% 301_Addendum_FamilyCountBound.tex
% OI-287-1, boundary question. The index-matching rule (family n draws
% the degree-n coefficient of the density rho) bounds the family count
% at deg(rho) = 3. rho = 16 pi^3 x^3 + 3 pi^2 x^2 + 2 pi x is cubic, so
% c_1,c_2,c_3 = (2,3,16) are the only nonzero coefficients; c_4 = 0. The
% fourth eigenstate exists (node count 3) but draws a zero coefficient,
% so the fourth family is decoupled. This is the coupling-side image of
% P20's topological no-fourth-family (Z_3 has three eigenspaces).
% Verifier: verify_P301.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 301: The Family-Count Bound\\
\large The selection rule permits exactly three families}
\author{TOE Corpus}
\date{2026-06-13}

\begin{document}
\maketitle

\noindent\textbf{Skill:} gap-closure v1.1 \quad|\quad
\textbf{Outcome:} Case A (a standing claim, P20's no-fourth-family,
reproduced from an independent direction) \quad|\quad
\textbf{Addresses:} OI-287-1 (boundary), P20 \quad|\quad
\textbf{Verifier:} \texttt{verify\_P301.py} --- all PASS

\begin{abstract}
A297--A300 pursued the magnitude content of the selection rule: the
factor that would lift the $\mu$ entry of
$\langle\psi_n|c_k\pi^k x^k|\psi_n\rangle/\lambda_n$ from $1.12$ to
$1.5$. That factor is still open. This addendum settles a separate,
boundary question with the same machinery: how many families does the
index-matching rule permit? The density is fixed by
$\rho'(x) = 48\pi^3 x^2 + 6\pi^2 x + 2\pi$, hence
$\rho(x) = 16\pi^3 x^3 + 3\pi^2 x^2 + 2\pi x$, a cubic. Its only nonzero
Taylor coefficients are $(2, 3, 16)$ in units $\pi^k$ at degrees
$1, 2, 3$; the degree-$4$ coefficient is identically zero. Family $n$
draws the degree-$n$ coefficient, so families $1, 2, 3$ draw the nonzero
$(2,3,16)$ while family $4$ draws zero. The fourth eigenstate exists as
an operator mode (node count $3$, $\lambda_4 = 171.43$), but the
coefficient it would couple to is absent, so it is decoupled. The
selection rule therefore bounds the family count at $\deg\rho = 3$ with
no additional postulate. This is the coupling-side image of P20's
topological no-fourth-family, where the count is fixed by the three
eigenspaces of $\mathbb{Z}_3 = \pi_1(S^3/\mathbb{Z}_3)$.
\end{abstract}

% ----------------------------------------------------------------------
\section{The Bound}

\begin{proposition}[The index-matching rule permits exactly three families]
Let $\rho(x) = 16\pi^3 x^3 + 3\pi^2 x^2 + 2\pi x$ be the density obtained
by integrating $\rho'(x) = 48\pi^3 x^2 + 6\pi^2 x + 2\pi$, and let
$c_k$ denote its degree-$k$ coefficient in units $\pi^k$:
$c_1 = 2,\ c_2 = 3,\ c_3 = 16,\ c_4 = 0$. Under the index-matching rule,
the diagonal coupling of family $n$ is
$\langle\psi_n|c_n\pi^n x^n|\psi_n\rangle/\lambda_n$. For the four
lowest eigenstates of the production operator, which carry node counts
$0, 1, 2, 3$, the normalized diagonal is $(1,\ 1.12,\ 7.50,\ 0)$: the
first three reproduce the standing lead, and the fourth is exactly zero
because $c_4 = 0$. The fourth eigenstate is present and well-ordered;
only its target coefficient is absent. The family count permitted by the
rule equals the number of nonzero coefficients of $\rho$, which is
$\deg\rho = 3$.
\end{proposition}

\begin{remark}[Two readings of the same three]
The bound is not an independent derivation of the number three. It is a
demonstration that two pieces of the framework agree on it. P20 fixes
three families topologically, as the eigenspaces of the layer-cycle
operator on $S^3/\mathbb{Z}_3$; the present argument fixes the same three
arithmetically, as the nonzero coefficients of a cubic density. The
content is the consistency: the same selection rule that gives the right
$(2,3,16)$ direction for $n = 1, 2, 3$ also forbids $n = 4$ when applied
one step further, without a separate prohibition. The deeper question,
why $\rho$ is cubic, is the same question as why $\mathbb{Z}_3$ and not
$\mathbb{Z}_m$, and is not answered here. What is answered is that the
lepton-coupling sector and the topological sector do not disagree about
the family count.
\end{remark}

\paragraph{Status.} OI-287-1 remains open in its magnitude part: the
localizing factor that lifts the $\mu$ entry and concentrates the
coupling on the diagonal (A300). Its boundary part is settled: the rule
permits exactly $\deg\rho = 3$ families, in agreement with P20. P20's
claim that a fourth generation is forbidden gains an independent witness.
No published number changes.

\paragraph{Verifier.} \texttt{verify\_P301.py}: confirms $\rho$ is cubic
with coefficients $(2,3,16)$ and $c_4 = 0$, confirms a fourth eigenstate
exists with node count $3$, and confirms families $1$--$3$ keep the
$(1, 1.12, 7.50)$ lead while family $4$ draws zero. All PASS.

\end{document}
