User:Passionpi

I hope this is readable.

Definition

$\beta (x)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)^{x}}}$

$\phi ={\frac {{\sqrt {5}}+1}{2}}$

$\gamma =0.577\cdots (1)$

$B(p,q)={\frac {\Gamma (p)\Gamma (q)}{\Gamma (p+q)}}$

$J_{3}\left(e^{-\pi }\right)={\frac {\pi ^{\frac {1}{4}}}{\Gamma \left({\frac {3}{4}}\right)}}$

$J_{4}\left(e^{-\pi }\right)={\frac {\pi ^{\frac {1}{4}}}{2^{\frac {1}{4}}\Gamma \left({\frac {3}{4}}\right)}}$

pi/4

$\sum _{m=1}^{\infty }{\frac {\sum _{n=1}^{3m-2}(-1)^{n+1}n^{2}}{\sum _{n=1}^{3m-2}n^{2}}}={\frac {\pi }{4}}$

Mixed

$\ln 2=16\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{n\pi }-1\right)}}-24\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{2n\pi }-1\right)}}+8\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{4n\pi }-1\right)}}$

$\ln {J_{3}\left(e^{-\pi }\right)}=2\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{n\pi }-1\right)}}-5\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{2n\pi }-1\right)}}+2\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{4n\pi }-1\right)}}$

Jacobi theta function

$\ln {J_{3}\left(e^{-\pi }\right)}=2\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{n\pi }+1\right)}}-\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{2n\pi }+1\right)}}$

$\ln {\frac {1}{J_{4}\left(e^{-\pi }\right)}}=2\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{n\pi }-1\right)}}-\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{2n\pi }-1\right)}}$

${\frac {\pi }{6}}+\ln {\frac {1}{J_{3}^{2}\left(e^{-\pi }\right)}}=3\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{n\pi }+1\right)}}+5\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{n\pi }-1\right)}}$

$\ln {\frac {1}{J_{4}^{2}\left(e^{-\pi }\right)}}=\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{n\pi }+1\right)}}+3\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{n\pi }-1\right)}}$

${\frac {\pi }{12}}+\ln {\frac {1}{J_{3}^{3}\left(e^{-\pi }\right)}}=2\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{2n\pi }+1\right)}}+5\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{2n\pi }-1\right)}}$

$\ln {\frac {1}{J_{3}\left(e^{-\pi }\right)J_{4}\left(e^{-\pi }\right)}}=\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{2n\pi }+1\right)}}+3\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{2n\pi }-1\right)}}$

$3\pi +\ln {\frac {J_{4}^{16}\left(e^{-\pi }\right)}{2^{11}}}=8\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{2n\pi }+1\right)}}+8\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{n\pi }+1\right)}}$

$\pi -\ln {2^{5}}=8\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{2n\pi }+1\right)}}-8\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{n\pi }+1\right)}}$

pi and Ln

$5\pi -\ln 2^{21}=24\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{n\pi }-1\right)}}+24\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{2n\pi }+1\right)}}$

Better approximation

${\frac {\pi }{4}}=\lim _{m\to \infty }\left[{\frac {(-1)^{m}}{4m}}+{\frac {(-1)^{m+1}}{(19m-55)^{2}}}+\sum _{n=1}^{m}{\frac {(-1)^{n-1}}{2n-1}}\right]$

pi

${\frac {\pi }{4}}=\tan ^{-1}\left({\frac {1}{{\sqrt {3}}+1}}\right)+\tan ^{-1}\left({\frac {1}{3{\sqrt {3}}+6}}\right)+2\tan ^{-1}\left({\frac {1}{{\sqrt {3}}+4}}\right)$

Perfect numbers

1st perfect number

$1+2^{1}+3=6\cdots (1)$

$1\times 2^{1}\times 3=6^{1}\cdots (2)$

$3_{terms}=2\times 1+1\cdots (3)$

2nd perfect number

$1+2+2^{2}+7+14=28\cdots (1)$

$1\times 2\times 2^{2}\times 7\times 14=28^{2}\cdots (2)$

$5_{terms}=2\times 2+1\cdots (3)$

3rd perfect number

$1+2+4+8+2^{4}+31+64+124+248=496\cdots (1)$

$1\times 2\times 4\times 8\times {2^{4}}\times 31\times 64\times 124\times 248=496^{4}\cdots (2)$

$9_{terms}=2\times 4+1\cdots (3)$

4th perfect number

$1+2+4+8+16+32+2^{6}+127+254+508+1016+2032+4064=8128\cdots (1)$

$1\times 2\times 4\times 8\times 16\times 32\times 2^{6}\times 127\times 254\times 508\times 1016\times 2032\times 4064=8128^{6}\cdots (2)$

$13_{terms}=2\times 6+1\cdots (3)$

5th perfect number

$1+2+4+8+16+32+64+...+2^{12}+...=33550336\cdots (1)$

$1\times 2\times 4\times 8\times 16\times 32{\cdots }\times 2^{12}\times {\cdots }=33550336^{12}\cdots (2)$

$25_{terms}=2\times 12+1\cdots (3)$

6th perfect number

$1+2+4+8+16+32+64+...+2^{16}+...=8589869056\cdots (1)$

$1\times 2\times 4\times 8\times 16\times 32{\cdots }\times 2^{16}\times {\cdots }=8589869056^{16}\cdots (2)$

$33_{terms}=2\times 16+1\cdots (3)$

nth perfect number

Perfect number = Pn

k = 1, 2, 4, 6, 12, 16, 18, ...

$\sum _{n=1}^{2k+1}P_{n}^{k}=\prod _{n=1}^{2k+1}P_{n}$

Curiosity of 123

$1^{1}+2^{3}=3^{2}\cdots (1)$

Is there any more of this type?

22

$a^{n-1}=a+2\times {b}$

$a^{n}+b^{2}=\left(a+b\right)^{2}$

Special case of Euler's constant

$z\gamma =\lim _{s\to \infty }\left(\sum _{n=1}^{s}{\frac {z}{n}}-\ln {\frac {\Gamma (n+z)}{\Gamma (n)}}\right)$

z = 1, it is a special of Euler's constant

$\gamma =\lim _{s\to \infty }\left(\sum _{n=1}^{s}{\frac {1}{n}}-\ln {n}\right)$

Ramanujan' series

$1-\left({\frac {1}{2}}\right)^{3}+\left({\frac {1\cdot 3}{2\cdot 4}}\right)^{3}-\cdots =\prod _{k=1}^{\infty }{\frac {\left(3!k\right)^{2}-\left(2!k-1!\right)^{2}}{\left(3!k\right)^{2}-\left(2!k-1!\right)^{2}+0!^{2}}}$

Ramanujan's problem

$2^{N}-7=X^{2}\cdots (1)$

Solution : {N: 3 , 4 , 5 , 7 , 15}

$2^{N}+7=X^{2}\cdots (2)$

Solution : {N: 1}

$3^{N}+13=X^{2}\cdots (3)$

Solution : {N: 1 , 5}

$5^{N}+11=X^{2}\cdots (4)$

Solution : {N: 1 , 2 , 5}

Harmonics

$\ln {\frac {2^{0}}{\pi }}+1=2\sum _{s=1}^{\infty }{\frac {\zeta (2s)}{2s+1}}-\sum _{s=1}^{\infty }{\frac {1}{s}}$

$\ln {\frac {2^{1}}{\pi }}+0=2\sum _{s=1}^{\infty }{\frac {\lambda (2s)}{2s+1}}-\sum _{s=1}^{\infty }{\frac {1}{s}}$

$\ln {\frac {2^{2}}{\pi }}-1=2\sum _{s=1}^{\infty }{\frac {\eta (2s)}{2s+1}}-\sum _{s=1}^{\infty }{\frac {1}{s}}$

Log of Ramanujan continued fractions and series

$\ln \left({\sqrt {\phi +2}}-\phi \right)e^{\frac {2\pi }{5}}=\sum _{n=1}^{\infty }{\frac {e^{2n\pi }-e^{4n\pi }-e^{6n\pi }+e^{8n\pi }}{n(1-e^{10n\pi })}}$

$\ln \left({\frac {\sqrt {5}}{1+\left[5^{\frac {3}{4}}(\phi -1)^{\frac {5}{2}}-1\right]^{\frac {1}{5}}}}-\phi \right)e^{\frac {2\pi }{\sqrt {5}}}=\sum _{n=1}^{\infty }{\frac {e^{2n\pi {\sqrt {5}}}-e^{4n\pi {\sqrt {5}}}-e^{6n\pi {\sqrt {5}}}+e^{8n\pi {\sqrt {5}}}}{n(1-e^{10n\pi {\sqrt {5}}})}}$

Ramanujan's series

${\sqrt {\frac {2}{\pi }}}\cdot {\frac {2}{\Gamma ^{2}\left({\frac {3}{4}}\right)}}=1+9\left({\frac {1}{4}}\right)^{4}+17\left({\frac {1\cdot 5}{4\cdot 8}}\right)^{4}+25\left({\frac {1\cdot {5}\cdot 9}{4\cdot {8}\cdot 12}}\right)^{4}+\cdots$

Log of it

$\ln \left({\sqrt {\frac {2}{\pi }}}\cdot {\frac {2}{\Gamma ^{2}\left({\frac {3}{4}}\right)}}\right)=\sum _{n=1}^{\infty }(-1)^{n-1}\left({\frac {\beta (n)}{n}}-\ln {\frac {n+1}{n}}\right)$

Almost integer & Approximation

${\frac {1}{\sqrt {2}}}\approx {{\frac {2}{3}}+{\frac {8}{3}}e^{{\frac {8}{e^{4\pi }+1}}-{\frac {4\pi }{3}}}}$

${\frac {1}{\sqrt {2}}}\approx {\frac {e^{{\frac {1}{e^{4\pi }+1}}-{\frac {\pi }{6}}}}{\left(3{\sqrt {2}}-4\right)^{\frac {1}{8}}}}$

$e^{\frac {\pi }{3}}\left(3{\sqrt {2}}-4\right)^{\frac {1}{4}}\approx {2.0000139...}$

${\frac {\pi ^{\frac {1}{4}}e^{{\frac {\pi }{2}}{\sqrt {\frac {3}{5}}}}}{2^{\frac {15}{8}}\Gamma \left({\frac {3}{4}}\right)}}\approx {0.999984...}$

$e^{5\pi }\approx {6635623.999...}$

$x=\left(\left({\sqrt {2}}^{\sqrt {2}}\right)^{{\sqrt {2}}^{\sqrt {2}}}\right)^{{\sqrt {2}}^{\sqrt {2}}}$

$e^{x^{\frac {1}{4}}}\approx {3.99962}$

${\frac {1}{2}}\ln \left({\frac {e{\sqrt {2e}}}{2}}\right)-{\frac {1}{2^{1}-2^{3}+2^{6}-2^{10}\times 2}}\approx {\gamma }$

${\frac {1}{\left(2^{1}+2^{3}\right)\left[e^{\pi {\sqrt {2\pi }}}-\left(2^{1}+2^{2}+2^{6}+2^{9}+2^{11}\right)\right]}}-\left({\frac {2^{4}-2^{0}}{2^{8}}}\right)^{2^{2}-2^{0}}\approx {\gamma }$

$\ln {\frac {e^{\frac {58}{25}}}{\sqrt {\frac {e^{\frac {2\times 58}{25}}-1}{\pi }}}}-{\frac {1}{\left(\ln 10\right)^{16}}}\approx {\gamma }$

$\left(2-e+2+{\frac {2}{4+{\frac {3}{6+{\frac {4}{5}}}}}}\right)^{2}\approx {2.9999947}$

$\pi \approx {\frac {357}{\left({{\sqrt {2}}^{{\sqrt {3}}^{\sqrt {5}}}}\right)^{4}}}+{\frac {1}{-1+2+3^{2}+4^{2}+5^{3}+6^{2}}}$

${\sqrt {2}}+{\sqrt {3}}\approx {\pi +{\frac {1^{2}}{2^{2}\left(3^{2}+4^{2}\right)\left(\pi -(5^{2})^{0}\right)}}}$

${\sqrt {3}}^{{\sqrt {3}}^{\sqrt {3}}}+{\frac {1}{1-2}}+{\frac {1}{3^{2}+4^{2}+5^{3}+6^{2}}}\approx {\pi }$

${\sqrt {3}}^{\sqrt {3}}\approx {2583.99902\times {10^{-4}}}$

$e^{\pi {\sqrt {3}}}\approx 2197.99087$

$e^{\pi {\sqrt {6+{\sqrt {6+{\sqrt {6}}}}}}}\approx {117.99965\times 10^{2}}$

$e^{\pi }-\pi \approx {\left(3\pi \right)^{\frac {4}{3}}+{\frac {1}{11-{\frac {1}{7^{2}}}}}}$

${\frac {3^{3^{2}}-3^{2^{2}}}{\pi {\sqrt {2^{2^{2}}-2^{3^{1}}}}}}-2^{2^{2}}-e^{-\pi ^{2}}\approx {3^{2^{2^{2}}-3^{2^{1}}}+2^{2^{1}}-1^{1^{1}}}$

Numbers from 1 to 10

$3\approx {\left({\frac {6}{5}}\phi ^{2}-\pi \right)^{-{\frac {1}{4}}}-\left({\frac {10}{\pi }}+10\right)^{-2}-e^{-\phi \pi {e}}-7^{-8}-9}$

$e\approx {3-\left({\frac {5}{7\times 9}}\right)^{\frac {1}{2}}}-{\frac {13^{0}}{\left(11^{0}+15^{2}\right)^{3}}}$

Two of each digits : 11,22,33,44,55,66

$\pi \approx {{\frac {355}{113}}-{\frac {66^{0}}{44^{2^{2}}}}}$

$\left(1+{\frac {1}{e^{2}}}\right)^{\left(1+{\frac {1}{e^{2}}}\right)^{\left(1+{\frac {1}{e^{2}}}\right)}}\approx \left({\frac {10}{11}}\right)^{2^{2^{2}}}-{\frac {16^{0}}{12\times 13\times 14\left(\left[15\left({\frac {17+3^{2}}{1^{2}+2^{2}}}\right)\right]^{\frac {1}{2^{3}}}+18^{0}\right)}}+{\frac {1}{2^{3^{3}}}}$

$\pi ^{2\times 3}\approx {\frac {2\times 3\times 5\times 7\times 11\times 13-17-2\times 3\times 5\times 7}{2\times 3\times 5+7^{0}}}$

${\frac {10^{2}}{\left(10-\pi ^{2}\right)^{2^{-3}}}}-10^{2}\approx {2^{2}+3^{2}+4^{2}+{\frac {1}{\left(2^{2}+3^{2}+4^{2}\right)^{3}-\left(2^{0}+2^{1}+2^{2}+2^{4}\right)\times {10^{3}}}}}$

$\left(1+{\frac {1}{\pi ^{2}}}\right)^{\left(1+{\frac {1}{\pi ^{2}}}\right)^{\left(1+{\frac {1}{\pi ^{2}}}\right)}}\approx \left({{\frac {1}{1^{2}}}+{\frac {1}{2^{2}+3^{2}\times 4^{2}}}}\right)^{2^{2^{2}}}\approx {\left[\left(1+{\frac {1}{e}}\right)^{\left(1+{\frac {1}{e}}\right)}\right]^{\frac {1}{2^{2}}}}\approx {e^{\frac {\phi {\pi }{e}}{2^{2^{0}}\times 2^{2^{1}}\times 2^{2^{2}}}}}$

$\left(1+{\frac {1}{\pi ^{2}}}\right)^{1+{\frac {1}{\pi ^{2}}}}\approx \left({{\frac {30}{2^{0}}}+{\frac {31^{-1}}{2^{2}+2^{2^{2}}}}}\right)^{{32}^{-1}}$

$e^{\frac {\pi }{e}}-\left({\frac {\pi }{e}}\right)^{\frac {\pi }{e}}\approx {1.99426...}$

${\frac {1}{3^{2}\pi }}-\left[\left(1+{\frac {1}{\pi }}\right)^{1+{\frac {1}{\pi }}}\right]^{4^{2}}+5^{3}+6^{3}\approx {0.99983...}$

$\pi {(10\phi )}^{3}-\left({\frac {\pi }{40}}\right)^{4}\approx {3\times 5\times 7\times 11\times 13}-\left(5+5+5+5^{2}+5^{0}\right)^{2}-\left(5^{2}+5^{0}\right)$

$\left[e^{\pi }+\left({\frac {\pi }{e}}\right)^{2}\right]^{2}\approx {5^{2^{2}}-5^{2}-5^{0}+{\frac {1}{\sqrt {7^{2}+8^{2}}}}}$

$e^{2\pi }-2\pi \approx {23^{2}+{\frac {1}{\sqrt {23}}}}$

$10^{2}\left[e^{\pi }-\left({\frac {\pi }{e}}\right)^{2}\right]^{\frac {1}{2^{3}}}-\left({\frac {2}{\pi }}\right)^{2^{4}}\approx {146.9999978...}$

$10e^{\frac {\pi \phi }{2}}\approx {126.99998...}$

${\frac {e^{\pi {\sqrt[{3}]{\phi }}}}{2\times 5}}\approx {3.997...}$

${\frac {1}{4^{2}+3^{2}}}\left({\frac {e^{4e^{\frac {\pi }{e}}}}{2^{2}\times {1^{2}}}}\right)\approx {3294.999662...}$

$\pi {\sqrt {\phi }}\approx {3.996...}$

$7\pi \approx {21.9911...}$

$12\approx {\left({\frac {6}{5}}\phi ^{2}-\pi \right)^{-2^{-2}}-\left({\frac {10}{\pi }}+10\right)^{-2}-e^{-\phi \pi {e}}}$

Difference of two square

$\pi =\lim _{n\to \infty }\left[\prod _{k=1}^{n}\left(1+{\frac {1}{2k-1}}\right)^{2}-\prod _{k=1}^{n-1}\left(1+{\frac {1}{2k-1}}\right)^{2}\right]$

${\frac {4}{\pi }}=\lim _{n\to \infty }\left[\prod _{k=1}^{n}\left(1+{\frac {1}{2k}}\right)^{2}-\prod _{k=1}^{n-1}\left(1+{\frac {1}{2k}}\right)^{2}\right]$

π/2

$\ln {\frac {\pi }{2}}-2+{\frac {\pi }{2}}=2\sum _{s=1}^{\infty }{\frac {1-\beta (2s+1)}{2s+1}}$

Alternating series

$\ln \left({\frac {2\sinh \pi }{\pi ^{2}}}\right)=\sum _{n=1}^{\infty }(-1)^{n-1}\left({\frac {\zeta (2n)}{n}}-\ln {\frac {n+1}{n}}\right)$

$\ln 1=\sum _{n=1}^{\infty }(-1)^{n-1}\left({\frac {\eta (n)}{n}}-\ln {\frac {n+1}{n}}\right)$

Beta function

$\pi =\lim _{n\to \infty }nB^{2}\left(n,{\frac {1}{2}}\right)$

Infinite product

${\frac {4}{\pi }}=\prod _{k=1}^{\infty }\left(1+{\frac {1}{4k}}\right)^{2}\left({\frac {2k+1}{2k+1+(-1)^{k-1}}}\right)^{(-1)^{k-1}}$

${\frac {2}{\pi }}\times {\frac {2}{\phi }}=\prod _{s=1}^{\infty }\left({\frac {s-{\frac {1}{\phi ^{2}}}}{s-{\frac {1}{\phi ^{2}}}+1}}\right)\left(1+{\frac {1}{2s}}\right)^{2}$

Symmetry patterns

$-{\frac {\pi }{4}}=\sum _{s=0}^{\infty }{\frac {(-1)^{s}}{2s+1}}\left({\frac {1}{2^{2s+1}}}+{\frac {2^{2s+1}}{1}}\right)-\sum _{s=0}^{\infty }(-1)^{s}{\frac {2^{2s+1}}{2s+1}}\zeta (4s+2)$

$+{\frac {\pi }{4}}=\sum _{s=0}^{\infty }{\frac {(-1)^{s}}{2s+1}}\left({\frac {1}{2^{2s+1}}}+{\frac {2^{2s+1}}{1}}\right)-\sum _{s=0}^{\infty }(-1)^{s}{\frac {2^{2s+1}}{2s+1}}\eta (4s+2)$

$+{\frac {0}{4}}=\sum _{s=0}^{\infty }{\frac {(-1)^{s}}{2s+1}}\left({\frac {1}{2^{2s+1}}}+{\frac {2^{2s+1}}{1}}\right)-\sum _{s=0}^{\infty }(-1)^{s}{\frac {2^{2s+1}}{2s+1}}\lambda (4s+2)$

Zeta function

$\ln 2-\gamma =\sum _{s=1}^{\infty }{\frac {\zeta (2s+1)}{(2s+1)2^{2s}}}$

$2\ln 2-\gamma -2\beta ^{'}(0)=\sum _{s=1}^{\infty }{\frac {\zeta (2s+1)}{(2s+1)2^{4s}}}$

$\beta ^{'}(0)=\sum _{s=0}^{\infty }{\frac {\eta (2s+1)}{(2s+1)2^{2s+1}}}$

Tanh(y/x)

$\tanh \left({\frac {y}{x}}\right)={\frac {e^{\frac {2y}{x}}-1}{e^{\frac {2y}{x}}+1}}={\frac {y}{1x+{\frac {y^{2}}{3x+{\frac {y^{2}}{5x+{\frac {y^{2}}{7x+\cdots }}}}}}}}$

Setting y = π and x = 2 Which gives Ramanujan equation

${\frac {e^{\pi }-1}{e^{\pi }+1}}={\frac {\pi }{2+{\frac {\pi ^{2}}{6+{\frac {\pi ^{2}}{10+{\frac {\pi ^{2}}{14+\cdots }}}}}}}}$

Setting y = π and x = 2π Which gives

${\frac {e-1}{e+1}}={\frac {\pi }{2\pi +{\frac {\pi ^{2}}{6\pi +{\frac {\pi ^{2}}{10\pi +{\frac {\pi ^{2}}{14\pi +\cdots }}}}}}}}$

ln4

$\ln 4=1+{\frac {1}{3}}+{\frac {1}{3^{3}-3}}+{\frac {1}{6^{3}-6}}+{\frac {1}{9^{3}-9}}+{\frac {1}{12^{3}-12}}+\cdots =1+{\frac {1}{3}}+\sum _{k=1}^{\infty }{\frac {1}{(3k)^{3}-3k}}$

Nested Radical

$2={\sqrt {2!+{\sqrt {2!+{\sqrt {2!+{\sqrt {2!+\cdots }}}}}}}}$

$3={\sqrt {3!+{\sqrt {3!+{\sqrt {3!+{\sqrt {3!+\cdots }}}}}}}}$

$2!+2=2^{2}\cdots (1)$

$3!+3=3^{2}\cdots (2)$

Is there any more of this type?

$n!+n=n^{2}\cdots (3)$

$n={\sqrt {n!+{\sqrt {n!+{\sqrt {n!+{\sqrt {n!+\cdots }}}}}}}}$

$2={\sqrt[{3}]{3!+{\sqrt[{3}]{3!+{\sqrt[{3}]{3!+{\sqrt[{3}]{3!+\cdots }}}}}}}}$

$3={\sqrt[{3}]{4!+{\sqrt[{3}]{4!+{\sqrt[{3}]{4!+{\sqrt[{3}]{4!+\cdots }}}}}}}}$

$5={\sqrt[{3}]{5!+{\sqrt[{3}]{5!+{\sqrt[{3}]{5!+{\sqrt[{3}]{5!+\cdots }}}}}}}}$

$9={\sqrt[{3}]{6!+{\sqrt[{3}]{6!+{\sqrt[{3}]{6!+{\sqrt[{3}]{6!+\cdots }}}}}}}}$

Strange pattern!

$3!+2=2^{3}\cdots (1)$

$4!+3=3^{3}\cdots (2)$

$5!+5=5^{3}\cdots (3)$

$6!+9=9^{3}\cdots (4)$

Is there more of this type?

$x!+y=y^{3}\cdots (5)$

$y={\sqrt[{3}]{x!+{\sqrt[{3}]{x!+{\sqrt[{3}]{x!+{\sqrt[{3}]{x!+\cdots }}}}}}}}$

Equation

Equation 1

$n!+n=n^{2}\cdots (1)$

Equation 2

$x!+Y=Y^{3}\cdots (2)$

Prime

p = {2,3,5,7,11,13,...}

${\frac {7}{6}}=\prod _{p}^{\infty }\left({\frac {p^{4}+1}{p^{4}-1}}\right)$

${\frac {123}{122}}=\prod _{p}^{\infty }\left({\frac {p^{8}+1}{p^{8}-1}}\right)$

Arctan(π/4)

${\frac {\pi }{2}}=\sum _{n=1}^{\infty }Arctan\left({\frac {1}{n^{2}}}\right)$

${\frac {\pi }{4}}=\sum _{n=1}^{\infty }Arctan\left({\frac {1}{2n^{2}}}\right)$

${\frac {\pi }{4}}=\sum _{n=1}^{\infty }Arctan\left({\frac {1}{n^{2}+n+1}}\right)$

Cube of Fibonacci series

$1^{3}+1^{3}+2^{3}+3^{3}+5^{3}+\cdots {+}F_{n}^{3}={\frac {F_{n}F_{n+1}^{2}+(-1)^{n+1}F_{n-1}+1}{2}}$

π

$\pi =24\sum _{k=0}^{\infty }{\frac {1}{2k+1}}\left[{\frac {2}{e^{(2k+1)\pi }+1}}+{\frac {1}{e^{(2k+1)\pi }-1}}\right]+24\sum _{k=1}^{\infty }{\frac {1}{2k}}\left[{\frac {2}{e^{2k\pi }-1}}+{\frac {1}{e^{2k\pi }+1}}\right]$

${\frac {\pi }{2^{3}\cdot {3^{2}}}}=\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{k}}\left[{\frac {1}{e^{k\pi }-1}}-{\frac {1}{3(e^{2k\pi }-1)}}+{\frac {1}{6(e^{4k\pi }-1)}}\right]$

$\pi =30\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{n\pi }-1\right)}}+42\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{n\pi }+1\right)}}-12\sum _{n=1}^{\infty }{\frac {1}{n\left(e^{2n\pi }+1\right)}}$

Logarithms of 2

$\ln 2=\sum _{n=1}^{\infty }{\frac {1}{n}}\left({\frac {1}{2^{2n}}}+{\frac {2}{3^{2n}}}+{\frac {2}{4^{2n}}}+{\frac {1}{5^{2n}}}\right)$

$\ln 2=12\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{k(e^{k\pi }-1)}}+4\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{k(e^{k\pi }+1)}}$

$\ln 2=\sum _{k=1}^{\infty }{\frac {1}{k}}\left[n\zeta (2k)-\sum _{s=1}^{n}{\frac {n}{s^{2k}}}-\sum _{s=1}^{n-1}{\frac {n-s}{(n+s)^{2k}}}\right]$

n ≥ 1

$\ln 2=8\sum _{k=0}^{\infty }{\frac {1}{2k+1}}\left[{\frac {1}{e^{(2k+1)\pi }-1}}+{\frac {1}{e^{(2k+1)+1}}}\right]$

$\ln 2={\frac {1}{2^{k}-(1+k)}}\cdot \sum _{n=1}^{\infty }{\frac {1}{n}}\left[\sum _{s=1}^{2^{k}-2}{\frac {2^{k}-(1+s)}{(1+s)^{2n}}}\right]$

Where k ≥ 2

$\ln 2={\frac {1}{k}}\cdot \sum _{n=1}^{\infty }{\frac {1}{n}}\left[(2^{k}-1)\zeta (2n)-\sum _{s=1}^{2^{k}-1}{\frac {2^{k}-s}{s^{2n}}}\right]$

Where k ≥ 1

Euler's Constant

$1=\lim _{x\to \infty }\left[{\frac {\gamma }{x}}+e^{-{\frac {\gamma }{x}}}\prod _{y=1}^{\infty }\left({\frac {xy}{1+xy}}\right)e^{\frac {1}{xy}}\right]$

$\gamma =\sum _{n=1}^{\infty }\left[{\frac {2}{2n-1}}-\ln \left({\frac {4n+1}{4n-3}}\right)\right]$

$\ln \left({\frac {2\pi }{e^{2\gamma }}}\right)=\sum _{n=1}^{\infty }{\frac {2}{2^{2n}}}\left[{\frac {\zeta (2n)}{2n}}+{\frac {\zeta (2n+1)}{2n+1}}\right]$

Alternating symmetric formula

$2\times {\frac {4}{\pi }}={\frac {\prod _{n=1}^{\infty }\left(1+{\frac {1}{4n^{2}-1}}\right)^{(-1)^{n+1}}}{\sum _{n=1}^{\infty }{\frac {1}{4n^{2}-1}}}}$

${\frac {\pi }{4}}={\frac {\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}k}{2k-1}}}{\sum _{k=1}^{\infty }{\frac {1}{(2k)^{2}-1}}}}$

Square Root of 2

Algorithms

$F_{n+1}={\frac {F_{n}}{a}}+{\frac {b}{F_{n}}}={\sqrt {\frac {ab}{a-1}}}$

Setting a = 2 and b = 1

$F_{n+1}={\frac {F_{n}}{2}}+{\frac {1}{F_{n}}}={\sqrt {2}}$

$F_{n+1}={\frac {F_{n}}{a}}+{\frac {b}{F_{n}}}+{\frac {F_{n}}{c}}={\sqrt {\frac {abc}{ac-(a+c)}}}$

Fo is an estimate

${\sqrt {2}}+{\frac {1}{\sqrt {2}}}={\sqrt {2+{\frac {\sqrt {2}}{2^{0}}}{\sqrt {2+{\frac {\sqrt {2}}{2^{1}}}{\sqrt {2+{\frac {\sqrt {2}}{2^{2}}}{\sqrt {2+{\frac {\sqrt {2}}{2^{3}}}{\sqrt {2+{\frac {\sqrt {2}}{2^{4}}}{\sqrt {2}}+\cdots }}}}}}}}}}$

${\frac {1}{\sqrt {2}}}={\sqrt {2-{\frac {\sqrt {2}}{2^{0}}}{\sqrt {2-{\frac {\sqrt {2}}{2^{1}}}{\sqrt {2-{\frac {\sqrt {2}}{2^{2}}}{\sqrt {2-{\frac {\sqrt {2}}{2^{3}}}{\sqrt {2-{\frac {\sqrt {2}}{2^{4}}}{\sqrt {2}}-\cdots }}}}}}}}}}$

${\sqrt {2}}=\prod _{s=1}^{\infty }{\frac {(4s)^{2}-1^{2}}{(4s)^{2}-2^{2}}}$

K = 0.9159655... (Catalan's Constant)

${\sqrt {2}}=e^{1-{\frac {2K}{\pi }}}\prod _{s=1}^{\infty }\left({\frac {4s-1}{4s+1}}\right)^{4s}e^{2}$

Continued fraction

${\sqrt[{16}]{p}}={\sqrt {k}}+{\frac {1}{{\sqrt {k}}+{\frac {1}{{\sqrt {k}}+{\frac {1}{{\sqrt {k}}+...}}}}}}$

${\sqrt {{\sqrt {4}}+{\sqrt {3}}}}-{\sqrt {2}}={\frac {1}{{\sqrt {2}}+{\frac {1}{{\sqrt {2}}+{\frac {1}{{\sqrt {2}}+...}}}}}}$

${\phi ^{2}}={\sqrt {5}}+{\frac {1}{{\sqrt {5}}+{\frac {1}{{\sqrt {5}}+{\frac {1}{{\sqrt {5}}+...}}}}}}$

Continued fraction of phi

$\phi ={\frac {{\sqrt {5}}+1}{2}}$

${\frac {1}{\phi ^{0}}}={\frac {1}{{\sqrt {1}}+{\frac {1}{{\sqrt {1}}+{\frac {1}{{\sqrt {1}}+...}}}}}}+{\frac {1}{{\sqrt {5}}+{\frac {1}{{\sqrt {5}}+{\frac {1}{{\sqrt {5}}+...}}}}}}$

${\frac {1}{\phi }}={\frac {1}{{\sqrt {1}}+{\frac {1}{{\sqrt {1}}+{\frac {1}{{\sqrt {1}}+...}}}}}}-{\frac {0}{{\sqrt {5}}+{\frac {1}{{\sqrt {5}}+{\frac {1}{{\sqrt {5}}+...}}}}}}$

${\frac {1}{\phi ^{2}}}={\frac {1}{{\sqrt {5}}+{\frac {1}{{\sqrt {5}}+{\frac {1}{{\sqrt {5}}+...}}}}}}-{\frac {0}{{\sqrt {1}}+{\frac {1}{{\sqrt {1}}+{\frac {1}{{\sqrt {1}}+...}}}}}}$

${\frac {1}{\phi ^{3}}}={\frac {1}{{\sqrt {1}}+{\frac {1}{{\sqrt {1}}+{\frac {1}{{\sqrt {1}}+...}}}}}}-{\frac {1}{{\sqrt {5}}+{\frac {1}{{\sqrt {5}}+{\frac {1}{{\sqrt {5}}+...}}}}}}$

${\frac {1}{\phi ^{4}}}={\frac {2}{{\sqrt {5}}+{\frac {1}{{\sqrt {5}}+{\frac {1}{{\sqrt {5}}+...}}}}}}-{\frac {1}{{\sqrt {1}}+{\frac {1}{{\sqrt {1}}+{\frac {1}{{\sqrt {1}}+...}}}}}}$

${\frac {1}{\phi ^{5}}}={\frac {2}{{\sqrt {1}}+{\frac {1}{{\sqrt {1}}+{\frac {1}{{\sqrt {1}}+...}}}}}}-{\frac {3}{{\sqrt {5}}+{\frac {1}{{\sqrt {5}}+{\frac {1}{{\sqrt {5}}+...}}}}}}$

${\frac {1}{\phi ^{6}}}={\frac {5}{{\sqrt {5}}+{\frac {1}{{\sqrt {5}}+{\frac {1}{{\sqrt {5}}+...}}}}}}-{\frac {3}{{\sqrt {1}}+{\frac {1}{{\sqrt {1}}+{\frac {1}{{\sqrt {1}}+...}}}}}}$

${\frac {1}{\phi ^{7}}}={\frac {5}{{\sqrt {1}}+{\frac {1}{{\sqrt {1}}+{\frac {1}{{\sqrt {1}}+...}}}}}}-{\frac {8}{{\sqrt {5}}+{\frac {1}{{\sqrt {5}}+{\frac {1}{{\sqrt {5}}+...}}}}}}$

and so on ...

Unity

$1=\phi ^{2}-1\phi \cdots (1)$

$1=2\phi ^{2}-1\phi ^{3}\cdots (2)$

$1=2\phi ^{4}-3\phi ^{3}\cdots (3)$

$1=5\phi ^{4}-3\phi ^{5}\cdots (4)$

$1=5\phi ^{6}-8\phi ^{5}\cdots (5)$

$1=13\phi ^{6}-8\phi ^{7}\cdots (6)$

$1=13\phi ^{8}-21\phi ^{7}\cdots (7)$

$1=34\phi ^{8}-21\phi ^{9}\cdots (8)$

and so on ...

Integers

$\phi ={\frac {{\sqrt {5}}+1}{2}}$

$1=\phi ^{2}-\phi \cdots (1)$

$2=-\phi ^{3}+3\phi ^{2}-\phi \cdots (2)$

$3=2\phi ^{4}-4\phi ^{3}+3\phi ^{2}-\phi \cdots (3)$

$4=-3\phi ^{5}+7\phi ^{4}-4\phi ^{3}+3\phi ^{2}-\phi \cdots (4)$

$5=5\phi ^{6}-11\phi ^{5}+7\phi ^{4}-4\phi ^{3}+3\phi ^{2}-\phi \cdots (5)$

and so on ...

$n=\left(\sum _{m=1}^{n}(-\phi )^{m}L_{m+1}\right)+(-\phi )^{n+1}F_{n}$

Approximation

$a\phi ^{\frac {3}{2}}+b\phi ^{\frac {1}{2}}-c\phi \approx {d}$

$(a+c)\phi ^{\frac {3}{2}}+(b+d)\phi ^{\frac {1}{2}}-(a+b+c)\phi \approx (a+d)$

Example

$1\phi ^{\frac {3}{2}}+3\phi ^{\frac {1}{2}}-3\phi \approx 1$

$4\phi ^{\frac {3}{2}}+4\phi ^{\frac {1}{2}}-7\phi \approx 2$

$11\phi ^{\frac {3}{2}}+6\phi ^{\frac {1}{2}}-15\phi \approx 6$

$26\phi ^{\frac {3}{2}}+12\phi ^{\frac {1}{2}}-32\phi \approx 17$

and so on ...

Square root

${\frac {\sqrt {2}}{{\sqrt {1}}+{\sqrt {3}}}}={\sqrt {{\sqrt {4}}-{\sqrt {3}}}}$

Prime number

$5^{2}+5^{4}=17^{2}+19^{2}\cdots (1)$

Is there any more of this kind?

$p{_{1}}^{2}+p{_{1}}^{4}=p_{2}^{2}+p_{3}^{2}$

${\sqrt {4n^{2}+4n-7}}=2{\sqrt {-1+{\frac {2n-1}{2^{0}}}{\sqrt {1+{\frac {2n-1}{2^{1}}}{\sqrt {1+{\frac {2n-1}{2^{2}}}{\sqrt {1+...}}}}}}}}$

${\sqrt {4n^{2}+4n-7}}$

is prime from n = 2 to 12 only

n = 2, gives

${\sqrt {17}}=2{\sqrt {-1+{\frac {3}{2^{0}}}{\sqrt {1+{\frac {3}{2^{1}}}{\sqrt {1+{\frac {3}{2^{2}}}{\sqrt {1+...}}}}}}}}$

2

$2=\prod _{n=1}^{\infty }\left[1+{\frac {1}{e^{(2n-1)\pi }-1}}\right]^{2}\left[\left(1+{\frac {1}{e^{n\pi }}}\right)\left(1-{\frac {1}{e^{n\pi }}}\right)^{2}\right]^{4}$