% 285_Addendum_IntegerTenAudit.tex
% OI-282-1 audited: the constant C simplifies to 10 alpha^-1/(1+mu1
% alpha^2) — the 10 is data-pinned to 6e-6 but underived. Candidates
% enumerated (metric components d(d+1)/2, bulk x quintic, C(5,2));
% the derivation requirement stated crisply. Case C, honest.
% Verifier: verify_P285.py.
% Copyright Léon Fernando Vlegels -- CC BY 4.0

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\title{Addendum 285: The Integer Ten\\
\large $C = 10\,\alpha^{-1}/(1+\mu_1\alpha^2)$: pinned, simplified,\\
candidates named, underived (OI-282-1 audit)}
\author{TOE Corpus}
\date{2026-06-10}

\begin{document}
\maketitle

\noindent\textbf{Skill:} gap-closure v1.1 \quad|\quad
\textbf{Outcome:} Case C (named missing theory; one simplification;
candidates enumerated, none promoted) \quad|\quad
\textbf{Addresses:} OI-282-1 (kept open, final form) \quad|\quad
\textbf{Verifier:} \texttt{verify\_P285.py} --- all PASS

\begin{abstract}
A282 reduced the mass-formula's free content to the integer $10$ and
the correspondence postulate. This audit extracts what can be extracted
without new theory. \emph{Simplification:} P03's constant collapses to
\begin{equation*}
C \;=\; \frac{10\,\mu_0^3}{\mu_0^2+\mu_1}
\;=\; \frac{10\,\alpha^{-1}}{1+\mu_1\alpha^2},
\end{equation*}
so the correspondence postulate reads, in its cleanest form:
\emph{geometric flow per QED flow equals ten inverse couplings times
the hierarchy span, dressed} ---
$\beta_{\rm geom}/\beta_{\rm QED} = 10\,\alpha^{-1}
L/(1+\mu_1\alpha^2)$. \emph{Pinning:} treating the prefactor as a free
real $a$ and solving against the measured span gives
$a = 10.00006$: the integer is data-pinned at $5.5\times10^{-6}$ ---
whatever the derivation is, it must produce exactly $10$, not
approximately. \emph{Candidates}, recorded with their texture and
without promotion: (i) $d(d+1)/2 = 10$ for $d = 4$ --- the independent
components of a metric tensor on $B^4$; the reading ``the
electron-to-Planck span is gravity's domain, and the correspondence
counts one $\alpha^{-1}$ per metric component'' has physical texture
and no derivation; (ii) $10 = 2\times5$, the even prime of the bulk
coefficient $16 = 2^4$ times the quintic prime (P12/P13);
(iii) $10 = \binom{5}{2}$, pairs from the five-fold convergence.
\emph{Requirement:} a closing derivation must produce, from corpus
primitives, the full identity $L = \tfrac{3\pi}{20}\mu_1(1 +
\mu_1\alpha^2)$ --- the integer, the specific moment $\mu_1$, and the
dressing --- not merely the numeral. OI-282-1 stands in this final
form; nothing here closes it, and saying so is the content.
\end{abstract}

\begin{proposition}[Simplification and pinning]
$10\mu_0^3/(\mu_0^2+\mu_1) = 10\mu_0/(1+\mu_1/\mu_0^2) =
10\alpha^{-1}/(1+\mu_1\alpha^2)$ identically ($\mu_0 = \alpha^{-1}$).
Solving $\beta_{\rm geom}/\beta_{\rm QED} = a\,\alpha^{-1}
L_{\rm meas}/(1+\mu_1\alpha^2)$ for $a$ gives $a = 10.000055$;
the nearest integers $9$ and $11$ are excluded at the $10\%$ level
while $10$ holds at $5.5$ ppm --- the constant is an exact integer or
the correspondence is numerology, with nothing in between surviving
the precision.
\end{proposition}

\begin{remark}[Why the candidates are recorded and not argued]
Each candidate explains the numeral; none explains the formula. The
metric-component reading is the most textured --- it gives the $10$ a
job (counting gravitational degrees of freedom over a span that ends at
the Planck mass) --- but it predicts nothing else and connects to no
corpus operator. Promotion criteria (A245/A180): a candidate earns
identification only when it derives the accompanying structure, here
the moment $\mu_1$ and the dressing. None does. Case C is the correct
grade, and the audit's value is the simplified target: future work aims
at one line, $L = \frac{3\pi}{20}\mu_1(1+\mu_1\alpha^2)$, knowing half
of it ($3\pi/2$, QED) is already external.
\end{remark}

\paragraph{Verifier.} \texttt{verify\_P285.py}: the simplification at
40 digits; the pinning ($a = 10.000055$, integer-exclusion bounds); the
candidate arithmetic ($d(d+1)/2$, $2\times5$, $\binom{5}{2}$); the
requirement statement. All PASS.

\end{document}
