Problem packetResearch packetR76
The candidate uses the classical Baum-Sweet convention
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Recorded status: established
Recorded scope: the full Baum-Sweet coefficient sequence
Complete recorded scope and conditions
{
"kind": "universal",
"statement": "the full Baum-Sweet coefficient sequence"
}Originating problem: Nonvanishing of Baum-Sweet Hankel determinants
Authored record and scope
- Authored title
- The candidate uses the classical Baum-Sweet convention
- Record type
- claim
- Stored status
- established
- Evidence grade
- sourced
- Recorded scope data
- { "kind": "universal", "statement": "the full Baum-Sweet coefficient sequence" }
2Authored explanation
The recurrence is \[ b_0=1,\qquad b_{2m+1}=b_m,\qquad b_{4m}=b_m,\qquad b_{4m+2}=0. \] It produces \(1,1,0,1,1,0,0,1,0,1,0,0,1,0,0,1,\ldots\) and is equivalent to the candidate's binary-block definition. For \(B(z)=\sum b_nz^n\), splitting indices gives \(B(z)=B(z^4)+zB(z^2)\). In characteristic two, this reduces to \(B^3+zB+1=0\). Baum and Sweet's original work and the later continued-fraction papers concern this algebraic series.
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3Evidence
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Verification source: annals.math.princeton.edu ↗, Leonard E. Baum and Melvin M. Sweet, Continued Fractions of Algebraic Power Series in Characteristic 2, Annals of Mathematics 103 (1976), pages 593-610
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"slug": "bsh-claim-classical-cubic",
"type": "claim",
"title": "The candidate uses the classical Baum-Sweet convention",
"summary": "Its generating series satisfies \\(B(z)=B(z^4)+zB(z^2)\\), and over \\(\\mathbb F_2\\) it is the Baum-Sweet cubic.",
"relevance": "For Nonvanishing of Baum-Sweet Hankel determinants, record bsh-claim-classical-cubic (“The candidate uses the classical Baum-Sweet convention”) records a bound, answer, status fact, or structural consequence. The record states: Its generating series satisfies \\(B(z)=B(z^4)+zB(z^2)\\), and over \\(\\mathbb F_2\\) it is the Baum-Sweet cubic.",
"relevance_source": "recorded",
"body": "The recurrence is\n\\[\nb_0=1,\\qquad b_{2m+1}=b_m,\\qquad b_{4m}=b_m,\\qquad b_{4m+2}=0.\n\\]\nIt produces \\(1,1,0,1,1,0,0,1,0,1,0,0,1,0,0,1,\\ldots\\) and is equivalent to the candidate's binary-block definition. For \\(B(z)=\\sum b_nz^n\\), splitting indices gives \\(B(z)=B(z^4)+zB(z^2)\\). In characteristic two, this reduces to \\(B^3+zB+1=0\\). Baum and Sweet's original work and the later continued-fraction papers concern this algebraic series.",
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"kind": "universal",
"statement": "the full Baum-Sweet coefficient sequence"
},
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"kind": "claim",
"citation": {
"url": "https://annals.math.princeton.edu/1976/103-3/p12",
"locator": "Leonard E. Baum and Melvin M. Sweet, Continued Fractions of Algebraic Power Series in Characteristic 2, Annals of Mathematics 103 (1976), pages 593-610"
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"source": {
"url": "https://annals.math.princeton.edu/1976/103-3/p12",
"locator": "Leonard E. Baum and Melvin M. Sweet, Continued Fractions of Algebraic Power Series in Characteristic 2, Annals of Mathematics 103 (1976), pages 593-610"
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"relations": [
{
"slug": "R77",
"title": "Non-apwenian does not mean that a determinant vanishes",
"object_type": "claim",
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{
"slug": "baum-sweet-hankel-nonvanishing",
"title": "baum sweet hankel nonvanishing",
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}6Provenance
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