A negative result, cleanly stated. Apéry’s analog circuit for $\zeta(3)$ uses a beautiful trick: read the target out of the ratio of two power-series companions to the same Picard–Fuchs ODE at the conifold. The natural attempt to port this to Ramanujan’s $\pi$ series fails — and not because the construction is hard. It fails because the Ramanujan Picard–Fuchs operator’s formal-power-series kernel is exactly one-dimensional, so a second companion to ratio against does not exist. This post records the obstruction and what actually does work for $\pi$.

The temptation

Apéry’s $\zeta(3)$ analog circuit, in the framing of an earlier post, hinges on the following observation. The Picard–Fuchs operator

(third-order, with regular singular points at $z = 0$, the conifold $z = 1$, and $\infty$) admits two linearly independent formal-power-series solutions at $z = 0$:

where $A_n$ are the Apéry numbers, $B_n$ are the second Apéry sequence (starting $B_0 = 0$, $B_1 = 6, \dots$). The miracle Apéry exploited is that both are integer (or near-integer) sequences satisfying the same three-term recurrence

Because the recurrence is three-term, the space of formal-series solutions is two-dimensional, and $A, B$ form a basis. At the conifold,

a fact that turns into an exponentially-converging analog readout: a PIVP coordinate $\zeta$ tracking $B''/A''$ approaches $\zeta(3)$ as the state is driven toward the conifold.

Ramanujan’s 1914 identity for $1/\pi$,

is built from the holonomic coefficient sequence $a_k$, which is the power-series coefficient of $\,_3F_2(1/4, 1/2, 3/4;\, 1, 1;\, 256 z)\,$ at $z = 0$. It satisfies a third-order Picard–Fuchs ODE

with regular singular points at $z = 0$, the conifold $z_c = 1/256$, and $\infty$.

The natural temptation: hunt for a second power-series companion $B(z) = \sum_n b_n z^n$, $\,P_{\text{Ram}}[B] = 0\,$, with $\,B''(z_c)/A''(z_c)\,$ a rational multiple of $\pi$. Then read $\pi$ out of the ratio in the same way Apéry reads $\zeta(3)$. Plug the construction into the BAC framework, get a fast analog computation of $\pi$.

The obstruction

Theorem. The formal-power-series kernel of $P_{\text{Ram}}$ is one-dimensional, spanned by $A(z)$.

The proof is short. Substitute $B(z) = \sum_n b_n z^n$ into $P_{\text{Ram}}$ and read off the coefficient of $z^m$:

Setting this to zero gives the recurrence

which is two-term: $b_{m+1}$ is determined by $b_m$ alone, with no contribution from $b_{m-1}$. Hence the formal-series kernel is spanned by a single sequence, recovered up to a scalar by $b_0$, which is just $a_m$ up to that scalar. $\square$

Compare to Apéry’s three-term recurrence: there, $b_{m+1}$ depends on both $b_m$ and $b_{m-1}$, so the kernel is two-dimensional, and a second companion exists.

The third-order Ramanujan PF does have three linearly independent solutions on the punctured disk $0 < |z| < 1/256$ — the missing two are not formal power series. They involve $\log z$ and $z \log^2 z$ terms, the higher Frobenius solutions at the maximally unipotent monodromy point $z = 0$. They cannot be encoded as $\,\sum b_n z^n\,$ with rational $b_n$, and so they are not directly usable as PIVP companions in the Apéry style.

What about the log-Frobenius companion?

Subtract the log term off and what remains is a power series:

Could $\widetilde B'' / A''$ at the conifold give $\pi$?

A high-precision check (mp.dps = 100, $N = 2000$, $z = (1 - 10^{-7}) \cdot z_c$) shows the conifold ratio converges to $8 \log 2 = \log(1/z_c)$. A trivial constant — the conifold’s own logarithmic scale, with no $\pi$ content. The log-Frobenius companion encodes only the conifold geometry, not period information about $\pi$.

This is consistent with the structural picture. The conifold of $P_{\text{Ram}}$ is a true singularity of the operator — $\pi$ enters only through the value of $A$ at a CM (complex multiplication) point inside the disk, not through any conifold limit ratio.

What survives: read against the large quantity

Apéry’s circuit is fast because the inverter target $\zeta(3) \approx 1.20$ is order-1, so the natural inverter relaxation rate is order-1 too. For Ramanujan, the natural target is

and Ramanujan’s identity rearranges to $\pi = 9801 / (2\sqrt 2 M_\infty)$.

Two ways to read out $\pi$ from this:

(A) Encode $1/\pi$ first, then apply the inverter $\dot P = 1 - (1/\pi) P$. Fixed point $\pi$, asymptotic rate $1/\pi \approx 0.318$.

(B) Encode $M_\infty$ directly (it’s the natural state of the Ramanujan series circuit), then apply the inverter $\dot P = 9801 - 2\sqrt 2\, M\, P$. Same fixed point $\pi$, asymptotic rate $2 \sqrt 2 M_\infty = 9801 / \pi \approx 3120$.

The two ODEs differ by the constant factor 9801: $9801 - 2\sqrt 2 M P = 9801 \cdot (1 - (2 \sqrt 2 M / 9801) P)$, so (B) is just (A) with time rescaled by 9801. In abstract PIVP this is a free trivial rescaling. In the Bounded Analog Complexity time-modulus — which counts time in units of the natural $\tau$ of the encoding ODE — it is a real factor-9801 reduction in time-modulus, paid for only by absorbing rational constants into the ODE coefficients.

Combined with the rate-$k$ logistic inverter from the previous post, the total relaxation rate becomes

The two speedups stack multiplicatively. There is no third obvious algebraic gain on top: an exhaustive check (Frobenius parameter variation $\partial_\rho^j M(z_0, 0)$ for $j = 1, 2, 3, 4$, PSLQ against a basis of $\{1, \log 2, \log 99, \log 396, \pi^2, \zeta(3), \psi(k/4) + \gamma\}$ to coefficient bound $10^{10}$) finds no closed-form expression for the variation derivatives, so they yield no further $\pi$-companion via algebraic rearrangement.

What the obstruction actually says

There is a tendency to view a two-line analog construction as something like a free lunch: write down the target as a generating-function value and trust that the right circuit will fall out. The Ramanujan-vs-Apéry gap is a useful corrective.

For $\zeta(3)$, the ratio readout exists because the Picard–Fuchs operator has a coincidence: its three-term recurrence makes the formal-series kernel two-dimensional, and the second basis element happens to be near-integer. This is special.

For $\pi$ via Ramanujan, the analogous coincidence does not hold: the recurrence is two-term, the kernel is one-dimensional, and the “natural place where $\pi$ lives” is a CM-point period of the single solution $A$, not a conifold ratio of two of them.

The result is that $\pi$’s analog complexity is set by the value $M_\infty \approx 1103$ — the rescaling factor, not the conifold geometry — and the only structural gains available are constant-scaling rearrangements of that value plus the rate-$k$ logistic inverter on top.

The 9801$\times$ speedup from rerouting the inverter is not just a nicety. It is the entire structural margin Ramanujan’s identity gives the analog computer over the naive “compute $1/\pi$ and invert it” approach, and it is a margin that exists because $1103 + 26390 k$ has the magic prefactor 9801, not because the Picard–Fuchs ODE has any second power-series solution to ratio against.

Code: experiments/ramanujan_apery_pi_v2.py (formal proof of the 2-term recurrence), experiments/log_frobenius_high_prec.py (the $8 \log 2$ check), experiments/ramanujan_subtraction.py (the 9801× speedup), experiments/ramanujan_variation_residual.py (Frobenius variation, PSLQ search). Written as part of an overnight investigation, 2026-04-24.