Problem packetResearch packetR1352
Current checked status and unresolved remainder
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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
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Verification source: mathoverflow.net ↗, Dataset references and independent 2026-08-01 status search.
4How it connects
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Replaced by
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Recorded for
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"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.",
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"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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