Keyboard shortcuts

Press ← or → to navigate between chapters

Press ? to show this help

Press Esc to hide this help

Metadata 
slug: oeis-a391810-laurent-first-moment
bibkey: hanna2025a391810
doi: null
url: https://oeis.org/A391810/internal
triage: theorem
motivation_gids:
  - D5/S3/ArithSums/HannaLaurentRowSums.series_coefficient
  - D5/S3/ArithSums/HannaLaurentRowSums.row_degree

The first weighted moment of A391810

Problem

Paul D. Hanna’s OEIS A391810, revision 39, formula (3), states:

(3) Sum_{k=0..floor((n+1)/2)^2} k * T(n,k) = 1 for n >= 1 (conjecture).

The original definition is:

Irregular triangle, read by rows, where row n lists the coefficients of nonpositive powers of x in the doubly infinite series Sum_{k=-oo..oo} x^k * (1 - x^k)^(n+k) for n >= 1.

Thus, for every positive natural n, expand S_n(x)=Sum_{j in Z} x^j*(1-x^j)^(n+j) formally at zero and let T(n,k)=[x^(-k)]S_n(x). The outer sum includes every integer j; negative integer powers use Laurent-field inverses. The row includes its constant coefficient at k=0 and its exact inclusive final index floor((n+1)/2)^2. Coefficients and the weighted row sum are signed integers. The literal Laurent realization uses rational coefficients and the exact integer-to-rational coefficient bridge. The shared library note’s claim quotation concerns formula (2); the quoted formula (3) is the distinct target here.

Motivation

This is a Tier 1 explicitly published universal coefficient identity. The existing coefficient bridge and degree bound turn its infinite-series definition into an exact finite polynomial question without changing the row or endpoint. The exact event and bounded qualification are registered at https://github.com/the-omega-institute/trureturing/issues/24461. The already established formula (2) is reused and is not another resolution.

Gap

The caller-supplied source and reuse qualification reports no settlement of this exact formula within its successful bounded searches. This is a bounded qualification, not a claim of global absence or priority. The current official OEIS export at https://raw.githubusercontent.com/oeis/oeisdata/master/seq/A391/A391810.seq contains the quoted definition and formula (3), with revision identifier 39. Its SHA256 is 49bbf5e8e42f91b6a3d838b2ce5c311eca58af74a83a4f6f684b09e1ddf0da83. The missing mathematical step beyond the reused coefficient and support results is the first weighted moment; the result below supplies that step for every positive natural n. No corresponding theory volume was identified by the qualified exact-source searches.

Route

Use the existing polynomial P_n(y)=-1+Sum_{j=1..n} y^j*(1-y^j)^(n-j). The frozen theorem D5/S3/ArithSums/HannaLaurentRowSums.series_coefficient identifies its coefficient at k with the literal Laurent coefficient at -k, including k=0. The frozen degree bound makes the published finite interval contain every polynomial coefficient.

Differentiate only this finite polynomial. At n=1 the derivative at y=1 is one. For n>=2, every differentiated summand with j<=n-2 vanishes at one. The last two differentiated terms contribute -(n-1) and n, respectively, so P_n'(1)=1. The polynomial derivative coefficient identity gives Sum_{k=0..floor((n+1)/2)^2} k*T(n,k)=1 in the integers. Transfer that same equality to the rational literal Laurent coefficients. The bilateral series itself is never evaluated at x=1.

Falsifier

A positive natural n whose exact signed weighted sum over the full published inclusive interval differs from one would refute formula (3). A computation that drops the constant coefficient, truncates the bilateral family before establishing the coefficient bridge, changes the endpoint, or takes absolute values would address a different question and is not such a falsifier.

Evidence

D5/S3/ArithSums/HannaLaurentFirstMoment.result states both the integer weighted equality and its literal rational Laurent counterpart, with the full positive-natural telescope and exact inclusive endpoint. Its two private helpers establish the derivative evaluation and the integer moment and are consumed in that proof. The retained Reg mirror reconstructs this same full statement and retains both conjuncts.

Every authored mathematical declaration has proof_shape: bind-only. The public result uses admission_basis: open-problem-resolution for the single preregistered external formula (3); the helpers have their actual proof consumers. There is no separate escape-witness claim, finite-instance delivery, new axiom, or second claim for formula (2).

Triage

The exact source event is a universal theorem question, with its definition, signs, coefficient domain and endpoint retained.

What the settlement shows

  • Proved (D5/S3/ArithSums/HannaLaurentFirstMoment.result): the decisive mechanism is P_n'(1)=1, obtained from the two surviving differentiated boundary terms, together with the existing coefficient bridge. The result covers every n>=1 without parity restrictions or a finite ceiling, in both integer row entries and literal rational Laurent coefficients. It needs no analytic convergence assumption or evaluation of the bilateral sum.
  • Proved (D5/S3/ArithSums/HannaLaurentRowSums.row_degree): the published endpoint contains the full support. This is an upper bound; equality with the degree or optimality for each n is not asserted by the first-moment proof.
  • Open: extensions to nonpositive row indices, other expansion conventions or coefficient fields, and sharpness of the endpoint are not established by this delivery. The original positive-n domain is retained throughout.
  • Proved (D5/S3/ArithSums/HannaLaurentFirstMoment.result): formula (3) supplies exactly its weighted-sum premise to any argument that uses it. The previously proved formula (2) retains its independent proof and scope. No other source argument or conjecture is certified merely by this premise.
  • Open candidate, formula (4): the source also asks whether Sum_{k=0..floor((n+1)/2)^2} k^2*T(n,k)=-(2*n-5) for every n>=2. A second-derivative investigation would be a separate target; no such identity or extension is proved here.
  • Open candidate, formula (5): the source also asks whether Sum_{k=0..floor((n+1)/2)^2} -(-1)^k*T(n,k)=2*round(2^n/3)=2*A001045(n) for every n>=1, with A001045 the Jacobsthal numbers. Evaluation at minus one is a possible separate route, not a consequence certified here.

ASSUMED-UNVERIFIED

Global prior solvedness and priority outside the supplied successful bounded screens remain ASSUMED-UNVERIFIED. The conjecture label alone supplies no global absence evidence. Independent final delivery reviews, required CI, publication and merge to dev remain separate obligations. The neighboring formulas (4) and (5) are candidates only; they and other A391810 or A391811 questions are not settled by this result. The audit’s continuation is unknown and entails no undecidability or inexhaustibility claim.