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I read somewhere that

$$e^{\pi\sqrt{163}}$$

is almost an integer and strangely enough this isn't just a random coincidence but rather there exists some general theory

http://en.wikipedia.org/wiki/Heegner_number

behind the occurences of these almost integers (and their relation to other areas of number theory)

Surely there are many other strange identities such as:

$$\sqrt{2} \approx \frac{3}{5} + \frac{\pi}{7 -\pi}$$

I'm guessing that this "coincidence" is probably similar to the earlier example a special case of some general theory that relates rational expressions of pi to algebraic integers.

Can someone point me in the right direction if not explain it here itself?

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You mean $e^{\pi\sqrt{163}}$. Is there a reason for your guess? – anon Feb 5 '14 at 4:17
    
see: [ wolframalpha.com/input/?i=Heegner+number ] verbatim: The Heegner numbers have a number of fascinating connections with amazing results in prime number theory. In particular, the j-function provides stunning connections between e, pi, and the algebraic integers. They also explain why Euler's prime-generating polynomial n^2-n+41 is so surprisingly good at producing primes. – janmarqz Feb 5 '14 at 4:21
1  
Relevant: Why is $e^{\pi \sqrt{163}}$ almost an integer? – Ben Feb 5 '14 at 5:44
    
Your identity may be rewritten as $$\pi\approx\frac{392-175\sqrt{2}}{46}\approx3.1415(7)$$ – Jaume Oliver Lafont Jan 22 at 1:01
    
You can get more correct decimals using less digits: $$\pi\approx\frac{192-98\sqrt{2}}{17}\approx3.141592(4)$$ – Jaume Oliver Lafont Jan 22 at 1:10

The approximation $$\sqrt{2} \approx \frac{3}{5} + \frac{\pi}{7 -\pi}$$

may be rewritten as $$\sqrt{2}-\frac{3}{5} \approx \frac{1}{\frac{7}{\pi} -1}$$

After some manipulation, this is found to be equivalent to $$\pi\approx\frac{392-175\sqrt{2}}{46},$$ so, at least for this case, some theory relating $\pi$ to algebraic integers would suffice.

A possible useful direction is shown by the following series, which is related to a similar approximation to $\pi$:

$$\sum_{k=0}^{\infty} \frac{15!(k+1)}{(8k+1)_{15}}=\frac{15}{8}\left(1716-7\left(99\sqrt{2}-62\right)\pi\right)\approx 1$$

where $(a)_n$ is a rising factorial or Pochhammer symbol $a\times(a+1)\times...\times (a+n-1)$.

(see http://math.stackexchange.com/a/1657416/134791)

This gives the approximation

$$\pi \approx \frac{3676}{15(99\sqrt{2}-62)}=\frac{1838(62+99\sqrt{2})}{118185}$$

with eight correct decimal digits.

A general series that might provide an explanation for this approximation, as well as others of the form $a+b\sqrt{2}$ for rational $a$ and $b$, is given by

$$\sum_{k=0}^\infty \frac{c}{\prod_{i=1}^{7}((8k+i)(8k+16-i))^{w_i}} \approx 1,$$ with constant $c$ and binary weighting exponents $w_i$ taking values either $0$ or $1$. The example provided above is the particular case $w_i=1$ for all $i$ from $1$ to $7$.

The numerator $c$ may be set by letting the first term of the series equal $1$.

$$\sum_{k=0}^\infty \prod_{i=1}^7 \left(\frac{i(16-i)}{(8k+i)(8k+16-i)}\right)^{w_i} \approx 1,$$

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I don't see how this is an answer to the question. – Gerry Myerson Mar 29 at 8:42
    
@GerryMyerson This explains an approximation of the same form as the one in the question, so hopefully it points in some useful direction, as it was asked. I cannot discard that a similar series with a denominator of the form $\prod_{i=1}^{15} (8k+i)^{w_i}$, where coefficients $w_i$ take the value $0$ or $1$, directly explains the approximation. – Jaume Oliver Lafont Mar 29 at 11:47
    
How is $\pi\approx a+b\sqrt2$ of the same form as $\pi/(a-\pi)\approx b+\sqrt2$? – Gerry Myerson Mar 29 at 21:46
1  
@GerryMyerson They are of the same form, but $a$ and $b$ are not the same. Solving for $\pi$ in $$\frac{\pi}{a-\pi} \approx b+\sqrt{2}$$ leads to $$\pi\approx a'+b'\sqrt{2}$$. If I made no mistake [here](math.stackexchange.com/questions/664175/…) $a'=\frac{392}{46}=\frac{196}{23}$ and $b'=\frac{175}{46}$ from the approximation by the OP. – Jaume Oliver Lafont Mar 30 at 7:41

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