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Problem packetResearch packetR1352

R1352Sourced evidence

Current checked status and unresolved remainder

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Authored summary

UNKNOWN as of 2026-07-27. The MathOverflow answer identifies the question with the open problem of proving unbounded partial quotients for \(\pi\). Current explicit upper bounds for the irrationality measure of \(\pi\) do not establish this target.

The record cites sources for its explanation.

Recorded status: reported

Recorded scope: No scope is recorded.

Originating problem: Unbounded continued-fraction coefficients of pi

Authored record and scope
Authored title
Current checked status and unresolved remainder
Record type
claim
Stored status
reported
Evidence grade
sourced

2Authored explanation

A dated independent review on 2026-08-01 checked the structured sources below, the complete visible source discussion, exact-title and equivalent-formulation searches, and the current TheoremDB corpus. On 2026-07-27 the MathOverflow answer and all comments were checked. They reduce the sine liminf to a classical Diophantine-approximation question and give no proof of unbounded partial quotients. When \(n\) lies near \(k\pi\), the quantities \(n\lvert\sin n\rvert\) and \(k^2\lvert\pi-n/k\rvert\) differ by bounded factors. Continued-fraction theory therefore gives the stated equivalence. Zeilberger and Zudilin, arXiv:1912.06345, prove an explicit finite upper bound for the irrationality measure of \(\pi\). Such a bound allows irrationality exponents above two and does not rule out bad approximability. Computing large partial quotients supplies witnesses only for finite thresholds. Each certified convergent and error interval remains reusable, while no finite list proves unboundedness. Trap: irrationality of \(\pi\), an irrationality-measure bound, or one unusually large partial quotient does not imply the required unbounded sequence.

A complete resolution must satisfy: Prove that for every \(C>0\) there is a positive integer \(n\) with \(n\lvert\sin n\rvert<C\). An equivalent proof may show that the continued-fraction partial quotients of \(\pi\) are unbounded, with the equivalence to the displayed limit justified.

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3Evidence

Replay package: source only

A verification source is cited. This record has no executable replay attached.

Verification source: mathoverflow.net ↗, Dataset references and independent 2026-08-01 status search.

4How it connects

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  "ref": "R1352",
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  "type": "claim",
  "title": "Current checked status and unresolved remainder",
  "summary": "UNKNOWN as of 2026-07-27. The MathOverflow answer identifies the question with the open problem of proving unbounded partial quotients for \\(\\pi\\). Current explicit upper bounds for the irrationality measure of \\(\\pi\\) do not establish this target.",
  "relevance": "Records the strongest checked neighboring results and the exact remainder future work must settle.",
  "relevance_source": "recorded",
  "body": "A dated independent review on 2026-08-01 checked the structured sources below, the complete visible source discussion, exact-title and equivalent-formulation searches, and the current TheoremDB corpus. On 2026-07-27 the MathOverflow answer and all comments were checked. They reduce the sine liminf to a classical Diophantine-approximation question and give no proof of unbounded partial quotients. When \\(n\\) lies near \\(k\\pi\\), the quantities \\(n\\lvert\\sin n\\rvert\\) and \\(k^2\\lvert\\pi-n/k\\rvert\\) differ by bounded factors. Continued-fraction theory therefore gives the stated equivalence. Zeilberger and Zudilin, arXiv:1912.06345, prove an explicit finite upper bound for the irrationality measure of \\(\\pi\\). Such a bound allows irrationality exponents above two and does not rule out bad approximability. Computing large partial quotients supplies witnesses only for finite thresholds. Each certified convergent and error interval remains reusable, while no finite list proves unboundedness. Trap: irrationality of \\(\\pi\\), an irrationality-measure bound, or one unusually large partial quotient does not imply the required unbounded sequence.\n\nA complete resolution must satisfy: Prove that for every \\(C>0\\) there is a positive integer \\(n\\) with \\(n\\lvert\\sin n\\rvert<C\\). An equivalent proof may show that the continued-fraction partial quotients of \\(\\pi\\) are unbounded, with the equivalence to the displayed limit justified.",
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6Provenance

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A statement this project treats as settled at the recorded evidence grade, with the work that backs it.

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