Problem packetResearch packetR688
Two residue classes are settled 2-adically
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Recorded status: supported
Recorded scope: all positive orders congruent to 0 or 1 modulo 3
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{
"kind": "family",
"statement": "all positive orders congruent to 0 or 1 modulo 3",
"family": "n congruent to 0 or 1 modulo 3"
}Originating problem: Nonvanishing of Rudin-Shapiro Hankel determinants
Authored record and scope
- Authored title
- Two residue classes are settled 2-adically
- Record type
- claim
- Stored status
- supported
- Evidence grade
- sourced
- Recorded scope data
- { "kind": "family", "statement": "all positive orders congruent to 0 or 1 modulo 3", "family": "n congruent to 0 or 1 modulo 3" }
2Authored explanation
Put \(u_n=(1-r_n)/2\) and \(v_k=u_k+u_{k+2}\pmod2\). Elementary row and column operations give \[ \frac{H_n(r)}{(-2)^{n-1}}\equiv H_{n-1}(v)\pmod2. \] The generating series \(V(x)=\sum v_kx^k\) obeys \[ x^2(1+x)V^2+(1+x)^2V+x=0 \] over \(\mathbb F_2\). Its periodic Hankel continued fraction has valuation parameters \((1,0)^*\). Han's Theorem 2.1 then says that \(H_m(v)\) is nonzero exactly when \(m\not\equiv1\pmod3\). Thus the displayed quotient is odd for \(n\equiv0,1\pmod3\).
This reduction is an editorial derivation from Han's theorem and should receive independent proof review. It reduces the original question to orders \(n=3m+2\), beginning with \(n=11\).
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3Evidence
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Verification source: irma.math.unistra.fr ↗, Guo-Niu Han, Hankel continued fraction and its applications, Theorem 2.1 and Algorithm 3.3; reduction derived for this entry
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"slug": "rsh-claim-two-residue-classes",
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"title": "Two residue classes are settled 2-adically",
"summary": "The published 2-adic formula proves \\(H_n\\ne0\\) for \\(n\\equiv0,1\\pmod3\\), while the remaining universal case \\(H_{3m+2}\\ne0\\) for every \\(m\\ge3\\) remains open despite reported modular certificates through order 5,000.",
"relevance": "For Nonvanishing of Rudin-Shapiro Hankel determinants, record rsh-claim-two-residue-classes (“Two residue classes are settled 2-adically”) records a bound, answer, status fact, or structural consequence. The record states: The published 2-adic formula proves \\(H_n\\ne0\\) for \\(n\\equiv0,1\\pmod3\\), while the remaining universal case \\(H_{3m+2}\\ne0\\) for every \\(m\\ge3\\) remains open despite reported modular certificates through order 5,000.",
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"body": "Put \\(u_n=(1-r_n)/2\\) and \\(v_k=u_k+u_{k+2}\\pmod2\\). Elementary row and column operations give\n\\[\n\\frac{H_n(r)}{(-2)^{n-1}}\\equiv H_{n-1}(v)\\pmod2.\n\\]\nThe generating series \\(V(x)=\\sum v_kx^k\\) obeys\n\\[\nx^2(1+x)V^2+(1+x)^2V+x=0\n\\]\nover \\(\\mathbb F_2\\). Its periodic Hankel continued fraction has valuation parameters \\((1,0)^*\\). Han's Theorem 2.1 then says that \\(H_m(v)\\) is nonzero exactly when \\(m\\not\\equiv1\\pmod3\\). Thus the displayed quotient is odd for \\(n\\equiv0,1\\pmod3\\).\n\nThis reduction is an editorial derivation from Han's theorem and should receive independent proof review. It reduces the original question to orders \\(n=3m+2\\), beginning with \\(n=11\\).",
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"url": "https://irma.math.unistra.fr/~guoniu/papers/p94hfrac.pdf",
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{
"slug": "R687",
"title": "Published binary formulas do not settle the signed determinant",
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{
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"title": "Exact signed determinant sweep through order 110",
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{
"slug": "R686",
"title": "Settle the remaining orders congruent to 2 modulo 3",
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{
"slug": "rudin-shapiro-hankel-nonvanishing",
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}6Provenance
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