Ramanujan's series for $1/\pi$ has four numbers in it that look like they were found in a drawer. Three of them are the same number. The fourth can be recovered by anyone with a calculator.
Here is the thing itself. Ramanujan wrote it down in 1914 and did not say where it came from.
$$\frac{1}{\pi} \;=\; \frac{2\sqrt{2}}{9801}\sum_{n=0}^{\infty}\frac{(4n)!}{(n!)^4}\cdot\frac{1103 + 26390\,n}{396^{4n}}$$It works. Take the first term alone — $n = 0$, so the factorials are all 1 — and you get $2\sqrt{2}\cdot 1103/9801$, which is $\pi^{-1}$ to six decimal places. Take eight terms and you have sixty-odd digits. It is one of the fastest things anyone had written down for $\pi$ until the 1980s.
And it is full of numbers nobody can account for. 9801. 396. 1103. 26390. They have the texture of things found rather than derived. A student meeting this formula is entitled to ask where they came from, and the usual answer — modular forms — is true and unsatisfying, in the way that physics is a true and unsatisfying answer to why the sky is blue.
So let us take the four numbers one at a time and see how far we get. The honest answer turns out to be: three of them all the way down, and one of them most of the way, and then a wall that this chapter can describe precisely but not climb.
Start with the easiest, because it is not really there.
$396^4 = 24{,}591{,}257{,}856$. That is an ugly number. But watch:
Exactly, in integers. So the $396^{4n}$ in the denominator is $256^n \cdot 9801^{2n}$, and 9801 was already on the page. The only new thing 396 brings is that factor of $256^n$.
And $256^n$ is not new either. There is a standard identity sitting inside the factorial:
$$\frac{(4n)!}{(n!)^4} \;=\; 256^{\,n}\,\frac{(1/4)_n\,(1/2)_n\,(3/4)_n}{(n!)^3}$$The $256^n$ in the numerator and the $256^n$ in the denominator cancel. Write the series in that form and 396 disappears entirely:
$$\frac{1}{\pi} \;=\; \frac{2\sqrt{2}}{9801}\sum_{n=0}^{\infty}\frac{(1/4)_n(1/2)_n(3/4)_n}{(n!)^3}\cdot\frac{1103+26390\,n}{9801^{2n}}$$The series is a series in $1/9801^2$. Every 396 in the classical statement is $4\cdot 99$ wearing a fourth power, and it is there only because $(4n)!/(n!)^4$ is easier to compute than three Pochhammer symbols. Three numbers left.
Now the interesting one. Put the series down for a moment and go somewhere that has nothing to do with $\pi$.
Solve $x^2 - 29y^2 = \pm 1$ in integers. This is Pell's equation, it is older than calculus, and the smallest solution is found by running the Euclidean algorithm on $\sqrt{29}$. You get $x=5$, $y=1$, with $5^2 - 29 = -4$; in the ring of integers of $\mathbb{Q}(\sqrt{29})$ the right object is
$$\varepsilon \;=\; \frac{5+\sqrt{29}}{2}$$the fundamental unit. Cube it:
Square that:
There it is. 9801 is the rational part of the sixth power of the fundamental unit of $\mathbb{Q}(\sqrt{29})$.
And because $\varepsilon^6$ has norm $+1$, its conjugate is its inverse: $\varepsilon^{-6} = 9801 - 1820\sqrt{29}$. Add them and the irrational parts cancel:
$$\varepsilon^{6} + \varepsilon^{-6} \;=\; 19602 \;=\; 2\cdot 9801$$So 9801 is half a trace. It is the most natural rational number attached to $\varepsilon^6$ — the one you get by adding the unit to its own inverse and halving.
Why $\varepsilon^6$, and why 29? Because the series lives at $N = 58$, and $58 = 2\cdot 29$. The bridge is a Weber class invariant, $g_{58}$, which is defined out of theta functions and has no visible connection to Pell at all:
Compute the singular modulus $k_{58} = k(e^{-\pi\sqrt{58}})$ from theta functions, form $g_{58}^{24} = (1-\alpha)^2/(4\alpha)$ with $\alpha = k_{58}^2$, take the square root, and compare with $9801 + 1820\sqrt{29}$ computed from the Euclidean algorithm. They agree to 200 decimal places.
A transcendental construction on the left. Integer arithmetic on the right. This is complex multiplication doing what it always does: an analytic function, evaluated at one special point, lands on an algebraic number — and here that number is a unit.
∎So 9801 is not a number Ramanujan found. It is a number the field $\mathbb{Q}(\sqrt{29})$ hands you. Two numbers left.
Worth a paragraph, because it turns a piece of folklore into arithmetic.
People say the series is fast. The rate is exactly computable: each term shrinks by the factor $256/396^4$, which we now know is $1/9801^2$. So the number of correct digits per term is
$$2\log_{10} 9801 \;=\; 7.9825\ldots$$Eight digits a term. And the reason it is eight and not two is that 9801 is large, and 9801 is large because $\varepsilon^6$ is large, and $\varepsilon^6$ is large because $\mathbb{Q}(\sqrt{29})$ has a big regulator. The series converges fast for an arithmetic reason. Nothing about it is lucky.
(The measured rate between the first two terms is $8.37$, not $7.98$. The asymptotic rate is a ceiling the series climbs toward from below, because $c_{n+1}/c_n$ only reaches 256 in the limit — at $n=1$ it is 105. Stating the limit as though it described term one would be the sort of smoothing this corpus tries not to do.)
Now 26390. Factor it and the field is still there:
13 and 70 are the coefficients of $\varepsilon^3 = 70 + 13\sqrt{29}$. And $29\cdot 1820/2 = 26390$ as well, where 1820 is the $\sqrt{29}$-part of $\varepsilon^6$. That is suggestive, but a factorisation is not a derivation — you can factor anything, and this corpus has a rule about reading meaning into that.
So here is the statement that is not a factorisation. Take the whole coefficient of $n$ in the series, prefactor and all:
Both sides expand to $\sqrt{2}\cdot 52780/9801$ once you substitute $\tanh(6\log\varepsilon) = (\varepsilon^6-\varepsilon^{-6})/(\varepsilon^6+\varepsilon^{-6})$ and use $\varepsilon^6 \pm \varepsilon^{-6} \in \{3640\sqrt{29},\,19602\}$. It is an identity in rational arithmetic, not a numerical coincidence.
∎Read it out loud: the coefficient of $n$ is $\sqrt{58}$, times the hyperbolic tangent of six times the regulator. And $\tanh$ of anything large is very nearly 1, so the coefficient is very nearly $\sqrt{58} = 7.61577\ldots$ It misses by $5.2\times 10^{-9}$, and the miss is exactly $-2/(\varepsilon^{12}+1)$.
That is 26390 accounted for: it is $\sqrt{58}$ bent by the unit, and the bend is the whole difference between a formula that works and one that nearly does. One number left.
Write the series abstractly. Put
$$F=\sum_{n\ge 0} c_n x^n, \qquad D=\sum_{n\ge 0} n\,c_n x^n, \qquad c_n = \frac{(4n)!}{(n!)^4},\quad x=\frac{1}{396^4}$$so that Ramanujan's claim is $aF + bD = T$ with $a=1103$, $b=26390$, and
$$T \;=\; \frac{9801}{2\sqrt{2}\,\pi} \;=\; 1103.0000268319745\ldots$$Stop and look at that. $T$ is already 1103, to five decimal places, and you can compute $T$ on a calculator without knowing anything about the series. That is the first half of the answer.
The second half is a bound. $D$ is small — its leading term is $24x$, about $9.76\times 10^{-10}$ — so $|bD| < 1/2$ for every integer $b$ with
$$|b| \;<\; \frac{1}{2D} \;=\; 5.123\times 10^{8}$$For any such $b$, the required $a = (T - bD)/F$ sits within $1/2$ of $T$, and $T$ sits within $3\times 10^{-5}$ of the integer 1103. So $a$ is 1103 or $a$ is not an integer at all. There is no choice in it:
The worst deviation across that entire range is $0.488$ — under a half, and only just, at the ends. Fix $a=1103$ and the remaining equation determines $b$ outright: $(T - 1103F)/D = 26390$, exactly.
So: given only that the coefficients are integers of reasonable size, the pair $(1103, 26390)$ is the only pair there is. No modular forms were used. The bound is the argument.
What is proved is bounded uniqueness: the only integer pair with $|b|\le 5\times 10^8$. Unbounded uniqueness would follow from $F/D$ being irrational, and that cannot be tested here — the computed $F$ and $D$ are truncations, hence rational by construction, so a rationality test on them tests the truncation.
A live example of why that matters: $F/D$ prints as $1024635736.25$ and is not $1024635736.25$. It misses by $8\times 10^{-9}$. Eyeballing the digits would have concluded the wrong thing.
The last section assumed the coefficients are integers. That assumption is doing real work, and nothing above earns it. Ramanujan's theorem earns it; the modular theory produces the series with algebraic coefficients and then a separate argument makes them rational, and then integral. This chapter cites that and does not prove it.
Which leaves 1103 in a different position from 9801 and 26390. Those two were recovered — you can build them from $\varepsilon$ and nothing else. 1103 was only cornered.
But there is a signpost, and it is sharp. The elliptic alpha function $\alpha(N) = E'/K - \pi/4K^2$ is the object the whole theory of these series is built on. It is algebraic at every $N$. At $N=58$ it lies in $\mathbb{Q}(\sqrt{2},\sqrt{29})$, and on the basis $\{1,\sqrt{29},\sqrt{2},\sqrt{58}\}$ its expression is unique. Here it is:
and $8824 = 8\cdot 1103$, with 1103 prime.
∎1103 is inside the alpha function, multiplying $\varepsilon^6$. It is not a stranger to 9801 and 26390 after all — all three live in the same object.
The base rate matters here and the corpus requires it stated. This is not a search that turned something up. The expression of an element on a basis is unique, so there was one candidate and it was checked to 250 places. The only reading-in is writing 8824 as $8\cdot 1103$, and since $8824 = 2^3\cdot 1103$ with 1103 prime, that is the only factorisation available. What is not shown is that this is the mechanism rather than a consequence of one — which is exactly the open door. Running the derivation the other way, from $\alpha(58)$ to the coefficient, would make 1103 as derived as the other two now are.
Finally, two things this chapter declines to call identities, because they are not.
The first is the famous near-integer $e^{\pi\sqrt{58}}$, missing $396^4 - 104$ by two parts in $10^9$. The second says the ratio of Ramanujan's two coefficients is nearly $\pi\sqrt{58}$ — near to seven digits, and no further.
Both are the same phenomenon, and it is the phenomenon this whole chapter has been about: $e^{\pi\sqrt{58}}$ is close to an integer because $j(\sqrt{-58})$ is an algebraic integer of small degree, which is another way of saying the class group is small, which is the same fact that made $\varepsilon^6$ the right unit. The near-misses are shadows of the exactness, not competitors to it.
So it is worth saying plainly which is which. $g_{58}^{12} = \varepsilon^6$ — exact. $2\sqrt{2}\cdot 26390/9801 = \sqrt{58}\tanh(6\log\varepsilon)$ — exact. $396^4 = 256\cdot 9801^2$ — exact, in integers. $26390/1103 \approx \pi\sqrt{58}$ — not exact, and the corpus does not get to promote it.
The verification script records seven gaps. The two that matter:
Integrality is cited, not shown. Section 6 forces the pair given that it is integral. Nothing here proves it must be.
One value of $N$. Everything above is at $N=58$. The family has other members — $N = 22, 37, 142$ and more — and until the same descriptions are checked there, “this is the mechanism” rests on one data point. That is the next thing to do, and it is a clean piece of work: the same script, a loop over $N$.
What can be said now: of the four numbers in Ramanujan's series, one is an artifact of notation, two are the fundamental unit of $\mathbb{Q}(\sqrt{29})$ read two different ways, and the fourth is pinned within $3\times 10^{-5}$ by a calculation any reader can do — and then sits, unexplained but not unlocated, inside $\alpha(58)$.
Producing script: book7/ch-1103-and-26390-verify.py — 10 sections, 7 gaps, all checks at 250 places or better. Cross-references: Book IV · the Euclidean algorithm (where $\varepsilon^6 = 9801+1820\sqrt{29}$ is worked by hand), Book IV · modular equations and $\pi$ (where $g_{58}^{12}$ and its five-parts-per-million near-miss are the subject), Chapter R, Chapter $\phi$R.