Negative-index generalized Fibonacci factorization

Abstract

Mane’s negative-index factorization holds for all k >= 2 and 2 <= s <= k+1.

Definition 1.1 (Conjecture 6.4).

$$claim\iff (\forall k\in \mathrm{Nat},(2\le k)\Rightarrow (\forall s\in \mathrm{Nat},(2\le s)\Rightarrow ((s\le k+1)\Rightarrow (F\left(k,-(\mathrm{Nat}.\mathrm{cast}\left((s\cdot k:\mathrm{Nat})\right):\mathbb{Z})\right)=-X\cdot (X^k+1{)}^{s-2}\cdot (X^k+C\left((\mathrm{Nat}.\mathrm{cast}\left(s\right):\mathbb{Z})\right))))))$$

Formalization. D5/S3/Zeros/NegativeIndexFibonacciFactorization.claim (✓ std3).

Citation. S. R. Mane (2026). Identically vanishing k-generalized Fibonacci polynomials. URL: https://arxiv.org/abs/2507.11596v4.

Commentary.

Mane, p. 19, Section 6: “Conjecture 6.4. Numerical calculations and symbolic manipulations yield the following. For k ≥ 2 and n = −sk, where s ∈ [2, k + 1] (the pattern fails for s ≥ k + 2), 𝓕_{−sk,k}(x) = −x(x^k + 1)^{s−2}(x^k + s). (6.4) Hence 𝓕_{−sk,k}(x) has roots x^k = −s. The case s = 2 was derived in the enumerated list following eq. (3.2).” Here k and s are natural numbers; the negative index is the integer negative of their natural product. F is the polynomial in ℤ[X] defined in NegativeIndexTribonacciAttainmentRefutation, with F k n = a k (1−n).toNat and a k m = 𝓕_{1−m,k}. The initial values and backward recurrence are those of (1.2)–(1.3). X is the polynomial indeterminate, C is Polynomial.C, and the coefficient s is cast to ℤ. The exponent s−2 uses natural truncated subtraction; 2 ≤ s makes it the ordinary nonnegative difference. The formula encodes the identity throughout the stated range; the neighbouring root and failure sentences describe its consequences and boundary.

Theorem 1.2 (The factorization holds throughout the conjectured range).

$$claim$$

Proof. Machine-checked in Lean as D5/S3/Zeros/NegativeIndexFibonacciFactorization.result (✓ std3). ∎

Resolves. Problems/mane-2026-negative-index-fibonacci-factorization (proved) by D5/S3/Zeros/NegativeIndexFibonacciFactorization.result.

Source. Repository-derived.

Acknowledgement. S. R. Mane (2026). Identically vanishing k-generalized Fibonacci polynomials. URL: https://arxiv.org/abs/2507.11596v4.

Commentary.

For the backward sequence, the generating function is (1−X^k z^k)/(1−(1+X^k)z^k+Xz^{k+1}). A finite geometric sum computes its coefficient at z^{sk+1}, because the omitted tail starts in degree k(s+1). The r-th summand has degrees between kr and (k+1)r. With s ≤ k+1, only r=s contributes to degree sk+1, and only r=s−1 contributes to degree (s−1)k+1. Subtracting X^k times the second coefficient from the first gives the displayed factorization for every permitted k and s.

References

• Truth anchor: D5/S3/Zeros/NegativeIndexFibonacciFactorization.claim
• Truth anchor: D5/S3/Zeros/NegativeIndexFibonacciFactorization.result
• Dependency: D5/S3/Zeros/NegativeIndexTribonacciAttainmentRefutation