Fourth edit: I decided to start an infinite formulas challenge based on the formulas below to be interpreted as music:
#InfinitePiChallenge:
Rules:
https://www.reddit.com/r/algorithmicmusic/comments/10cifzg/a_little_challenge/
Here is my interpretation of this challenge: https://www.youtube.com/watch?v=RdCDsSfe2_E
Third edit:
For those interested in the Sagemath-code to produce your own formula, given three natural numbers $x<y<z$, it can be found here. I am sharing those formulas in public domain, for the benefit of all, if there is any:
https://github.com/githubuser1983/generate_formulas_for_pi/blob/main/formulas_for_pi.ipynb
First edit: I have found a method which allows one to plug in some numbers $a,b,c$ in this formula and get formulas for $\pi$:
$$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( a \right )^{2k+1}+\left ( b \right )^{2k+1}+\left ( c \right )^{2k+1}}{4^k(2k+1)}$$
Some formulas generated this way are:
$$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( -\frac{4}{65} \, \sqrt{13} \sqrt{5} \right )^{2k+1}+\left ( \frac{17}{26} \, \sqrt{2} \right )^{2k+1}+\left ( \frac{9}{130} \, \sqrt{13} \sqrt{5} \sqrt{2} \right )^{2k+1}}{4^k(2k+1)}$$
$$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( -\frac{261}{47965} \, \sqrt{181} \sqrt{53} \sqrt{2} \right )^{2k+1}+\left ( \frac{2071}{47965} \, \sqrt{53} \sqrt{5} \sqrt{2} \right )^{2k+1}+\left ( \frac{1309}{47965} \, \sqrt{181} \sqrt{5} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( -\frac{931}{96050} \, \sqrt{113} \sqrt{85} \right )^{2k+1}+\left ( \frac{2061}{19210} \, \sqrt{85} \right )^{2k+1}+\left ( \frac{1781}{19210} \, \sqrt{113} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( -\frac{19}{2210} \, \sqrt{85} \sqrt{65} \sqrt{2} \right )^{2k+1}+\left ( \frac{32}{1105} \, \sqrt{85} \sqrt{13} \right )^{2k+1}+\left ( \frac{53}{2210} \, \sqrt{65} \sqrt{13} \sqrt{2} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( -\frac{96}{12665} \, \sqrt{149} \sqrt{85} \right )^{2k+1}+\left ( \frac{437}{25330} \, \sqrt{85} \sqrt{17} \sqrt{2} \right )^{2k+1}+\left ( \frac{351}{25330} \, \sqrt{149} \sqrt{17} \sqrt{2} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( -\frac{283}{3770} \, \sqrt{145} \right )^{2k+1}+\left ( \frac{1037}{3770} \, \sqrt{13} \right )^{2k+1}+\left ( \frac{413}{18850} \, \sqrt{145} \sqrt{13} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( -\frac{464}{3485} \, \sqrt{41} \right )^{2k+1}+\left ( \frac{1161}{6970} \, \sqrt{17} \sqrt{2} \right )^{2k+1}+\left ( \frac{889}{34850} \, \sqrt{41} \sqrt{17} \sqrt{2} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( -\frac{704}{40001} \, \sqrt{181} \sqrt{13} \right )^{2k+1}+\left ( \frac{3781}{80002} \, \sqrt{17} \sqrt{13} \sqrt{2} \right )^{2k+1}+\left ( \frac{925}{80002} \, \sqrt{181} \sqrt{17} \sqrt{2} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( -\frac{320}{42601} \, \sqrt{145} \sqrt{113} \right )^{2k+1}+\left ( \frac{697}{85202} \, \sqrt{113} \sqrt{65} \sqrt{2} \right )^{2k+1}+\left ( \frac{3069}{426010} \, \sqrt{145} \sqrt{65} \sqrt{2} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( -\frac{18751}{1380634} \, \sqrt{113} \sqrt{41} \right )^{2k+1}+\left ( \frac{10323}{1380634} \, \sqrt{149} \sqrt{113} \right )^{2k+1}+\left ( \frac{17475}{1380634} \, \sqrt{149} \sqrt{41} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( -\frac{21131}{3505970} \, \sqrt{181} \sqrt{65} \sqrt{2} \right )^{2k+1}+\left ( \frac{25023}{3505970} \, \sqrt{149} \sqrt{65} \sqrt{2} \right )^{2k+1}+\left ( \frac{10272}{1752985} \, \sqrt{181} \sqrt{149} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( -\frac{12879}{2152090} \, \sqrt{181} \sqrt{145} \right )^{2k+1}+\left ( \frac{27721}{2152090} \, \sqrt{145} \sqrt{41} \right )^{2k+1}+\left ( \frac{24769}{2152090} \, \sqrt{181} \sqrt{41} \right )^{2k+1}}{4^k(2k+1)}$$
Other formulas are:
$$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( -\frac{1}{21} \, \sqrt{7} \sqrt{3} \right )^{2k+1}+\left ( \frac{4}{21} \, \sqrt{7} \sqrt{3} \right )^{2k+1}+\left ( \frac{2}{3} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( -\frac{1}{26} \right )^{2k+1}+\left ( \frac{3}{26} \, \sqrt{13} \sqrt{3} \right )^{2k+1}+\left ( \frac{3}{26} \, \sqrt{13} \sqrt{3} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( \frac{1}{133} \, \sqrt{19} \sqrt{3} \right )^{2k+1}+\left ( \frac{31}{399} \, \sqrt{21} \sqrt{3} \right )^{2k+1}+\left ( \frac{5}{133} \, \sqrt{21} \sqrt{19} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( -\frac{1}{91} \, \sqrt{13} \sqrt{7} \right )^{2k+1}+\left ( \frac{9}{91} \, \sqrt{13} \sqrt{7} \right )^{2k+1}+\left ( \frac{3}{7} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( \frac{1}{399} \, \sqrt{19} \sqrt{7} \right )^{2k+1}+\left ( \frac{4}{57} \, \sqrt{21} \sqrt{7} \right )^{2k+1}+\left ( \frac{10}{399} \, \sqrt{21} \sqrt{19} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( -\frac{1}{273} \, \sqrt{21} \sqrt{13} \right )^{2k+1}+\left ( \frac{16}{273} \, \sqrt{21} \sqrt{13} \right )^{2k+1}+\left ( \frac{4}{13} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( -\frac{31}{546} \, \sqrt{13} \sqrt{7} \right )^{2k+1}+\left ( \frac{109}{546} \, \sqrt{7} \sqrt{3} \right )^{2k+1}+\left ( \frac{73}{546} \, \sqrt{13} \sqrt{3} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( -\frac{4}{133} \, \sqrt{19} \sqrt{7} \right )^{2k+1}+\left ( \frac{9}{133} \, \sqrt{19} \sqrt{7} \right )^{2k+1}+\left ( \frac{6}{7} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( -\frac{37}{798} \, \sqrt{21} \sqrt{3} \right )^{2k+1}+\left ( \frac{101}{798} \, \sqrt{19} \sqrt{3} \right )^{2k+1}+\left ( \frac{25}{798} \, \sqrt{21} \sqrt{19} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( -\frac{223}{5187} \, \sqrt{21} \sqrt{13} \right )^{2k+1}+\left ( \frac{311}{5187} \, \sqrt{19} \sqrt{13} \right )^{2k+1}+\left ( \frac{235}{5187} \, \sqrt{21} \sqrt{19} \right )^{2k+1}}{4^k(2k+1)}$$
My questions are:
Q1) Are these formulas known or the method to generate them known?
Q2) Is any of this formula of some usage to someone? :-) Whatever that means.
Q3) Also if anyone knows of a way to further "simplify" the $a^{2k+1}+b^{2k+1}+c^{2k+1}$ to get some "nicer" formulas, that would also be nice to share.
Thanks for your help.
If there is interest, I can share the code and the method.
Edit: Here are the "top ten" formulas sorted by "velocity of convergence" to $\pi$:
$$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( \frac{1}{4} \right )^{2k+1}+\left ( \frac{1}{4} \, \sqrt{3} \sqrt{2} \right )^{2k+1}+\left ( \frac{1}{4} \, \sqrt{3} \sqrt{2} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( \frac{2}{15} \, \sqrt{5} \sqrt{3} \right )^{2k+1}+\left ( \frac{1}{6} \, \sqrt{3} \right )^{2k+1}+\left ( \frac{3}{10} \, \sqrt{5} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( \frac{1}{6} \, \sqrt{3} \right )^{2k+1}+\left ( \frac{2}{15} \, \sqrt{5} \sqrt{3} \right )^{2k+1}+\left ( \frac{3}{10} \, \sqrt{5} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( \frac{1}{8} \, \sqrt{3} \sqrt{2} \right )^{2k+1}+\left ( \frac{5}{28} \, \sqrt{7} \right )^{2k+1}+\left ( \frac{3}{28} \, \sqrt{7} \sqrt{3} \sqrt{2} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( \frac{1}{5} \, \sqrt{5} \right )^{2k+1}+\left ( \frac{1}{10} \, \sqrt{5} \sqrt{2} \right )^{2k+1}+\left ( \frac{1}{2} \, \sqrt{2} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( \frac{1}{10} \, \sqrt{5} \sqrt{2} \right )^{2k+1}+\left ( \frac{1}{5} \, \sqrt{5} \right )^{2k+1}+\left ( \frac{1}{2} \, \sqrt{2} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( \frac{1}{12} \, \sqrt{5} \sqrt{3} \right )^{2k+1}+\left ( \frac{7}{132} \, \sqrt{11} \sqrt{6} \right )^{2k+1}+\left ( \frac{1}{44} \, \sqrt{11} \sqrt{6} \sqrt{5} \sqrt{3} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( \frac{4}{91} \, \sqrt{13} \sqrt{7} \right )^{2k+1}+\left ( \frac{1}{14} \, \sqrt{7} \sqrt{3} \right )^{2k+1}+\left ( \frac{3}{26} \, \sqrt{13} \sqrt{3} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( \frac{5}{51} \, \sqrt{17} \right )^{2k+1}+\left ( \frac{1}{3} \right )^{2k+1}+\left ( \frac{3}{17} \, \sqrt{17} \right )^{2k+1}}{4^k(2k+1)}$$ $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( \frac{2}{77} \, \sqrt{21} \sqrt{11} \right )^{2k+1}+\left ( \frac{1}{22} \, \sqrt{11} \sqrt{5} \right )^{2k+1}+\left ( \frac{1}{14} \, \sqrt{21} \sqrt{5} \right )^{2k+1}}{4^k(2k+1)}$$
Second edit: Since requested, here are a few formulas where $a,b,c,L$ are integers, sorted by "error" with $N=10$ and the top ten I could find: where we have $\epsilon, x,y,z$ and $x,y,z$ are the numbers to generate the formula,$\epsilon$ is the error to $\pi$ for $N=10$:
4.91689293680153e-6 1 2 4 $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( 17 \right )^{2k+1}+\left ( 63 \right )^{2k+1}+\left ( 63 \right )^{2k+1}}{\left ( 98 \right )^{2k+1} 4^k(2k+1)}$$ 0.0000153102607067801 1 2 3 $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( 39 \right )^{2k+1}+\left ( 11 \right )^{2k+1}+\left ( 46 \right )^{2k+1}}{\left ( 66 \right )^{2k+1} 4^k(2k+1)}$$ 0.0000209005609743684 1 2 6 $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( 1353 \right )^{2k+1}+\left ( 2987 \right )^{2k+1}+\left ( 4062 \right )^{2k+1}}{\left ( 5742 \right )^{2k+1} 4^k(2k+1)}$$ 0.0000342678923108686 1 2 8 $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( 609 \right )^{2k+1}+\left ( 1159 \right )^{2k+1}+\left ( 1731 \right )^{2k+1}}{\left ( 2394 \right )^{2k+1} 4^k(2k+1)}$$ 0.0000369975262812794 1 2 5 $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( 78 \right )^{2k+1}+\left ( 34 \right )^{2k+1}+\left ( 111 \right )^{2k+1}}{\left ( 153 \right )^{2k+1} 4^k(2k+1)}$$ 0.0000467633890810504 1 2 10 $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( 5763 \right )^{2k+1}+\left ( 9917 \right )^{2k+1}+\left ( 15813 \right )^{2k+1}}{\left ( 21573 \right )^{2k+1} 4^k(2k+1)}$$ 0.0000565682056867800 1 2 7 $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( 43 \right )^{2k+1}+\left ( 23 \right )^{2k+1}+\left ( 68 \right )^{2k+1}}{\left ( 92 \right )^{2k+1} 4^k(2k+1)}$$ 0.0000787760002207705 1 2 12 $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( 8451 \right )^{2k+1}+\left ( 12089 \right )^{2k+1}+\left ( 21826 \right )^{2k+1}}{\left ( 29106 \right )^{2k+1} 4^k(2k+1)}$$ 0.0000850493982005318 1 2 11 $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( 267 \right )^{2k+1}+\left ( 175 \right )^{2k+1}+\left ( 474 \right )^{2k+1}}{\left ( 630 \right )^{2k+1} 4^k(2k+1)}$$ 0.0000904264442818103 1 2 9 $$\pi = 2 \cdot \sum_{k=0}^{\infty}\binom{2k}{k} \frac{\left ( 309 \right )^{2k+1}+\left ( 209 \right )^{2k+1}+\left ( 559 \right )^{2k+1}}{\left ( 741 \right )^{2k+1} 4^k(2k+1)}$$
Starting with my comment.
$\sum_{k=0}^{\infty} x^k\binom{2k}{k} = \dfrac1{\sqrt{1-4x}}$
$\sum_{k=0}^{\infty} x^{2k}\binom{2k}{k} = \dfrac1{\sqrt{1-4x^2}} $
$\sum_{k=0}^{\infty} \dfrac{x^{2k+1}}{2k+1}\binom{2k}{k} = \int_0^x \dfrac{dt}{\sqrt{1-4t^2}} =\dfrac{\arcsin(2x)}{2} $
$\sum_{k=0}^{\infty} \dfrac{x^{2k+1}}{2^{4k+1}(2k+1)}\binom{2k}{k} = \int_0^{x/2} \dfrac{dt}{\sqrt{1-4t^2}} =\dfrac{\arcsin(x)}{2} $
$\dfrac12\sum_{k=0}^{\infty} \dfrac{x^{2k+1}}{4^{2k}(2k+1)}\binom{2k}{k} = \int_0^{x/2} \dfrac{dt}{\sqrt{1-4t^2}} =\dfrac{\arcsin(x)}{2} $
$\sum_{k=0}^{\infty} \dfrac{x^{2k+1}}{4^{2k}(2k+1)}\binom{2k}{k} =\arcsin(x) $
$2\sum_{k=0}^{\infty} \dfrac{a^{2k+1}+b^{2k+1}+c^{2k+1}}{4^{2k}(2k+1)}\binom{2k}{k} =2(\arcsin(a)+\arcsin(b)+\arcsin(c)) $
so we want $\dfrac{\pi}{2} =\arcsin(a)+\arcsin(b)+\arcsin(c) $