Proving the high precision of the series correct to at least half a billion digits

Question

I recently learned about this high-precision series. It is claimed that it is correct to at least half a billion digits. I am curious to know how it works.

$$ \sum_{n=1}^\infty\frac{\left\lfloor n e^{\frac\pi3\sqrt{163}}\right\rfloor}{2^n} = 1280640 $$

• It reminds me of Ramanujan's constant, $$ e^{\pi\sqrt{163}}\approx640320^3+744 $$ which is an almost integer.
• Using this approximation and $$ \left\lfloor n\sqrt[3]{640320^3+744}\right\rfloor\approx 640320n, $$ It can be shown that the sum is very close to $1280640$.
• But my question is, how is it so accurate to at least half a billion digits?
• A floor function is involved in it, increasing its accuracy and making it challenging to prove that it is so accurate.

Any proof demonstrating its accuracy is highly appreciated. Thank you!

Answer 1

Define the real valued function

$$ f(x) := \sum_{n=1}^\infty\frac{\lfloor n x\rfloor}{2^n}. \tag1 $$

Since $ \lfloor x+k\rfloor = k+\lfloor x\rfloor $ for all integer $k$, this implies

$$ f(k+x) = \sum_{n=1}^\infty\frac{\lfloor n (k+x)\rfloor}{2^n} = \sum_{n=1}^\infty\frac{nk+\lfloor n x\rfloor}{2^n} = 2k+f(x). \tag2 $$

If $0<x$ and $0<n$ integer, then $0\le\lfloor nx\rfloor\le nx.$

If $0<x<\frac1k$ and $0<n<k$ integer, then $0=\lfloor nx\rfloor$. This implies

$$ 0 \le f(x)\le\sum_{n=k}^\infty\frac{\lfloor n x\rfloor}{2^n} \le \sum_{n=k}^\infty\frac{n x}{2^n} = 2x\frac{1+k}{2^k}. \tag3 $$

Define the constants

$$ c := e^{\frac\pi3\sqrt{163}}, \,\, s := 640320, \,\, r := c-s \approx 6\times 10^{-10}<\frac1{10^9}. \tag4 $$

Now use equation $(2)$ to get

$$ f(c) = 2s + f(r) = 1280640 + f(r). \tag5 $$

Use equation $(3)$ to bound $f(r)$ as

$$ 0 < f(r) < 2r\frac{10^9}{2^{10^9}}<\frac2{2^{10^9}}. \tag6 $$

More precisely

$$ \frac1r = 1653264929.0322\dots, \,\, \log_{10}\frac1{f(r)}=497682334.4082\dots \tag7 $$

which is almost half a billion. This answers the question

But my question is, how is it so accurate to at least half a billion digits?