; n = 0 gives F(2) = 1, which is a Fibonacci number, and is outside the range.

Motivation

The frozen theorem D5/S3/Arith/FriedFibonacciShiftNonFibonacci.result settles the conjecture affirmatively in the equivalent form the source itself supplies: for every n at least one and every index m, F(m) differs from F(n + 2) + 2nF(n + 1).

Gap

The source reaches the conjecture from the closed-form description of A248982 and offers no line of attack, writing only that a proof “likely proceeds along similar lines as the previous theorem” — the previous theorem being a rounding-formula induction with no Fibonacci content. The obstacle is that F(n + 2) + 2nF(n + 1) grows like 2n times a Fibonacci number, so it is not close to any single Fibonacci number in ratio, and the usual identities for F(a) ± F(b) do not apply. The recorded proof works instead with the exact position of the number between two consecutive Fibonacci numbers of shifted index.

Route

Write X n = F(n + 2) + 2nF(n + 1). The recurrence gives X n = F(n) + (2n + 1)F(n + 1).

Suppose F(m) = X n for some n ≥ 1.

Small n. For n ≤ 7 the values of X n are 4, 11, 23, 48, 93, 177, 328, all below F(14) = 377. Since F is monotone, m < 14, leaving finitely many pairs (n, m), none of which is a solution.

Large n. For n ≥ 8 first note X n ≥ 3F(n + 1) > F(n + 1), so m ≥ n + 2; write m = j + n + 1 with j ≥ 1. The addition formula F(j + n + 1) = F(j)F(n) + F(j + 1)F(n + 1) turns F(m) = X n into

F(j)F(n) + F(j + 1)F(n + 1) = F(n) + (2n + 1)F(n + 1).

Since F(j) ≥ 1, the left side is at least F(n) + F(j + 1)F(n + 1), so F(j + 1) ≤ 2n + 1 and the equation rearranges, with subtraction staying inside the natural numbers, to

(F(j) − 1)F(n) = (2n + 1 − F(j + 1))F(n + 1).

If F(j) = 1 then j ≤ 2, so F(j + 1) ≤ 2 while the right factor 2n + 1 is at least seventeen; comparing the two sides of the original equation after cancelling F(n) gives F(j + 1)F(n + 1) = (2n + 1)F(n + 1) with F(j + 1) ≤ 2 < 3 ≤ 2n + 1, which is impossible because F(n + 1) > 0.

Otherwise F(j) ≥ 2, so F(j) − 1 > 0. The displayed identity shows that F(n + 1) divides (F(j) − 1)F(n), and F(n) and F(n + 1) are coprime, so F(n + 1) divides F(j) − 1 and hence F(n + 1) ≤ F(j) − 1. The left side of the identity is then at least F(n + 1)F(n), while its right side is at most 2nF(n + 1) because F(j + 1) ≥ 1. Cancelling the positive factor F(n + 1) gives F(n) ≤ 2n, which contradicts 2n < F(n) for n ≥ 8. That last bound is an induction: it holds at n = 8 since F(8) = 21 > 16, and the step adds F(n − 1) ≥ F(7) = 13 > 2 to F(n).

Falsifier

A single n ≥ 1 with F(n + 2) + 2nF(n + 1) equal to some F(m) refutes the conjecture. Inside the recorded proof, the statement fails if F(n) and F(n + 1) share a common factor for some n, if 2n ≥ F(n) for some n ≥ 8, or if the addition formula F(j + n + 1) = F(j)F(n) + F(j + 1)F(n + 1) fails for some pair.

Evidence

Every step of the argument was checked numerically for n = 1 through n = 2999 before the proof was recorded: the rewriting X n = F(n) + (2n + 1)F(n + 1), the bracketing of X n between consecutive Fibonacci numbers, the addition formula at the relevant indices, the divisibility consequence, and the bound 2n < F(n). No violation occurred. Independently, X n was tested against the Fibonacci numbers for n = 0 through n = 4000; the only value that is a Fibonacci number is X 0 = 1, which is outside the range of the conjecture and is exactly the reason the source indexes its two sets from one.

The values of X n for n = 1, …, 7 are 4, 11, 23, 48, 93, 177, 328; the Fibonacci numbers in that range are 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377.

Triage

theorem; Tier 1 named external open question, preregistered in issue 9382 before the proof was recorded. The admission basis is escape-witness with proof_shape: content; the declared escape content is the bound 2n < F(n) for n ≥ 8 obtained by induction, the rearranged identity (F(j) − 1)F(n) = (2n + 1 − F(j + 1))F(n + 1), and the divisibility consequence F(n + 1) ≤ F(j) − 1. The computational use is none: the delivered statement is a universally quantified theorem with no bounded enumeration, checker, numeric reduction or certified instance among its declarations.

ASSUMED-UNVERIFIED

The literature screen is bounded: the journal article and its abstract page, the OEIS entries A248982 and A007502, and the withdrawn preprint arXiv:2509.26138 were opened. That preprint, whose Theorem 23 and Proposition 24 addressed this conjecture, was withdrawn by its author on 2025-11-06 with the comment “This paper has been withdrawn by Duc Hieu Le”; the withdrawal was confirmed directly on the arXiv abstract page. The Journal of Integer Sequences assigns no DOI and citation-index result pages were not reachable, so no worldwide priority claim is made.

The closed-form description of A248982 for n ≥ 10, which the source states alongside the conjecture, is not settled here. Disjointness of the two displayed sets is one ingredient of it; the remaining content is the identification of (a_n) with those values.