slug: mane-2026-negative-index-fibonacci-factorization
bibkey: mane2026vanishing
doi: null
url: https://arxiv.org/abs/2507.11596v4
triage: theorem
motivation_gids:
  - D5/S3/Zeros/NegativeIndexFibonacciFactorization.result

Mane’s negative-index factorization

Problem

S. R. Mane, Identically vanishing k-generalized Fibonacci polynomials, arXiv:2507.11596v4, p. 19, Section 6, Conjecture 6.4, states:

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 recurrence (1.2)–(1.3) is

$${\mathcal{F}}_{n,k}(x)=\sum _{j=1}^k{x}^{k-j}{\mathcal{F}}_{n-j,k}(x),{\mathcal{F}}_{1,k}=1,{\mathcal{F}}_{n,k}=0(-(k-2)\le n\le 0).$$

The frozen definitions a and F in D5/S3/Zeros/NegativeIndexTribonacciAttainmentRefutation.lean encode $a_k(m)={\mathcal{F}}_{1-m,k}$ and F k n = a k (1-n).toNat. The claim is an identity in $\mathbb{Z}[X]$ with natural $k,s$, $k\ge 2$ and $2\le s\le k+1$; the index is the integer $-sk$.

Motivation

Preregistration #14688 selects MANE-2, the full factorization identity. The settling declaration is D5/S3/Zeros/NegativeIndexFibonacciFactorization.result : claim. It provides the exact restricted attainment range related to Conjecture 6.6.

Gap

This is a tier-1 externally published named conjecture. The source and literature readings recorded in #14688 inspect arXiv versions, author papers, classical root papers and citation indexes; no settlement was found in that scope. Finite searches do not establish exhaustive absence of prior work.

Route

Put $t=X^k$, $A(z)={\sum }_{m\ge 0}{a}_k(m)z^m$ and $B(z)=z^k((1+t)-Xz)$. The recurrence gives

$$A(z)(1-B(z))=1-t{z}^k.$$

A finite geometric sum gives the coefficients below degree $kL$ without an analytic convergence assumption. At degree $sk+1$, with $s\le k+1$, only $B^s$ contributes; at degree $(s-1)k+1$, only $B^{s-1}$ contributes. Their coefficients are $-sX(1+t{)}^{s-1}$ and $-(s-1)X(1+t{)}^{s-2}$. Subtracting $t$ times the latter gives $-X(1+t{)}^{s-2}(t+s)$.

The public surface consists of claim and result; the private definitions and all sixteen private theorems have live consumers. Every theorem is bind-only after expansion through same-delivery helpers to frozen and pinned Mathlib declarations. Admission is open-problem-resolution (#14688; Proved), with no escape witness. Utility is none: this is a universal polynomial identity, with no fixed-parameter numerical certificate in the delivered Lean module. Information-escape registration is paused under CLAUDE.md §3.9.

Falsifier

A mismatch between the recurrence, index convention, quantifier range or polynomial factors and the source invalidates fidelity. A failing in-range instance refutes the claim. A prior published settlement invalidates the literature gap. The computations below are checks, not proof premises.

Evidence

The Lean module and its corresponding Blueprint Scribe establish the exact MANE-2 claim. The settling Scribe node has OpenProblemResolutionClaim(Proved) and cites Library/Zeros/mane2026vanishing.md. The axiom closure of every public declaration is contained in $\{\mathrm{propext},Classical.choice,Quot.sound\}$.

Triage

What the settlement shows

Proved in this module — the full permitted range. result establishes (6.4) for every $k\ge 2$ and $2\le s\le k+1$. Its private A_identity, A_trunc, G_coeff_diag and a_factor implement the coefficient argument.

Proved by the following paper argument — why the pattern stops at $s=k+1$. The $r$th summand $B^r$ has degree support in $[kr,(k+1)r]$. For $r<s$, its largest possible endpoint is $(k+1)(s-1)$, which is strictly below $sk+1$ exactly when $s\le k+1$. For $r>s$, its lower endpoint is at least $k(s+1)>sk+1$ for $k\ge 2$. At $s=k+2$, the $r=s-1=k+1$ upper endpoint equals $sk+1$; its leading coefficient is $(-X{)}^{k+1}$. The shifted numerator contribution at degree $(s-1)k+1$ still has only the diagonal summand, because $s-1=k+1$. Consequently an additional nonzero term $(-X{)}^{k+1}$ appears. This is a paper derivation of the sharp next boundary, not an additional Lean theorem. The kernel-checked window exclusion used within the permitted range is G_coeff_diag.

Proved by the following paper argument — roots and multiplicities. Over $\mathbb{C}$, the factors $x$, $x^k+1$ and $x^k+s$ have disjoint root sets for $s\ge 2$. Every root of either binomial is nonzero; its derivative $k{x}^{k-1}$ is nonzero since $k\ge 2$. Thus every root with $x^k=-s$ has multiplicity one, every root with $x^k=-1$ has multiplicity $s-2$ (and is absent at $s=2$), and zero has multiplicity one. Complex $k$th roots of $-s$ exist and have $|x{|}^k=s$; all other nonzero roots have $|x{|}^k=1$. Hence ${\zeta }_{-sk,k}=s$ throughout the proved range, including the restricted attainment range of Conjecture 6.6 for $k\ge 3$. These are algebraic consequences of result proved here on paper; no separate Lean multiplicity or supremum theorem is delivered.

Computed — $s=k+2$. For $k=2,\dots ,7$, exact symbolic recurrence calculations give six of six matches for

$${\mathcal{F}}_{-k(k+2),k}(x)=-x{Q}_k(x^k),Q_k(t)=(t+k+2)(1+t{)}^k+(-1{)}^{k}t.$$

The required orchestrator script python3 /tmp/op-mane/check64.py, exit $0$, gives 27 of 27 in-range matches and six of six failures of (6.4) at $s=k+2$, for $k=2,\dots ,7$. The supplemental self-contained script python3 /tmp/op-mane/stageb/boundary64.py, exit $0$, additionally checks the displayed $Q_k$ identity in six cases and the root-factor multiplicities in all 27 in-range cases. These computations establish only their finite scope; the uniform boundary argument is the paper proof above, with its Lean formalization open here.

Open — the source’s broader simplicity sentence. The source says “All such roots appear to be simple” for $j\in [2,k+1]$ with $x^k=-j$ at the additional negative indices discussed on p. 19. The in-range $j=s$ roots are simple by the paper argument above and have exact symbolic multiplicity one in all 27 tested cases. Simplicity at other indices is open here. The $x^k=-1$ roots are outside $j>1$ and can have multiplicity $s-2$; they do not refute that sentence.

Open — neighbouring questions and dependence. Conjecture 6.5, the universal upper-bound clause and asymptotic branch claims of Conjecture 6.6 are not settled here. The companion settlement #14674 refutes unrestricted attainment. This identity supplies attainment only in the stated restricted range. Any source use of (6.4) within that range can use the proved identity; extrapolations beyond it gain no premise from this result. In particular, the p. 20 inference of unbounded growth at fixed $k$ is not justified by (6.4): its parameter $s$ is bounded by $k+1$, and the $s=1$ case cited there is outside the conjecture’s $s\ge 2$ range.

Reproducible symbolic checks

The scripts require Python 3 and SymPy. Save the following exact UTF-8 bytes at the named command paths, or run equivalent local paths. The computations are exact polynomial expansions and gcd/division checks, with no numerical tolerance.

Command: python3 /tmp/op-mane/check64.py. Exit code: $0$. SHA-256: e7ee83f451689b57f35da62752803542cf7acab552b81e0c6980ffe2eeb6e214.

import sympy as sp
x=sp.symbols('x')
def F(k, nmin):
    vals={1:sp.Integer(1)}
    for n in range(-(k-2),1): vals[n]=sp.Integer(0)
    n=1
    while n-k>=nmin:
        vals[n-k]=sp.expand(vals[n]-sum(x**(k-j)*vals[n-j] for j in range(1,k)))
        n-=1
    return vals
ok=0; bad=0
for k in range(2,8):
    v=F(k,-k*(k+2))
    for s in range(2,k+2):
        lhs=v[-s*k]; rhs=sp.expand(-x*(x**k+1)**(s-2)*(x**k+s))
        if sp.expand(lhs-rhs)==0: ok+=1
        else: bad+=1; print("FAIL",k,s)
    # s=k+2 fails?
    s=k+2; print(k, "s=k+2 matches pattern:", sp.expand(v[-s*k]+x*(x**k+1)**(s-2)*(x**k+s))==0)
print("pattern holds", ok, "fails", bad)

Command: python3 /tmp/op-mane/stageb/boundary64.py. Exit code: $0$. SHA-256: 535eb1b77cb46fcff894c432957caac480507c9d1e764d9e75bfff399f4b4e63.

import sympy as sp
x=sp.symbols('x')
def F(k, nmin):
    vals={1:sp.Integer(1)}
    for n in range(-(k-2),1): vals[n]=sp.Integer(0)
    n=1
    while n-k>=nmin:
        vals[n-k]=sp.expand(vals[n]-sum(x**(k-j)*vals[n-j] for j in range(1,k)))
        n-=1
    return vals
ok=0; bad=0
for k in range(2,8):
    v=F(k,-k*(k+2))
    for s in range(2,k+2):
        lhs=v[-s*k]; rhs=sp.expand(-x*(x**k+1)**(s-2)*(x**k+s))
        if sp.expand(lhs-rhs)==0: ok+=1
        else: bad+=1; print("FAIL",k,s)
    # s=k+2 fails?
    s=k+2; print(k, "s=k+2 matches pattern:", sp.expand(v[-s*k]+x*(x**k+1)**(s-2)*(x**k+s))==0)
print("pattern holds", ok, "fails", bad)

q_matches=0; root_matches=0
def multiplicity(p,q):
 count=0
 while True:
  quo,rem=sp.div(p,q,x)
  if rem!=0:return count
  p=quo;count+=1
for k in range(2,8):
 vals=F(k,-k*(k+2)); s=k+2
 q=(x**k+k+2)*(1+x**k)**k+(-1)**k*x**k
 assert sp.expand(vals[-k*s]+x*q)==0
 q_matches+=1
 for s in range(2,k+2):
  f=vals[-k*s]
  assert multiplicity(f,x)==1
  assert multiplicity(f,x**k+s)==1
  assert multiplicity(f,x**k+1)==s-2
  assert sp.gcd(x**k+s,sp.diff(x**k+s,x))==1
  assert sp.gcd(x**k+1,sp.diff(x**k+1,x))==1
  root_matches+=1
print('Boundary Q_k identity matches:',q_matches,'/ 6')
print('Root-factor multiplicities match:',root_matches,'/ 27')

ASSUMED-UNVERIFIED

Exhaustive absence of a prior settlement is not established by the finite literature search. The general root-multiplicity, amplitude and sharp-boundary consequences have paper proofs here but no separate Lean declarations. The broader simplicity assertion and neighbouring conjectures identified above remain open here.