slug: goel-2026-pisano-rank-ratio-oq4 bibkey: goel2026sophiegermain doi: null url: https://arxiv.org/abs/2604.17847v3 triage: theorem motivation_gids:

D5/S3/Arith/GoelPisanoRatioRefutation.result

Goel’s OQ4: Fibonacci period-to-rank ratios omit five

Problem

Aradhya Goel, Sophie Germain Primes and the Totient of Fibonacci Numbers, arXiv:2604.17847v3, Section 10, OQ4, p. 10, asks:

OQ4. Values of $π (q) / z (2 q + 1)$ . Is ${π (q) / z (2 q + 1) : q > 5 with z (2 q + 1) ∣ π (q)} = {odd integers}$ ?

The versioned TeX source reads verbatim:

\item[OQ4.] \textbf{Values of $\pi(q)/z(2q+1)$.} Is $\{\pi(q)/z(2q+1) : q > 5 \text{ with } z(2q+1) \mid \pi(q)\} = \{\text{odd integers}\}$?

The Introduction defines $z (p)$ at primes as the least positive index $k$ with $p ∣ F_{k}$ , and $π (n)$ as the period of the Fibonacci residues modulo $n$ . The standing domain in Sections 5–7 is Sophie Germain primes: $q$ and $2 q + 1$ are both prime. The assertion here uses $q > 5$ and $z (2 q + 1) ∣ π (q)$ . The right-hand side is the positive odd integers; reading it as all odd integers also fails because the ratios are positive.

Motivation

The frozen declaration D5/S3/Arith/GoelPisanoRatioRefutation.result has type ¬ claim, where claim is the displayed equality of natural-number sets with the complete standing prime domain. The resolution kind is Refuted.

Gap

Preregistration #13841 identifies this as a Tier 1 numbered external question. Its literature screen checked the versioned source, MathDB queries “Pisano period rank apparition”, “Sophie Germain totient Fibonacci” and “Goel Fibonacci”, and repository owners. No resolution of OQ4 was found in that searched scope. This is a bounded negative search result, not a claim of exhaustive worldwide priority.

Route

Put $p = 2 q + 1$ and $ε (r) = (5/ r)$ , the Legendre symbol at a prime. The frozen rank and apparition theorems give $z (p) ∣ p - ε (p)$ and the usual return-pair bounds on $π (q)$ . If $ε (p) = 1$ , then $z (p)$ divides both $2 q$ and $q^{2} - 1$ , hence $2$ . If $ε (q) = 1$ after $ε (p) = - 1$ , then $z (p)$ divides both $q - 1$ and $2 (q + 1)$ , hence $4$ . Both contradict $z (p) \geq 5$ for $p > 11$ . Thus $ε (q) = - 1$ , so $q \equiv 2$ or $3 (mod 5)$ and $π (q) ∣ 2 (q + 1)$ . Consequently $5 ∤ π (q)$ and $5 ∤ π (q) / z (p)$ . The odd number $5$ cannot belong to the value set.

Falsifier

A Sophie Germain prime $q > 5$ satisfying $z (2 q + 1) ∣ π (q)$ and $5 ∣ π (q) / z (2 q + 1)$ would contradict the exclusion on the proof path. A finite search with no such value does not establish this universal exclusion; the kernel-checked result establishes the negation of OQ4.

Evidence

The formal definitions use Nat.fib, natural-number divisibility, and the least-positive-period infimum $\operatorname{sInf}{k\in\mathbb N:0<k\land\forall m\in\mathbb N, F_{m+k}\bmod n=F_m\bmod n}$. The rank is the existing FibonacciAtomic.TimeSampling.zeroRank with its literal least-positive-zero expression. result has no hypotheses. Its axiom closure is contained in {propext, Classical.choice, Quot.sound}. The source’s Lemma 6.3 and Theorem 7.1 also imply the exclusion on paper; the formal proof uses existing frozen Fibonacci declarations and consumed private bridges for the literal period definition.

Triage

Tier 1; admission_basis: open-problem-resolution with preregistration #13841; proof_shape: bind-only; escape_witness: none. The only public theorem is result. All private auxiliaries are consumed on its proof path. The symbolic argument has utility: none. Information-escape registration is paused under CLAUDE.md §3.9.

What the settlement shows

Proved in this module: OQ4 is false (D5/S3/Arith/GoelPisanoRatioRefutation.result). The consumed private steps epsilon_q_negative, five_not_dvd_pi and ratio_avoids_five establish the failure mechanism uniformly in every eligible prime: the negative quadratic character forces a period bound coprime to $5$ . In particular, no odd multiple of $5$ is attained. These private steps are proof components, not additional public settlement declarations.
Open: whether the value set is exactly the positive odd integers coprime to $5$ . Computed: for $7 \leq q < 200000$ , among 2055 Sophie Germain primes, 489 satisfy the divisibility condition; all 489 ratios are odd and coprime to $5$ , and all corresponding primes satisfy $q \equiv 8 (mod 15)$ . The observed values and multiplicities are [[1, 317], [3, 128], [7, 3], [9, 19], [11, 4], [13, 1], [17, 2], [21, 3], [27, 1], [33, 2], [39, 1], [43, 1], [63, 1], [69, 2], [81, 1], [97, 1], [357, 1], [1377, 1]]. First occurrences in the same window (the smallest eligible $q$ for each ratio) are:

ratio first $q$

1 83

3 23

7 16673

9 20393

11 75503

13 95393

17 117503

21 40823

27 85103

33 1583

39 161303

43 16253

63 97523

69 65963

81 59453

97 91373

357 147083

1377 126683

In that window $19$ and $23$ do not occur. No claim of global nonattainment follows from this bounded absence. For $q = 23$ , $π (23) = 48$ and $z (47) = 16$ , giving ratio $3$ .
Computed: extending the least-positive-zero definition to composite arguments changes the question. At $q = 241$ , $2 q + 1 = 483$ is composite, $π (q) = 240$ , $z (483) = 24$ , and the ratio is the even number $10$ . At $q = 30661$ , $2 q + 1 = 61323$ is composite, $π (q) = 10220$ , $z (61323) = 2044$ , and the ratio is $5$ . Each period and rank is also checked by direct recurrence iteration. These are examples for the extended domain; they do not contradict the prime-domain exclusion. Open: classification of the extended-domain values.
Computed: the quotient lower bound in the source’s Theorem 6.2 is odd for every eligible prime in the stated window. The value of $∣ S (q) ∣$ itself is not computed here. Proved on this module’s proof path: the lower bound $π (q) / z (2 q + 1)$ is also coprime to $5$ . This does not imply that $∣ S (q) ∣$ is coprime to $5$ : being a lower bound supplies no divisibility relation. Open: a Lean formalization of the full $S (q)$ counting statement and any further restriction on $∣ S (q) ∣$ . The source’s Theorems 6.2 and 7.1 require no equality with all odd integers, so the refutation supplies no refutation of those theorems. OQ4 is the closing question being settled.

ratio	first $q$
1	83
3	23
7	16673
9	20393
11	75503
13	95393
17	117503
21	40823
27	85103
33	1583
39	161303
43	16253
63	97523
69	65963
81	59453
97	91373
357	147083
1377	126683

Reproducible computations

All items marked computed above use the following self-contained script, with Python 3 and SymPy 1.14.0. Save the block as /tmp/goel-oq4-triage.py. Command: python3 /tmp/goel-oq4-triage.py. Exit code: 0. SHA256 of the exact UTF-8 script, including its final newline: ad9c92e63ba8647cbdb7efc9fef0716ad4bf2af35ddba6d291a77afd48dc6dd5. The prime-period divisor search uses the classical bound $π (p) ∣ p^{2} - 1$ for $p \neq = 5$ , and $π (5) = 20$ . The composite rank search uses the least common multiple of the prime-power period bounds. Direct iteration independently checks the two extended-domain examples and the $q = 23$ example.

import json
from collections import Counter
from math import lcm
from sympy import divisors, factorint, isprime, primerange

def fibpair(n, modulus):
    if n == 0:
        return 0, 1 % modulus
    a, b = fibpair(n // 2, modulus)
    c = a * (2 * b - a) % modulus
    d = (a * a + b * b) % modulus
    return (d, (c + d) % modulus) if n % 2 else (c, d)

def prime_period(p):
    bound = 20 if p == 5 else p * p - 1
    return next(k for k in divisors(bound) if fibpair(k, p) == (0, 1 % p))

def rank(n):
    bound = 1
    for p, e in factorint(n).items():
        bound = lcm(bound, prime_period(p) * p ** (e - 1))
    return next(k for k in divisors(bound) if fibpair(k, n)[0] == 0)

def direct(n):
    a, b, k, first_zero = 0, 1, 0, None
    while True:
        a, b = b, (a + b) % n
        k += 1
        if a == 0 and first_zero is None:
            first_zero = k
        if (a, b) == (0, 1):
            return k, first_zero

bound = 200000
eligible, total = [], 0
for q in primerange(7, bound):
    if not isprime(2 * q + 1):
        continue
    total += 1
    period, z = prime_period(q), rank(2 * q + 1)
    if period % z == 0:
        eligible.append((int(q), int(period), int(z), int(period // z)))
assert len(eligible) == 489
assert all(q % 15 == 8 and r % 2 == 1 and r % 5 != 0 for q, _, _, r in eligible)
counts = Counter(r for _, _, _, r in eligible)
first_q = {}
for q, _, _, r in eligible:
    first_q.setdefault(r, q)
composite_examples = []
for q in [241, 30661]:
    n = 2 * q + 1
    period, z = prime_period(q), rank(n)
    assert isprime(q) and not isprime(n)
    assert direct(q)[0] == period and direct(n)[1] == z
    assert period % z == 0
    composite_examples.append(dict(q=q, n=n, period=int(period), rank=int(z), ratio=int(period // z)))
assert composite_examples[0]['ratio'] == 10
assert composite_examples[1]['ratio'] == 5
assert direct(23)[0] == 48 and direct(47)[1] == 16
print(json.dumps(dict(bound_exclusive=bound, sophie_germain_count=total,
    eligible_count=len(eligible), ratio_counts=sorted(counts.items()), first_q=first_q,
    residues_mod15=sorted(set(q % 15 for q, _, _, _ in eligible)),
    missing_odd_coprime5_through199=[r for r in range(1, 200, 2) if r % 5 and r not in counts],
    q23=dict(period=48, rank47=16, ratio=3), composite_examples=composite_examples,
    sympy_version=__import__('sympy').__version__), sort_keys=True))

ASSUMED-UNVERIFIED

The literature screen is bounded to the sources and queries recorded above. The computation is finite experimental evidence, not a Lean certificate for its table or for global attainment. The corrected and composite-domain classification questions remain open. No new axiom or assumption is used by the settling Lean declaration.

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