Problem packetResearch packetR58
Every Mersenne input attains the proposed bound
Link to a section
The author reports this result.
Recorded status: supported
Recorded scope: every integer n=2^m-1 with m at least 1
Complete recorded scope and conditions
{
"kind": "family",
"statement": "every integer n=2^m-1 with m at least 1",
"family": "Mersenne inputs n=2^m-1"
}Originating problem: Sharp multipliers for balanced binary products
Authored record and scope
- Authored title
- Every Mersenne input attains the proposed bound
- Record type
- claim
- Stored status
- supported
- Evidence grade
- self_reported
- Recorded scope data
- { "kind": "family", "statement": "every integer n=2^m-1 with m at least 1", "family": "Mersenne inputs n=2^m-1" }
2Authored explanation
Fix \(m\geq1\), put \(n=2^m-1\), and let \(u=k-1\). The identity \[ k(2^m-1)=u2^m+(2^m-1-u) \] expresses the product as two binary blocks of width \(m\): the padded \(m\)-bit expansion of \(u\), followed by its bitwise complement. The padded word has length \(2m\) and exactly \(m\) ones.
If \(1\leq k\leq2^{m-1}\), then \(u<2^{m-1}\). Its padded block begins with at least one zero. Ordinary binary notation removes those leading zeros while retaining all \(m\) ones in the two-block word. The resulting word has fewer than \(2m\) bits and still has \(m\) ones, so it is unbalanced.
For \(k=2^{m-1}+1\), one has \(u=2^{m-1}\). The first block now begins with 1, so the ordinary expansion has all \(2m\) bits. The two complementary blocks contain \(m\) ones in total, making the product balanced. Every smaller positive multiplier has already been excluded, which proves \[ k(2^m-1)=2^{m-1}+1. \] This proves the sharp family in the candidate and covers the forward direction of its equality statement.
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Replay material: source only
3Evidence
A verification source is cited. This record has no executable replay attached.
Verification source: arxiv.org ↗, Complete elementary complement-block proof supplied in this record
4How it connects
Informs
- claim
Supported by
- attempt
Recorded for
- problem
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Cite the original sources separately.
Machine-readable record
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"title": "Every Mersenne input attains the proposed bound",
"summary": "For every m at least 1, k(2^m-1)=2^(m-1)+1.",
"relevance": "For Sharp multipliers for balanced binary products, record bbmb-claim-mersenne-sharpness (“Every Mersenne input attains the proposed bound”) records a bound, answer, status fact, or structural consequence. The record states: For every m at least 1, k(2^m-1)=2^(m-1)+1.",
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"body": "Fix \\(m\\geq1\\), put \\(n=2^m-1\\), and let \\(u=k-1\\). The identity\n\\[\nk(2^m-1)=u2^m+(2^m-1-u)\n\\]\nexpresses the product as two binary blocks of width \\(m\\): the padded \\(m\\)-bit expansion of \\(u\\), followed by its bitwise complement. The padded word has length \\(2m\\) and exactly \\(m\\) ones.\n\nIf \\(1\\leq k\\leq2^{m-1}\\), then \\(u<2^{m-1}\\). Its padded block begins with at least one zero. Ordinary binary notation removes those leading zeros while retaining all \\(m\\) ones in the two-block word. The resulting word has fewer than \\(2m\\) bits and still has \\(m\\) ones, so it is unbalanced.\n\nFor \\(k=2^{m-1}+1\\), one has \\(u=2^{m-1}\\). The first block now begins with 1, so the ordinary expansion has all \\(2m\\) bits. The two complementary blocks contain \\(m\\) ones in total, making the product balanced. Every smaller positive multiplier has already been excluded, which proves\n\\[\nk(2^m-1)=2^{m-1}+1.\n\\]\nThis proves the sharp family in the candidate and covers the forward direction of its equality statement.",
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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.