Anomalous Partial Quotients in the Continued Fraction of $\sqrt{\zeta(3)-S_N}$

Abstract

Let $S_N = \sum_{j=1}^N j^{-3}$ and $R_N = \zeta(3) - S_N$. The simple continued fraction of $\sqrt{R_N}$ has partial quotients of generic size $O(N)$. We prove that at the sequence of indices $N_k = (Q_{2k+1}-1)/2$, where $Q_{2k+1}$ are companion Pell numbers, the continued fraction begins \[ \sqrt{R_{N_k}} = \bigl[0;\; M_k-1,\; 1,\; 6M_k^3+12M_k-2,\; 1,\; \ldots\,\bigr], \] with $M_k = P_{2k+1}$ (Pell numbers), and the third partial quotient grows cubically while generic ones are linear. We determine all partial quotients through the fifth: \begin{align*} \PQ_0 &= M_k - 1, & \PQ_2 &= 6M_k^3 + 12M_k - 2, & \PQ_4 &= \Bigl\lfloor\frac{10M_k - 261}{261}\Bigr\rfloor, \PQ_1 &= 1, & \PQ_3 &= 1, & \PQ_5 &= \Bigl\lfloor\frac{261}{r_k}\Bigr\rfloor + \epsilon_k, \end{align*} where $r_k = (10M_k) \bmod 261$ satisfies the recurrence $r_{k+1} \equiv 6r_k - r_{k-1} \pmod{261}$, and $\epsilon_k = -1$ at the $k$ with $r_k \mid 261$ (the two residue classes $k \equiv 57, 62 \pmod{60}$), and $\epsilon_k = 0$ otherwise. All six formulas follow from the Euler--Maclaurin expansion of $1/\sqrt{R_{N_k}}$, carried to sufficient precision, combined with the Pell identity $Q_{2k+1}^2 - 2M_k^2 = -1$. The delicate first step, $\PQ_0 = M_k - 1$, is proved by rationalizing the irrational factor $\sqrt{2}$ in the Euler--Maclaurin expansion; we complement this proof with a heuristic derivation via Gosper's bihomographic continued-fraction algorithm that exposes the underlying mechanism. All claimed results have been formalized in LEAN with the aid of Aristotle.