Problem packetResearch packetR744
Each monotonicity comparison reduces to an integer inequality
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The author records a mathematical identity.
Recorded status: established
Recorded scope: every positive integer n and every consecutive pair of admissible indices
Complete recorded scope and conditions
{
"kind": "universal",
"statement": "every positive integer n and every consecutive pair of admissible indices"
}Originating problem: Eventual monotonicity in a signed subset-sum local limit
Recorded relationships: All 492 comparisons through n=1000 are strict increases
Authored record and scope
- Authored title
- Each monotonicity comparison reduces to an integer inequality
- Record type
- claim
- Stored status
- established
- Evidence grade
- mathematical_identity
- Recorded scope data
- { "kind": "universal", "statement": "every positive integer n and every consecutive pair of admissible indices" }
- Linked research record IDs
- R745
2Authored explanation
Let \[ F_n(x)=\prod_{k=1}^n(1+x^k)=\sum_s c_{n,s}x^s, \qquad T_n=\frac{n(n+1)}2. \] Choosing the indices assigned sign \(+1\) shows \[ S_n=0\quad\Longleftrightarrow\quad \sum_{k:\varepsilon_k=1}k=\frac{T_n}{2}. \] Hence, for admissible \(n\), \[ C_n=c_{n,T_n/2}=[x^{n(n+1)/4}]F_n(x), \qquad \Pr(S_n=0)=\frac{C_n}{2^n}. \] The coefficients obey the exact subset-sum recurrence \[ c_{n,s}=c_{n-1,s}+c_{n-1,s-n}, \] with \(c_{0,0}=1\). Also \[ Q_n=\sigma_n^2=\frac{n(n+1)(2n+1)}6. \] For consecutive admissible indices \(a<b\), all quantities are positive. Squaring and clearing denominators gives \[ A_b>A_a \quad\Longleftrightarrow\quad C_b^2Q_b-C_a^2Q_a4^{b-a}>0. \] This equivalence is the comparison used by the executable artifact. Every decision is an integer sign test.
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3Evidence
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Verification source: oeis.org ↗, A063865 gives the sign-count and central-coefficient interpretations; the recurrence and cleared comparison are derived directly here
4What was measured
5How it connects
Supports
- claim
Recorded for
- problem
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"title": "Each monotonicity comparison reduces to an integer inequality",
"summary": "A central coefficient recurrence gives the exact probability, and squaring clears every square root and power-of-two denominator.",
"relevance": "For Eventual monotonicity in a signed subset-sum local limit, record ssclt-claim-exact-coefficient-test (“Each monotonicity comparison reduces to an integer inequality”) records a bound, answer, status fact, or structural consequence. The record states: A central coefficient recurrence gives the exact probability, and squaring clears every square root and power-of-two denominator.",
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"body": "Let\n\\[\nF_n(x)=\\prod_{k=1}^n(1+x^k)=\\sum_s c_{n,s}x^s,\n\\qquad T_n=\\frac{n(n+1)}2.\n\\]\nChoosing the indices assigned sign \\(+1\\) shows\n\\[\nS_n=0\\quad\\Longleftrightarrow\\quad \\sum_{k:\\varepsilon_k=1}k=\\frac{T_n}{2}.\n\\]\nHence, for admissible \\(n\\),\n\\[\nC_n=c_{n,T_n/2}=[x^{n(n+1)/4}]F_n(x),\n\\qquad \\Pr(S_n=0)=\\frac{C_n}{2^n}.\n\\]\nThe coefficients obey the exact subset-sum recurrence\n\\[\nc_{n,s}=c_{n-1,s}+c_{n-1,s-n},\n\\]\nwith \\(c_{0,0}=1\\). Also\n\\[\nQ_n=\\sigma_n^2=\\frac{n(n+1)(2n+1)}6.\n\\]\nFor consecutive admissible indices \\(a<b\\), all quantities are positive. Squaring and clearing denominators gives\n\\[\nA_b>A_a\n\\quad\\Longleftrightarrow\\quad\nC_b^2Q_b-C_a^2Q_a4^{b-a}>0.\n\\]\nThis equivalence is the comparison used by the executable artifact. Every decision is an integer sign test.",
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}7Provenance
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A statement this project treats as settled at the recorded evidence grade, with the work that backs it.