In Mathworld's "*Pi Approximations*", (line 58), Weisstein mentions one by the mathematician Daniel Shanks that differs by a mere $10^{-82}$,

$$\pi \approx \frac{6}{\sqrt{3502}}\ln(2u)\color{blue}{+10^{-82}}\tag{1}$$

and says that $u$ is a product of four simple quartic units, but does not explicitly give the expressions. I managed to locate Shanks' *Dihedral Quartic Approximations and Series for Pi (1982)**Quartic Approximations for Pi* (1980) online before so,

$$u = (a+\sqrt{a^2-1})^2(b+\sqrt{b^2-1})^2(c+\sqrt{c^2-1})(d+\sqrt{d^2-1}) \approx 1.43\times10^{13}$$

where,

$$\begin{aligned} a &= \tfrac{1}{2}(23+4\sqrt{34})\\ b &= \tfrac{1}{2}(19\sqrt{2}+7\sqrt{17})\\ c &= (429+304\sqrt{2})\\ d &= \tfrac{1}{2}(627+442\sqrt{2}) \end{aligned}$$

(*Remark*: In Shank's paper, the expressions for $a,b$ are different since he didn't express them as squares.)

A small tweak to $(1)$ can *vastly* increase its accuracy to $10^{-161}$,

$$\pi \approx \frac{1}{\sqrt{3502}}\ln\big((2u)^6+24\big)\color{blue}{+10^{-161}}\tag{2}$$

I noticed the constant $u$ can also be expressed in terms of the *Dedekind eta function* $\eta(\tau)$ as,

$$u = \frac{1}{2}\left(\frac{\eta(\tfrac{1}{2}\sqrt{-3502})}{\eta(\sqrt{-3502})}\right)^4\approx 1.43\times 10^{13}$$

which explains why $24$ improves the accuracy. Note that the *class number* of $d = 4\cdot3502$ is $h(-d)=16$, and $u$ is an algebraic number of deg $16$. Mathworld has a list of class numbers of *d*. However, we can also use those with $h(-d)=8$ such as,

$$x = \frac{1}{\sqrt{2}}\left(\frac{\eta(\tfrac{1}{2}\sqrt{-742})}{\eta(\sqrt{-742})}\right)^2 \approx 884.2653\dots$$

which is a root of the 8th deg,

$$1 + 886 x + 1535 x^2 + 962 x^3 + 1628 x^4 - 962 x^5 + 1535 x^6 - 886 x^7 + x^8 = 0$$

** Question**:

*Analogous to $u$, how do we express $x$ as a product of*

**two**quartic units?