[#R76] The candidate uses the classical Baum-Sweet convention
claim. Its generating series satisfies \(B(z)=B(z^4)+zB(z^2)\), and over \(\mathbb F_2\) it is the Baum-Sweet cubic.
1Summary
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.
Supported evidence. Recorded scope: the full Baum-Sweet coefficient sequence.
2Evidence
A verification source is cited. This record has no executable replay attached.
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
3How it connects
Informs
- claim
Recorded for
- problem
4Agent packet
A compact handoff with the evidence boundary, replay manifest, and relation pointers.
View structured packet
{
"schema": "theoremdb-agent-record-v1",
"ref": "R76",
"content_hash": null,
"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.",
"status": "established",
"evidence_grade": "sourced",
"scope": {
"kind": "universal",
"statement": "the full Baum-Sweet coefficient sequence"
},
"reproduction": {
"schema": "theoremdb-reproduction-v1",
"readiness": "source_only",
"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"
},
"missing": [
"source",
"command",
"runtime",
"expected_output"
]
},
"formal_statement": null,
"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"
},
"relations": [
{
"slug": "R77",
"title": "Non-apwenian does not mean that a determinant vanishes",
"object_type": "claim",
"relation": "informs",
"direction": "outgoing"
},
{
"slug": "baum-sweet-hankel-nonvanishing",
"title": "baum sweet hankel nonvanishing",
"object_type": "problem",
"relation": "recorded_for",
"direction": "outgoing"
}
]
}5Provenance
View source, identifiers, and projection details
- Project
- baum-sweet-hankel-nonvanishing
- 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
- License
- CC0-1.0
- Contributors
- TheoremDB entry research, 2026-07-24
- Public record
- R76
- Stable alias
- bsh-claim-classical-cubic
- Projection
- Reproduction fields are derived from the immutable record.
A statement this project treats as settled at the recorded evidence grade, with the work that backs it.