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I found this very interesting so I tried to find a similar formula for the n-th digit of $\sqrt2$ in decimal. For that I used the following, already existing formula for the decimal expansion of $\sqrt2$ (from http://oeis.org/A002193):

$a(n)=-10 \cdot \lfloor 2^{-\frac{3}{2}+n} \cdot 5^{-2+n}\rfloor + \lfloor 2^{-\frac{1}{2}+n} \cdot 5^{-1+n} \rfloor$,

applying this on it:

$\lfloor x\rfloor =x-\frac{1}{2}+\frac{\sum _{k=1}^{\infty } \frac{\sin (2 k \pi x)}{k}}{\pi }$

and then following the same steps you did for base 2. This way I got the following formula:

$B_{10}(n) = \frac{9}{2}-\frac{10 \tan ^{-1}\left(\cot \left(\sqrt{2} \pi10^{n-2} \right)\right)}{\pi }+\frac{\tan ^{-1}\left(\cot \left(\sqrt{2} \pi 10^{n-1} \right)\right)}{\pi }$,

which was working as expected. At this point I had a suspicion for a generalized formula for the n-th digit in any base $b \geq 2$ :

$B_{b}(n) = \frac{b-1}{2}-\frac{b \tan ^{-1}\left(\cot \left(\sqrt{2} \pi \cdot b^{n-2} \right)\right)}{\pi }+\frac{\tan ^{-1}\left(\cot \left(\sqrt{2} \pi \cdot b^{n-1} \right)\right)}{\pi }$

which is the formula you found for $b=2$ and the one I found for $b=10$ and I also tried $b=16$ for Hexadecimal and it returned correct values. But at this point I had another suspicion, namely that you could generalize it so that you get the b-th base expansion for any positive real number $0 < u \in \mathbb{R}$:

$B_{b}(n) = \frac{b-1}{2}-\frac{b \tan ^{-1}\left(\cot \left( u \cdot \pi \cdot b^{n-2} \right)\right)}{\pi }+\frac{\tan ^{-1}\left(\cot \left(u \cdot \pi \cdot b^{n-1} \right)\right)}{\pi }$,

and indeed, this formula could return for example the binary and decimal expansions of $u=\pi$ or $u=\frac{1}{9}$. I am not sure how useful these formulae would be though because you need the value of $u$ to calculate the n-th digit with it.

I am interested in proving that the proportion of binary digits of $\sqrt{2}$ is exactly $\frac{1}{2}$, and I have worked on this for a long time. You wrote a great article using the Fourier series expansion for the fractional part function. I am looking at a simpler formula (I have yet to find one!) but have asked my question on StackExchange, here. For your convenience, my question is posted below:

Is it possible to find an expression of the form $$\lfloor 2^n \sqrt{2}\rfloor = \sum_{k=1}^r (\alpha_k + \beta_ki)\cdot\Big(a_k+b_k i\Big)^n, $$ where $i^2 = -1$ and $\alpha_k, \beta_k, a_k, b_k$ real numbers? Maybe with $r< 5$ if possible. I am trying to compute the mean value (average) of $\{ 2^n \sqrt{2}\}= 2^n \sqrt{2} - \lfloor 2^n \sqrt{2}\rfloor$ over $n=0, 1,2, \cdots$. Let us denote as $p$ this mean value: $p$ is the proportion of binary digits of $\sqrt{2}$ equal to one, and $p$ is widely believed to be equal to $\frac{1}{2}$.

Note that $$p=\lim_{n\rightarrow\infty} p_n, \mbox{ with } p_n = \frac{1}{n}\sum_{k=1}^n \Big( 2^k \sqrt{2} - \lfloor 2^k \sqrt{2}\rfloor\Big).$$ Assuming we have a simple closed form for $\lfloor 2^k \sqrt{2}\rfloor$ as suggested in the first formula, then we can have a closed form for $p_n$, involving a finite number of terms. I don't expect this to be true if you replace $\sqrt{2}$ by (say) $\pi$ or $\log 2$, but I would expect this to work with any quadratic irrational.

Of course you can always use the Fourrier series to represent the fractional part function, but I don't think it would be easy to handle: it involves an infinite number of terms ( $r=\infty$.)

Just checking: is this, then, some kind of spigot algorithm for sqrt(2)?

https://en.wikipedia.org/wiki/Spigot_algorithm

I think it is then a spigot algorithm, because you have extensions, like the famous one for Pi in base 16 which calculates the nth digit directly... Algorithm here can be read as function:

> A further refinement is an algorithm which can compute a single arbitrary digit, without first computing the preceding digits: an example is the Bailey-Borwein-Plouffe formula, a digit extraction algorithm for ? which produces hexadecimal digits.

https://en.wikipedia.org/wiki/Spigot_algorithm

Appendix: To simplify $B_n$ formula.

We have: $$B_n=\frac{i \left(-i \pi +2 \log \left(1-e^{-i 2^{-\frac{1}{2}+n} \pi }\right)-2 \log \left(1-e^{i 2^{-\frac{1}{2}+n} \pi }\right)-\log \left(1-e^{-i 2^{\frac{1}{2}+n} \pi }\right)+\log \left(1-e^{i 2^{\frac{1}{2}+n} \pi }\right)\right)}{2 \pi }$$

 sol2 = (I (-I \[Pi] + 2 Log[1 - E^(-I 2^(-(1/2) + n) \[Pi])] - 2 Log[1 - E^(I 2^(-(1/2) + n) \[Pi])] - Log[1 - E^(-I 2^(1/2 + n) \[Pi])] +  Log[1 - E^(I 2^(1/2 + n) \[Pi])]))/(2 \[Pi])
 sol5 = FullSimplify[Re[sol2] // ComplexExpand, Assumptions -> n \[Element] Integers]
 Table[sol5, {n, 1, 5}]

$\left\{\frac{\pi -4 \tan ^{-1}\left(\frac{\sin \left(\sqrt{2} \pi \right)}{1-\cos \left(\sqrt{2} \pi \right)}\right)+2 \tan ^{-1}\left(\frac{\sin \left(2 \sqrt{2} \pi \right)}{1-\cos \left(2 \sqrt{2} \pi \right)}\right)}{2 \pi },\frac{\pi -4 \tan ^{-1}\left(\frac{\sin \left(2 \sqrt{2} \pi \right)}{1-\cos \left(2 \sqrt{2} \pi \right)}\right)+2 \tan ^{-1}\left(\frac{\sin \left(4 \sqrt{2} \pi \right)}{1-\cos \left(4 \sqrt{2} \pi \right)}\right)}{2 \pi },\frac{\pi -4 \tan ^{-1}\left(\frac{\sin \left(4 \sqrt{2} \pi \right)}{1-\cos \left(4 \sqrt{2} \pi \right)}\right)+2 \tan ^{-1}\left(\frac{\sin \left(8 \sqrt{2} \pi \right)}{1-\cos \left(8 \sqrt{2} \pi \right)}\right)}{2 \pi },\frac{\pi -4 \tan ^{-1}\left(\frac{\sin \left(8 \sqrt{2} \pi \right)}{1-\cos \left(8 \sqrt{2} \pi \right)}\right)+2 \tan ^{-1}\left(\frac{\sin \left(16 \sqrt{2} \pi \right)}{1-\cos \left(16 \sqrt{2} \pi \right)}\right)}{2 \pi },\frac{\pi -4 \tan ^{-1}\left(\frac{\sin \left(16 \sqrt{2} \pi \right)}{1-\cos \left(16 \sqrt{2} \pi \right)}\right)+2 \tan ^{-1}\left(\frac{\sin \left(32 \sqrt{2} \pi \right)}{1-\cos \left(32 \sqrt{2} \pi \right)}\right)}{2 \pi }\right\}$

Now we need find a pattern.The first ingredient in the formula: We have: $$\left\{4 \tan ^{-1}\left(\frac{\sin \left(\sqrt{2} \pi \right)}{1-\cos \left(\sqrt{2} \pi \right)}\right),4 \tan ^{-1}\left(\frac{\sin \left(2 \sqrt{2} \pi \right)}{1-\cos \left(2 \sqrt{2} \pi \right)}\right),4 \tan ^{-1}\left(\frac{\sin \left(4 \sqrt{2} \pi \right)}{1-\cos \left(4 \sqrt{2} \pi \right)}\right),4 \tan ^{-1}\left(\frac{\sin \left(8 \sqrt{2} \pi \right)}{1-\cos \left(8 \sqrt{2} \pi \right)}\right),\tan ^{-1}\left(\frac{\sin \left(16 \sqrt{2} \pi \right)}{1-\cos \left(16 \sqrt{2} \pi \right)}\right)\right\}$$

 FindSequenceFunction[{1, 2, 4, 8, 16}, n]
 (*2^(-1 + n)*)

Finding the second ingredient in the formula is the same how I found at first:

 FindSequenceFunction[{2, 4, 8, 16, 32}, n]
 (*2^n*)

Substitution:

$$\frac{\pi -\left(4 \tan ^{-1}\left(\frac{\sin \left(n \sqrt{2} \pi \right)}{1-\cos \left(n \sqrt{2} \pi \right)}\right)\text{/.}\, n\to 2^{n-1}\right)+\left(2 \tan ^{-1}\left(\frac{\sin \left(n \sqrt{2} \pi \right)}{1-\cos \left(n \sqrt{2} \pi \right)}\right)\text{/.}\, n\to 2^n\right)}{2 \pi }$$

  FullSimplify[(\[Pi] - (4 ArcTan[Sin[n Sqrt[2] \[Pi]]/(1 - Cos[n Sqrt[2] \[Pi]])] /. n -> 2^(n - 1)) + (2 ArcTan[Sin[n Sqrt[2] \[Pi]]/(
  1 - Cos[n Sqrt[2] \[Pi]])] /. n -> 2^n))/(2 \[Pi])] // Expand

$$B_n = \frac{1}{2}-\frac{2 \tan ^{-1}\left(\cot \left(2^{-\frac{3}{2}+n} \pi \right)\right)}{\pi }+\frac{\tan ^{-1}\left(\cot \left(2^{-\frac{1}{2}+n} \pi \right)\right)}{\pi }$$

Yes, I found this formula.Is it new? Probably so. I searched in the web, but did not find a similar formula.

I tried to simplify, but I have not yet succeeded.