Ramanujan's series for $1/\pi$ carries 9801 in front and 396 underneath. Both are 99. The corpus had already computed 99 twice, in a chapter that did not know what it was for.
This chapter continues Chapter R, which established the singular moduli $\alpha_n = k(e^{-\pi\sqrt n})^2$ and ran Watson's algorithm from page 320 of the notebook. At $n = 58$ that algorithm returned $U = 1$, $V = 9801$, $W = 9801$, $S = 9802$. Nothing in that chapter used those numbers. This one says what they were for.
Every figure below is recomputed by ch-ramanujan-1pi-verify.py, which fails loudly rather than rounding to agreement. It failed three times while this chapter was being written. §6 says how.
From Ramanujan's 1914 paper, restated by Bailey and the Borweins:
$$\frac{1}{\pi} \;=\; \frac{2\sqrt 2}{9801}\sum_{n=0}^{\infty}\frac{(4n)!\,(1103 + 26390\,n)}{(n!)^4\,396^{4n}}$$Computed at 120-digit precision against a Machin $\pi$, it gains eight correct digits per term, flat — 8, 16, 24, 32, 40, 48, 56, and 104 by the twelfth term. The rate is not a coincidence of the first few terms; it is the ratio $396^4 = 24{,}591{,}257{,}856$ against the growth of $(4n)!/(n!)^4$.
Bailey and the Borweins state in one line that this series “is a specialization ($N = 58$)” of their Theorem 5. That is the whole hinge of this chapter, because $N = 58$ is a row this corpus had already computed.
Chapter R's closed form for the singular modulus at 58, recomputed here from the theta series and agreeing to 80 digits:
$$\alpha_{58} = (13\sqrt{58} - 99)^2\,(99 - 70\sqrt 2)^2$$Ninety-nine, twice, in a formula written down for an unrelated reason. And in the series:
| 9801, the prefactor | = 99² |
| 396, the base | = 4 · 99 |
| 396⁴, one term's worth | = 256 · 99⁴ = 24,591,257,856 |
| Watson's row at 58 | V = W = 9801 = 99² |
The 9801 standing in front of Ramanujan's series and the 9801 Watson's algorithm returns are the same 99, arrived at from two directions that had no reason to meet.
The class invariant is $g_N^{12} = (k'_N)^2/(2k_N)$. Computed from the theta series at $N = 58$ it is
$$g_{58}^{12} = 19601.99999489847\ldots$$which is near 19602 and near 19601 and equal to neither. It is not an integer at all. It is
$$g_{58}^{12} = \varepsilon^6 = 9801 + 1820\sqrt{29}, \qquad \varepsilon = \tfrac{5+\sqrt{29}}{2}$$the sixth power of the fundamental unit of $\mathbb{Q}(\sqrt{29})$. There is the 9801 again, exactly, as a rational part. And because $\varepsilon$ has norm $-1$, its sixth power has norm $+1$:
$$9801^2 - 29\cdot 1820^2 = 96{,}059{,}601 - 96{,}059{,}600 = 1$$Pell's equation, closing on the nose. The consequence is that the conjugate is the inverse, $g^{-12} = 9801 - 1820\sqrt{29}$, so the irrational parts cancel when the two are added:
$$g_{58}^{12} + g_{58}^{-12} = 19602 = 2\cdot 99^2 \qquad \text{exactly, to thirty decimal places of zeros}$$That is the exact integer the near-miss was gesturing at, and it is exact only because Pell closes. The corpus has met this equation before: Chapter H used $p^2 - 2q^2 = \pm 1$ to rank the convergents of $\sqrt 2$ after a first attempt ranked them wrongly by decimal distance. Same equation, different discriminant, and here its job is to make a transcendental quantity rational.
The copy of Bailey–Borwein–Borwein in hand has usable text everywhere except Theorem 5, where the OCR collapses into fragments — x2Nn;1, (1 + k2 )2, and so on. Transcribing that would have been guessing with a citation attached.
So it was not transcribed. Instead each candidate reading of $x_N$ was evaluated at $N = 58$ and compared against $1/396^4$, the value the series itself forces. A reading either lands on the integer or it does not:
| g¹² + g⁻¹² | ratio 4.82 × 10¹⁴ |
| 2 / (g¹² + g⁻¹²) | ratio 2.51 × 10⁶ |
| 1 / (g¹² + g⁻¹²)² | ratio 64.000000000 |
| 1 / (64 (g¹² + g⁻¹²)²) | ratio 1.000000000 — exact |
| g⁻²⁴ / 64 (the near-miss) | ratio 1.000000005205 |
One reading closes and the rest do not, and the one that closes does so because $64 \cdot 19602^2 = 396^4$ in integers — both sides being $256 \cdot 99^4$. The last row is the instructive one: using $g^{-24}$ instead of $(g^{12}+g^{-12})^{-2}$ agrees to eight figures and is still wrong, because $g^{12}$ alone is irrational.
The OCR is still unusable. The identity is not. This is a method the corpus can reuse: when a source is damaged, do not reconstruct it — enumerate the readings and let arithmetic pick.
Chudnovsky's series, same shape, larger discriminant:
$$\frac{1}{\pi} = 12\sum_{n=0}^{\infty}\frac{(-1)^n (6n)!\,(13591409 + 545140134\,n)}{(3n)!\,(n!)^3\,640320^{3n + 3/2}}$$Measured here: 14 digits per term against Ramanujan's 8 — 14, 28, 42, 56, 71, and 113 digits by the seventh term. The rate is $\log_{10}(640320^3) - \log_{10}(1728) = 17.419 - 3.238 = 14.18$, which is the figure the literature quotes.
Ramanujan's is the $N = 58$ case. Chudnovsky's is built on 163 — the last Heegner number, which ch-ramanujan-verify.py already recovered by exhaustively counting reduced binary quadratic forms, and which wp82-k0-floor-verify.py used for the class-number floor under Volume XI. Three scripts written for three purposes are looking at one object from three sides.
The producing script failed three times before it passed, and the failures are more useful than the passes.
The square. The theta routine returned $k$, not $\alpha = k^2$ — $(\theta_2/\theta_3)^2$ where it should have been the fourth power. Caught because the closed form disagreed at the fifth digit, not because the code was re-read. Chapter R's own theta routine should be re-checked for the same square.
The display. A string slice str(x)[:46] silently ate the exponent, printing 6.5063772... for a number that was $6.5 \times 10^{-10}$, and made two correct numbers look like a catastrophe. Formatting is not cosmetic when the formatter is also the instrument.
The near-integer, twice. $g_{58}^{12}$ was asserted to be 19601, then 19602. It is neither. Asserting a near-integer as an integer is the failure this corpus keeps meeting, and the fix was not a looser tolerance but the exact algebraic form.
And one in the descendant. Chudnovsky was first coded with $640320^3/24$, which belongs to the binary-splitting arrangement and not to this one. It gained 12 digits a term and then stalled. The convergence rate is what caught it — 12.8 against the quoted 14.18.
sorryAx. A clean axiom report is not a reading of the statement: per R20, a theorem can assume its conclusion and still report clean. Follow the link before citing one as evidence.