% 303_Addendum_DegreeFactor.tex
% OI-287-1. An eighth mechanism class is excluded: a degree-dependent
% factor, a fixed function w(k) of the monomial degree k applied as a
% per-column weight to the A299 base, does not supply the missing
% localizing factor. None of the four declared candidates (factorial
% 1/k!, inverse degree 1/k, pi-power strength, linear strength) peaks
% on the diagonal, and all four move the mu entry further from 1.5 than
% the A299 control. A pure column weight reshapes the diagonal away from
% the target rather than toward it. The localizing factor is not a fixed
% function of the monomial degree alone.
% Verifier: verify_P303.py.
% Copyright Léon Fernando Vlegels -- CC BY 4.0

\documentclass[11pt]{article}
\usepackage{amsmath,amssymb,amsthm}
\usepackage[margin=1in]{geometry}
\usepackage{hyperref}

\newtheorem{proposition}{Proposition}[section]
\newtheorem{remark}[proposition]{Remark}

\title{Addendum 303: The Degree Factor\\
\large An eighth excluded class, and why a column weight cannot localize}
\author{TOE Corpus}
\date{2026-06-13}

\begin{document}
\maketitle

\noindent\textbf{Skill:} gap-closure v1.1 \quad|\quad
\textbf{Outcome:} Case D (mechanism class excluded; OI-287-1 lead
unchanged) \quad|\quad
\textbf{Addresses:} OI-287-1 \quad|\quad
\textbf{Verifier:} \texttt{verify\_P303.py} --- all PASS

\begin{abstract}
A300 bracketed the factor missing from the A299 base
$B_{nk} = \langle\psi_n|c_k\pi^k x^k|\psi_n\rangle / \lambda_n$, whose
diagonal normalizes to $(1, 1.12, 7.50)$ against the target $(1, 1.5, 8)$,
and excluded the node count as a uniform mechanism: it lifts the $\mu$ and
$\tau$ rows by different relative amounts, so the factor is not a function
of the family index alone. A300's stated next step was to test the
complementary hypothesis, that the missing factor is a fixed function of
the monomial degree $k$, not of the family $n$. This addendum runs that
test. Four degree-dependent candidates are declared in advance and applied
as a per-column weight $w(k)$ on the A299 base: the Taylor factorial
$1/k!$, the inverse degree $1/k$, a degree-resolved strength
$\pi^{-k}\kappa\Omega$, and a linear strength $\kappa\Omega\,k$, with
$\kappa = \alpha^{5/4}$ and $\Omega = \alpha^{-1}$. None peaks on the
diagonal under the column-standardized criterion, and all four carry the
normalized diagonal away from $(1, 1.5, 8)$: the closest, inverse degree,
reaches $(1, 0.56, 2.50)$ at $L^2 = 5.58$, far worse than the A299 control
at $L^2 = 0.63$, and its $\mu$ entry moves to $0.56$, away from $1.5$ rather
than toward it. A degree-dependent column weight is therefore excluded as an
eighth mechanism class. The structural reason is simple and is the content
of the result: a weight that depends only on $k$ scales an entire column
uniformly, so it cannot change the position of the peak within a column and
cannot lift one diagonal entry relative to the column it sits in. The
missing factor is not a fixed function of the monomial degree alone.
\end{abstract}

% ----------------------------------------------------------------------
\section{The Probe}

\begin{proposition}[A degree-only column weight cannot localize]
Let $B_{nk} = \langle\psi_n|c_k\pi^k x^k|\psi_n\rangle / \lambda_n$ be the
A299 base, diagonal $(1, 1.12, 7.50)$. For a weight $w(k)$ depending only on
the degree $k$, the weighted matrix is $M_{nk} = w(k)\,B_{nk}$, a uniform
rescaling of each column. Such a rescaling leaves $\arg\max_n$ of every
column fixed, so $M$ peaks on the diagonal exactly when $B$ does, namely not
at all. Evaluated on the four declared weights, the normalized diagonals are
$(1, 0.56, 1.25)$ for $w = 1/k!$, $(1, 0.56, 2.50)$ for $w = 1/k$,
$(1, 0.36, 0.76)$ for $w = \pi^{-k}\kappa\Omega$, and $(1, 2.24, 22.50)$ for
$w = \kappa\Omega\,k$. None peaks, none lands within $20\%$ of $(1, 1.5, 8)$,
and the closest by $L^2$ is the inverse degree at $L^2 = 5.58$, against the
unweighted control's $L^2 = 0.63$. The pre-registered HIT criterion, a
diagonal peak together with all three entries within $20\%$, is met by no
candidate. A degree-only factor is excluded as an eighth mechanism class.
\end{proposition}

\begin{remark}[Why the column weight fails, and what it implies]
The failure is not numerical accident but a consequence of where the target
information lives. The diagonal entry of column $k$ is row $k$; a weight
$w(k)$ multiplies that entry and the two off-diagonal entries of the column
by the same number, so the ratio of the diagonal entry to its column is
invariant. Localization requires lifting the diagonal entry relative to the
others in its column, which a function of $k$ alone cannot do. The four
candidates merely re-weight the already-correct relative magnitudes between
columns, and since the A299 diagonal is already close in the cross-column
sense, any nontrivial $w(k)$ can only spoil it; that is why every candidate
increases the $L^2$ distance. Read together with A300, the two complementary
hypotheses are now both excluded: the factor is neither a function of the
family index $n$ alone (A300) nor of the monomial degree $k$ alone (here).
The honest narrowing is that the missing factor must couple $n$ and $k$
jointly, acting within a column to raise the $n = k$ entry above its
neighbors. The selection rule's localization is an off-diagonal-suppression
effect, not a row or column rescaling.
\end{remark}

\paragraph{Status.} OI-287-1 remains open. Its form is unchanged by this
addendum: the missing factor lifts the $\mu$ entry by roughly $34\%$ and the
$\tau$ entry by roughly $7\%$, and is now known to be neither a function of
the family index alone (A300) nor of the monomial degree alone (this
addendum). The next probe must test a jointly indexed kernel that suppresses
entries with $n \neq k$. No published number changes.

\paragraph{Verifier.} \texttt{verify\_P303.py}: recomputes the
eigenstructure (eigenvalues ordered, node counts $(0,1,2)$), confirms the
A299 base diagonal reproduces $(1, 1.12, 7.50)$, confirms each declared
candidate's diagonal and that none yields a valid diagonal peak, and
confirms the NULL conditions, that no candidate lands within $20\%$ and the
closest candidate's $\mu$ entry moves away from $1.5$. All PASS.

\end{document}
