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
$$\ln2=16\sum_{n=1}^{\infty}\frac{1}{n\left(e^{n\pi}-1\right)}- 24\sum_{n=1}^{\infty}\frac{1}{n\left(e^{2n\pi}-1\right)}+ 8\sum_{n=1}^{\infty}\frac{1}{n\left(e^{4n\pi}-1\right)}$$

$$\ln{J_3\left(e^{-\pi}\right)}=2\sum_{n=1}^{\infty}\frac{1}{n\left(e^{n\pi}-1\right)}- 5\sum_{n=1}^{\infty}\frac{1}{n\left(e^{2n\pi}-1\right)}+ 2\sum_{n=1}^{\infty}\frac{1}{n\left(e^{4n\pi}-1\right)}$$

Jacobi theta function
$$\ln{J_3\left(e^{-\pi}\right)}=2\sum_{n=1}^{\infty}\frac{1}{n\left(e^{n\pi}+1\right)}- \sum_{n=1}^{\infty}\frac{1}{n\left(e^{2n\pi}+1\right)}$$

$$\ln\frac{1}{J_4\left(e^{-\pi}\right)}=2\sum_{n=1}^{\infty}\frac{1}{n\left(e^{n\pi}-1\right)}- \sum_{n=1}^{\infty}\frac{1}{n\left(e^{2n\pi}-1\right)}$$

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

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

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

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

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

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

pi and Ln
$$5\pi-\ln2^{21}=24\sum_{n=1}^{\infty}\frac{1}{n\left(e^{n\pi}-1\right)}+ 24\sum_{n=1}^{\infty}\frac{1}{n\left(e^{2n\pi}+1\right)}$$

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

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

$$1\times2^{1}\times3=6^1\cdots(2)$$

$$3_{terms}=2\times1+1\cdots(3)$$

2nd perfect number

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

$$1\times2\times2^{2}\times7\times14=28^2\cdots(2)$$

$$5_{terms}=2\times2+1\cdots(3)$$

3rd perfect number

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

$$1\times2\times4\times8\times{2^4}\times31\times64\times124\times248=496^4\cdots(2)$$

$$9_{terms}=2\times4+1\cdots(3)$$

4th perfect number

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

$$1\times2\times4\times8\times16\times32\times2^{6}\times127\times254 \times508\times1016\times2032\times4064=8128^6\cdots(2)$$

$$13_{terms}=2\times6+1\cdots(3)$$

5th perfect number

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

$$1\times2\times4\times8\times16\times32{\cdots}\times2^{12}\times{\cdots}=33550336^{12}\cdots(2)$$

$$25_{terms}=2\times12+1\cdots(3)$$

6th perfect number

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

$$1\times2\times4\times8\times16\times32{\cdots}\times2^{16}\times{\cdots}= 8589869056^{16}\cdots(2)$$

$$33_{terms}=2\times16+1\cdots(3)$$

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
$$1-\left(\frac{1}{2}\right)^3+\left(\frac{1\cdot3}{2\cdot4}\right)^3-\cdots= \prod_{k=1}^{\infty}\frac{\left(3!k\right)^2-\left(2!k-1!\right)^2} {\left(3!k\right)^2-\left(2!k-1!\right)^2+0!^2}$$

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

$$\sqrt{\frac{2}{\pi}}\cdot{\frac{2}{\Gamma^2\left(\frac{3}{4}\right)}}= 1+9\left(\frac{1}{4}\right)^4+17\left(\frac{1\cdot5}{4\cdot8}\right)^4+ 25\left(\frac{1\cdot{5}\cdot9}{4\cdot{8}\cdot12}\right)^4+\cdots$$

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

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

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

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

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

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

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

$$\frac{1}{2}\ln\left(\frac{e\sqrt{2e}}{2}\right)-\frac{1}{2^1-2^3+2^6-2^{10}\times2} \approx{\gamma}$$

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

$$\ln\frac{e^{\frac{58}{25}}}{\sqrt{\frac{e^{\frac{2\times58}{25}}-1}{\pi}}}- \frac{1}{\left(\ln10\right)^{16}}\approx{\gamma}$$

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

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

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

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

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

$$e^{\pi\sqrt{3}}\approx2197.99087$$

$$e^{\pi\sqrt{6+\sqrt{6+\sqrt{6}}}}\approx{117.99965\times10^2}$$

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

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

Numbers from 1 to 10

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

$$e\approx{3-\left(\frac{5}{7\times9}\right) ^{\frac{1}{2}}}-\frac{13^0}{\left(11^0+15^2\right)^3}$$

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

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

$$\left(1+\frac{1}{e^2}\right)^{\left(1+\frac{1}{e^2}\right)^{\left(1+\frac{1}{e^2}\right)}} \approx \left(\frac{10}{11}\right)^{2^{2^2}}-\frac{16^0}{12\times13\times14\left(\left[15\left (\frac{17+3^2}{1^2+2^2}\right)\right]^{\frac{1}{2^3}} +18^0\right)}+\frac{1}{2^{3^3}}$$

$$\pi^{2\times3}\approx{\frac{2\times3\times5\times7\times11\times13- 17-2\times3\times5\times7}{2\times3\times5+7^0}}$$

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

$$\left(1+\frac{1}{\pi^2}\right)^{\left(1+\frac{1}{\pi^2}\right)^ {\left(1+\frac{1}{\pi^2}\right)}}\approx \left({\frac{1}{1^2}+\frac{1}{2^2+3^2\times4^2}}\right)^{2^{2^2}} \approx{\left[\left(1+\frac{1}{e}\right)^{\left(1+\frac{1}{e}\right)}\right]^{\frac{1}{2^2}}} \approx{e^{\frac{\phi{\pi}{e}}{2^{2^0}\times2^{2^1}\times2^{2^2}}}}$$

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

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

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

$$\pi{(10\phi)}^3-\left(\frac{\pi}{40}\right)^4 \approx{3\times5\times7\times11\times13}- \left(5+5+5+5^2+5^0\right)^2-\left(5^2+5^0\right)$$

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

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

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

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

$$\frac{e^{\pi\sqrt[3]\phi}}{2\times5}\approx{3.997...}$$

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

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

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

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

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

$$\ln1=\sum_{n=1}^{\infty}(-1)^{n-1} \left(\frac{\eta(n)}{n}-\ln\frac{n+1}{n}\right)$$

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
$$\ln2-\gamma=\sum_{s=1}^{\infty}\frac{\zeta(2s+1)}{(2s+1)2^{2s}}$$

$$2\ln2-\gamma-2\beta^{'}(0)=\sum_{s=1}^{\infty}\frac{\zeta(2s+1)}{(2s+1)2^{4s}}$$

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

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
$$\ln4=1+\frac{1}{3}+\frac{1}{3^3-3}+\frac{1}{6^3-6}+\frac{1}{9^3-9}+\frac{1}{12^3-12}+\cdots= 1+\frac{1}{3}+\sum_{k=1}^{\infty}\frac{1}{(3k)^3-3k}$$

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
$$\ln2=\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{2^{2n}}+\frac{2}{3^{2n}} +\frac{2}{4^{2n}}+\frac{1}{5^{2n}}\right)$$

$$\ln2=12\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k(e^{k\pi}-1)}+ 4\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k(e^{k\pi}+1)}$$

$$\ln2=\sum_{k=1}^{\infty}\frac{1}{k}\left[n\zeta(2k)-\sum_{s=1}^{n}\frac{n}{s^{2k}}- \sum_{s=1}^{n-1}\frac{n-s}{(n+s)^{2k}}\right]$$

n ≥ 1

$$\ln2=8\sum_{k=0}^{\infty}\frac{1}{2k+1}\left[\frac{1}{e^{(2k+1)\pi}-1} +\frac{1}{e^{(2k+1)+1}}\right]$$

$$\ln2=\frac{1}{2^k-(1+k)}\cdot\sum_{n=1}^{\infty}\frac{1}{n} \left[\sum_{s=1}^{2^k-2}\frac{2^k-(1+s)}{(1+s)^{2n}}\right]$$

Where k ≥ 2

$$\ln2=\frac{1}{k}\cdot\sum_{n=1}^{\infty}\frac{1}{n} \left[(2^k-1)\zeta(2n)-\sum_{s=1}^{2^k-1}\frac{2^k-s}{s^{2n}}\right]$$

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 ...

$$n=\left(\sum_{m=1}^n(-\phi)^mL_{m+1}\right)+(-\phi)^{n+1}F_n$$

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

$$1\phi^{\frac{3}{2}}+3\phi^{\frac{1}{2}}-3\phi\approx1$$

$$4\phi^{\frac{3}{2}}+4\phi^{\frac{1}{2}}-7\phi\approx2$$

$$11\phi^{\frac{3}{2}}+6\phi^{\frac{1}{2}}-15\phi\approx6$$

$$26\phi^{\frac{3}{2}}+12\phi^{\frac{1}{2}}-32\phi\approx17$$

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

$$2=\frac{\left[n^p+(n+r)^p\right]^2+\left[n^p+(n+r)^p+2n\right]^2} {n^2+\left[n^p+(n+r)^p+n\right]^2}$$

0
$$0=1^8+2^6-3^4+4^2\cdots(1)$$

Ramanujan's problem
$$2^N-7=X^2\cdots(1)$$

Solution N = 3, 4 , 5 , 7 , 15

Sum

$$3^2+4^2+5^2+7^2+15^2=18^2\cdots(1)$$

$$3+4+5+7+15=34\cdots(2)$$

$$2^{34}-2^{18}=131071^2\cdots(3)$$

$$2^{34}+2^{18}=131073^2\cdots(4)$$

$$2^{18}=\frac{131071^2+131073^2}{131072}\cdots(5)$$