Given the Dedekind eta function $\eta(\tau)$ with $\tau = \sqrt{-n}$. Define the Ramanujan $G_n$ and $g_n$ functions as,

$$G_n = 2^{-1/4}\frac{\eta^2(\tau)}{\eta(\tau/2)\,\eta(2\tau)}$$ $$g_n = 2^{-1/4}\frac{\eta(\tau/2)}{\eta(\tau)}$$ where $G_n$ and $g_n$ are for odd and even $n$, respectively.

There are exactly 9 primes $p = 4m+1$ such that $4p$ or $8p$ have class number 2 or 4, namely,

I. Class number 2

$$G^{24}_{5} \approx 2^3\sqrt5 $$ $$\color{blue}{g^{24}_{10}} \approx 2^4\cdot3^2\sqrt5 $$ $$G^{24}_{13} \approx 2^3\cdot3^2\cdot5\sqrt{13} $$ $$G^{24}_{37} \approx 2^3\cdot3^2\cdot5\cdot7^2\cdot29\sqrt{37} $$ $$\color{blue}{g^{24}_{58}} \approx 2^4\cdot3^4\cdot5\cdot7\cdot11^2\cdot13\sqrt{29} $$

II. Class number 4

$$G^{24}_{17} \approx 2^6\cdot5^2\sqrt{17} $$ $$\color{blue}{g^{24}_{34}} \approx 2^7\cdot3^5\cdot11\sqrt{17} $$ $$G^{24}_{73} \approx 2^6\cdot3^4\cdot5^2\cdot13\cdot17\cdot29\sqrt{73}$$ $$\color{blue}{g^{24}_{82}} \approx 2^8\cdot3^5\cdot5^2\cdot11\cdot17\cdot19\sqrt{41} $$ $$G^{24}_{97} \approx 2^6\cdot3^4\cdot5^2\cdot13\cdot17\cdot37\cdot41\sqrt{97} $$ $$G^{24}_{193} \approx 2^6\cdot3^4\cdot5^2\cdot13^2\cdot17\cdot41\cdot61\cdot73\cdot149\sqrt{193} $$

where the approximations are good to the nearest integer. Since,

$$\pi \approx \frac1{\sqrt{n}}\,\ln\big(2^6G^{24}_n\big) \\ \pi\approx \frac1{\sqrt{n}}\,\ln\big(2^6g^{24}_n\big)$$


$$\pi\sqrt{193} \approx 12\ln2 + 4\ln3 +\dots +\ln149 +\tfrac12\ln{193}$$ or a linear sum of the logarithms of a few small primes $ \leq 193$. Compare to other primes $p=4m+1$,

$$G^{24}_{89} \approx 23\cdot18253\cdot29347\sqrt{89} $$ $$G^{24}_{101} \approx 2\cdot5^2\cdot 1601407991\sqrt{101} $$ $$G^{24}_{109} \approx 2\cdot5\cdot 26279318873\sqrt{109} $$ and so on.

Q: So for those 9 primes, why is $G_n$ or $g_n$ highly factorable into a form involving just small primes $\leq n$?

  • $\begingroup$ Probably because its square can be expressed as a product over the $j(\frac{az+b}{d}),ad=n,b \bmod n$ $\endgroup$ – reuns Apr 21 at 17:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.