A five-step procedure whose only input is an integer $N$, and whose output is a series for $1/\pi$. It works for six values of $N$ and no others, and 58 is the last one.
Chapter 1103 took Ramanujan's series apart and found that three of its four constants are the fundamental unit of $\mathbb{Q}(\sqrt{29})$ read three different ways. It closed by admitting a weakness: everything was done at $N=58$. One value. And a mechanism demonstrated once is a story, not a mechanism.
So here is the same construction written as a procedure — input an integer, turn a crank, get a series — and run on every $N$ up to 600.
The interesting part is not that it works. The interesting part is how rarely it works, and what that says about 58.
Input: an integer $N$. Nothing else — no lookup table, no constants supplied from outside.
1. $q = e^{-\pi\sqrt{N}}$ — a transcendental number.
2. $k = \big(\theta_2(q)/\theta_3(q)\big)^2$, $\alpha = k^2$ — the singular modulus.
3. $g^{12} = \sqrt{(1-\alpha)^2/(4\alpha)}$ — the Weber class invariant.
4. $t = \tfrac12\!\left(g^{12} + g^{-12}\right)$. Keep $N$ only if $t$ is an integer.
5. $\displaystyle b = \frac{\sqrt{N}\,(g^{12}-g^{-12})}{4\sqrt{2}}, \quad x = \frac{1}{256\,t^2}, \quad a = \frac{t/(2\sqrt2\,\pi) - bD}{F}$
with $F=\sum c_n x^n$ and $D=\sum n c_n x^n$, $c_n = (4n)!/(n!)^4$. Output the series $\;1/\pi = P\sum c_n (A+Bn)x^n$.
Every piece of that came from the previous chapter. Step 4 is “9801 is half a trace” written generally. Step 5's $b$ is the $\sqrt{58}\tanh(6\log\varepsilon)$ identity with the $\tanh$ unfolded. Step 5's $x$ is the observation that 396 was notation and the real argument is $1/t^2$.
Step 4 is the gate, and it is the whole story.
$t$ is a transcendental expression. Asking it to be an integer is asking for a great deal. Over $N = 1\ldots 600$:
Six. That is the entire list, and $N=58$ is the largest member of it.
$N=2$ doesn't count. There $t=1$, so $256x = 1$ exactly — and $256$ is precisely the growth rate of $c_n = (4n)!/(n!)^4$. The procedure has handed back the radius of convergence. The terms behave like $n^{-3/2}(A+Bn) \sim n^{-1/2}$ and the series diverges; it does not merely converge slowly. Five usable values of $N$ remain.
Two of the six half-traces are squares of others on the list:
This corpus states that and does not explain it. There is no obvious map sending 6 to 10 and 22 to 58, and inventing one is exactly the kind of thing the placement rules exist to prevent.
Each line below was produced by the procedure from $N$ alone, then checked against $1/\pi$ to 120 decimal places or better.
The bottom line is Ramanujan's, written the way Chapter 1103 said to write it: $396^{4n}$ replaced by $(256\cdot 9801^2)^n$, which is the same number and an honest one.
These are not new series. They are believed to be in Ramanujan's 1914 list, and this chapter does not claim otherwise — the interest is that five steps and one integer produce them. Ramanujan supplied the series and no derivation. What is reconstructed here is not the results; it is the crank.
Digits per term is $2\log_{10} t$. That is the entire convergence story: it depends on $t$ and on nothing else. So:
Monotone in $t$, and $t$ is monotone in $N$. So the fastest series in this family is the one at the largest surviving $N$, and the largest surviving $N$ is 58.
Ramanujan's series is the fastest of its kind because 58 is the last entry on a list that stops.
That is a different kind of statement from “nobody has found a faster one.” Within this family, with this shape, the search is over before it starts. Eight digits per term is not a property of 58; it is a property of 58 being last.
And the list stopping is the same phenomenon as the Heegner numbers stopping at 163, which Chapter R deals with: the half-trace is an integer when $g_N^{12}$ is a unit in a real quadratic field, and that stops happening once class numbers grow. The finiteness is class-field theory, showing up as a short list.
The scan covers $N \le 600$. It does not prove that no larger $N$ has an integer half-trace. The reason to expect none — class numbers growing, so $g_N^{12}$ ceases to be a quadratic unit — is theory this chapter cites and does not show. Until that is closed, the last of six means the last below 600, and the title is a claim with a bound on it.
Two limits, stated because the procedure looks more autonomous than it is.
Step 5 still uses the series. The coefficient $a$ is obtained by solving the identity $aF + bD = t/(2\sqrt2\pi)$, not by an independent formula. So the procedure verifies a shape it was handed; it does not derive the shape. That is the same wall Chapter 1103 hit, and it has not moved: integrality is cited, not shown.
The formula for $b$ was read off one case. $b = \sqrt{N}(g^{12}-g^{-12})/4\sqrt2$ came from $N=58$ and is here shown to work at four more $N$. Five agreements is evidence. It is not a proof, and the script says so.
What has moved: Chapter 1103's gap G4 said the mechanism rested on one data point. It now rests on five, and the selection rule that picks those five is sharp enough to explain why the sixth-best series in the world is the best one there is.
Producing script: book7/ch-the-last-of-six-verify.py — 4 sections, 6 gaps, sweep over $N \le 600$ at 150 digits, each emitted series checked to 120 places. Cross-references: Chapter 1103 (the construction, at $N=58$), Book IV · the Euclidean algorithm, Chapter R (Heegner, and lists that stop).