$e^{\pi\sqrt N}$ is very close to an integer for some smallish $N$s. What about $\pi^{e\sqrt N}$?

Question

Heegner numbers (1, 2, 3, 7, 11, 19, 43, 67, 163 - let's use symbol $H_n$) are know for peculiar property that $e^{\pi\sqrt{H_n}}$ are almost integers:

$$e^{\pi \sqrt{19}} \approx 96^3+744-0.22$$ $$e^{\pi \sqrt{43}} \approx 960^3+744-0.00022$$ $$e^{\pi \sqrt{67}} \approx 5280^3+744-0.0000013$$ $$e^{\pi \sqrt{163}} \approx 640320^3+744-0.00000000000075$$

(given that they are all less than 200, it goes far beyond "chance" and "randomness")

Even stranger, related to the above:

$$19 = 3 \cdot 2 \cdot 3+1$$

$$43 = 7 \cdot 2 \cdot 3+1$$

$$67 = 11 \cdot 2 \cdot 3+1$$

$$163 = 27 \cdot 2 \cdot 3+1$$

and

$$96^3 =(2^5 \cdot 3)^3$$

$$960^3=(2^6 \cdot 3 \cdot 5)^3$$

$$5280^3=(2^5 \cdot 3 \cdot 5 \cdot 11)^3$$

$$640320^3=(2^6 \cdot 3 \cdot 5 \cdot 23 \cdot 29)^3$$

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Related to this, I would like to know:

Are there natural numbers N (of fairly similar range, so, lets say under 500) that produce "almost" integers in the expression $\pi^{e\sqrt{N}}$?

If yes, do they have other interesting properties, like Heegner numbers do?

If not, all right, one more reason to appreciate Heegners. :)

Answer 1

Up to 100000, the 10 best $N$ such that $e^{\pi\sqrt{N}}$ is almost an integer. The error $\delta$ is given such that the nearest integer is at $10^{\delta}$ from the result.

$$ \begin{array}{c|c} N & \delta \\\hline 163 & -12.12\\ 4\cdot163 & -9.79\\ 9\cdot163 & -8.01\\ 58 & -6.75\\ 16\cdot163 & -6.51\\ 67 & -5.87\\ 22905 & -5.61\\ 95041 & -5.55\\ 54295 & -5.37\\ 25\cdot163 & -5.2\\ \end{array} $$

As you can see, no $N$ beats 163 up to 100000. (For N = 4 x 163.)

For $\pi^{e\sqrt{N}}$, the behaviour is much more regular and you obtain :

$$ \begin{array}{c|c} N & \delta \\\hline 66972 & -5.03 \\ 85516 & -5.01 \\ 53204 & -4.91 \\ 46665 & -4.9 \\ 50237 & -4.8 \\ 93909 & -4.53 \\ 52970 & -4.4 \\ 10024 & -4.32 \\ 84702 & -4.17 \\ 6814 & -4.17 \\ \end{array}$$

So, it seems there is something strange in $e^{\pi\sqrt{N}}$ that makes that question interesting !

Answer 2

The phenomenon with the Heegner numbers can be generalized to,

$$e^{\pi\sqrt{-d}/a}\tag1$$

with discriminants $d=b^2-4ac$, of the quadratic,

$$P(n) = an^2+bn+c\tag2$$

These $d$ have very interesting properties.

I. Connection to prime-generating polynomials:

I'm sure you are familiar with Euler's,

$$P(n) =n^2+n+41\tag3$$

However, there are other optimum prime-generating polynomials with $a\neq1$,

$$P(n) =2n^2+29\tag4$$

$$P(n) =2n^2+2n+19\tag5$$

$$P(n) =3n^2+3n+23\tag6$$

$$P(n) =4n^2+163\tag7$$

$$P(n) =6n^2+6n+31\tag8$$

and others. Using the values of their $a,d$ into $(1)$, one gets,

$$\begin{aligned} &e^{\pi\sqrt{163}/1} = 640320^3 +743.999999\dots\\ &e^{\pi\sqrt{232}/2} = e^{\pi\sqrt{58}} = 396^4 -104.0000001\dots\\ &e^{\pi\sqrt{148}/2} = e^{\pi\sqrt{37}} = (84\sqrt{2})^4 +103.99997\dots\\ &e^{\pi\sqrt{267}/3} = 300^3 + 41.99997\dots\\ &e^{\pi\sqrt{10432}/4} = e^{\pi\sqrt{4\cdot163}} = (640320^3+744)^2 - 2\cdot \color{blue}{196883}.99999\dots\\ &e^{\pi\sqrt{708}/6} = 1060^2 + 9.99992\dots \end{aligned}$$

and so on.

1. What's 196884? (OEIS)
2. And here why $\log(196883) < 4\pi,\,$ or approx $12.19 < 12.56,$ is important to quantum gravity. (Huh?)

More on these prime-generating polynomials here.

II. Connection to pi formulas:

In addition, each one of these integer approximations can be used in a Ramanujan-Sato pi formula. The most famous of course, is with $d = 4\cdot58$,

$$\frac{1}{\pi} = \frac{2 \sqrt 2}{99^2} \sum_{k=0}^\infty \frac{(4k)!}{k!^4} \frac{58\cdot455k+1103}{396^{4k}}$$

III. Connection to Pell equations:

Furthermore, they are also connected to Pell equations. For example, the fundamental solution to,

$$x^2-3\cdot163y^2 = 1$$

$$x,\;y = 7592629975,\;343350596$$

hence the fundamental unit,

$$U =x+y\sqrt{489} =7592629975+343350596\sqrt{489} = \big(35573\sqrt{3}+4826\sqrt{163}\big)^2$$

Then,

$$\Big(3\sqrt{3}\big(U^{1/2}-U^{-1/2}\big)+6\Big)^3 = 640320^3$$

You can read more in this MO post. There is a lot more to $e^{\pi\sqrt{n}}$ than being near-integers, it seems.