No is not algorithm, it's better than that. It simple, it's a formula, or function.

An example:

To start calculating binary digits from, say, the 8th place only need substitute for n=8,9,10,11,12,13,14,15....That it's all.

formula =1/2 - (2 ArcTan[Cot[2^(-(3/2) + n) \[Pi]]])/\[Pi] + ArcTan[Cot[2^(-(1/2) + n) \[Pi]]]/\[Pi]

sol = Table[formula, {n, 8, 15}] // N // Round

${1, 0, 0, 0, 0, 0, 1, 0}$

and say from n=1....15:

 sol2 = Table[formula, {n, 1, 15}] // N // Round

${1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0}$

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 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

Hey, while trying to simplify the equation further I was able to get this:

The $n$-th digit in base $b$ of a number $u$ can be calculated as:

$$ B_b(n):=\frac{b-1}{2}-\frac{b\cdot\tan^{-1}(\cot(u\cdot\pi\cdot b^{n-1}))}{\pi}+\frac{\tan^{-1}(\cot(u\cdot\pi\cdot b^{n}))}{\pi} $$

Given that $\tan^{-1}(\cot(x))=\frac{\pi}{2}-(x \mod \pi)$ this can be simplified to: $$ B_b(n):=\frac{b-1}{2}-\frac{b\cdot(\frac{\pi}{2}-(u\cdot\pi\cdot b^{n-1}\mod \pi))}{\pi}+\frac{\frac{\pi}{2}-(u\cdot\pi\cdot b^{n}\mod \pi)}{\pi} $$

Given that $\frac{x\mod a}{a}=\frac{x}{a}-\lfloor\frac{x}{a}\rfloor$ this can be further simplified to:

$$ \begin{align} B_b(n):&=\frac{b-1}{2}-b\cdot(\frac{1}{2}-(u\cdot b^{n-1}-\lfloor u\cdot b^{n-1}\rfloor)) + \frac{1}{2}-u\cdot b^{n}+\lfloor u\cdot b^{n}\rfloor \ &= \frac{b-1}{2}-\frac{b}{2}+b\cdot u\cdot b^{n-1}-b\cdot\lfloor u\cdot b^{n-1}\rfloor + \frac{1}{2}-u\cdot b^{n}+\lfloor u\cdot b^{n}\rfloor \ &= -b\cdot\lfloor u\cdot b^{n-1}\rfloor +\lfloor u\cdot b^{n}\rfloor \ &= \lfloor u\cdot b^{n}\rfloor -b\cdot\lfloor u\cdot b^{n-1}\rfloor \ \end{align} $$

This one is nice and simple but I still don't think it would be useful