User:Deen314

22/4/21

Memorisation:

Surahs:

1. 108

Books:

1. The minor resurrection (22/4/21-29/4/21)

2. Juz Amma (29/4/21-)

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$$\int_{0}^{1}\int_{0}^{1}\frac{\mathrm dx \mathrm dy}{(1-xy)\sqrt{xy(1-y)}}=\pi^2$$

$$\int_{0}^{1}\int_{0}^{1}\frac{\mathrm dx \mathrm dy}{(1-xy)\sqrt{x(1-xy)}}=2^2$$ -

$$\int_{0}^{1}\int_{0}^{1}\frac{x^4}{\left(\frac{1}{4}-x+x^2\right)\left(\frac{1}{2}-x+x^2\right)\sqrt{y(1-y)}}\mathrm dx \mathrm dy=\pi^2$$ - $$\int_{0}^{1}\int_{0}^{1}\frac{x^4}{\left(\frac{1}{2}-x+x^2\right)^2\left(\frac{1}{4}-x+x^2\right)\sqrt{1-y^2}}\mathrm dx \mathrm dy=\pi^2$$ ---

$$\pi\approx \frac{22}{7}$$

$$\int_{0}^{1}\frac{x^4(1-x)^4}{x^2-2x+2}\mathrm dx=\frac{22}{7}-\pi$$


 * Symmetrical Wallis's formula**

$$\frac{\prod_{n=1}^{\infty}\left(\frac{2n}{2n-1}\cdot\frac{2n}{2n+1}\right)^{(-1)^{n+1}}}{\sum_{n=1}^{\infty}(-1)^{n+1}\frac{2n+2n}{(2n-1)(2n+1)}}=\frac{4}{\pi}$$


 * Euler's constant**

$$\lim_{n \to \infty}\left(\sum_{k=1}^{n-1}\frac{1}{k}-\sum_{k=n}^{(n-1)^2}\frac{1}{k}\right)=\gamma$$


 * $\pi\approx \frac{22}{7}$**

$$\sum_{n=1}^{\infty}\left(\frac{\sin(22n)}{7n}\right)^{2k+1}=\frac{1}{2}\left(\pi-\frac{22}{7}\right)^{2k+1}$$


 * Pythagoras's constant**

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


 * Natural Logarithm of $2$**

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


 * Cube of Fibonacci sequence**

$$1^3+1^3+2^3+3^3+5^3+8^3+\cdots+F_n^3=\frac{F_nF_{n+1}^2+(-1)^{n+1}F_{n-1}+1}{2}$$


 * Difference of two squares**

$$\lim_{n \to \infty}\left(\prod_{k=1}^{n}\left(1+\frac{1}{2k-1}\right)^{2}-\prod_{k=1}^{n-1}\left(1+\frac{1}{2k-1}\right)^{2}\right)=\pi$$


 * Dirichlet Beta Function**

$$2+2\sum_{k=1}^{\infty}\frac{1-\beta(2k+1)}{2k+1}=\frac{\pi}{2}+\ln \frac{\pi}{2}$$


 * zeta Function**

$$\lim_{n \to \infty}\left(\sum_{k=1}^{n}\frac{\zeta(2k)}{k}-\ln(2n)\right)=\gamma$$


 * Fibonacci Numbers and Golden ratio**

$$\sqrt{\frac{\phi F_{2n-1}}{F_{2n}}}=1+\frac{1}{\sqrt{\phi^{4n-1}F_{2n-1}F_{2n}}+\phi^{2n-1}F_{2n}}$$


 * Ramanujan's Type**

$$16\sum_{n=1}^{\infty}\frac{n^3}{e^{n\pi}-1}-16^2\sum_{n=1}^{\infty}\frac{n^3}{e^{4n\pi}-1}=1$$


 * arctan**

$$\sum_{k=0}^{\infty}\arctan\left(\frac{4}{k^2+4k+1}\right)=\pi$$

$$\sum_{n=0}^{2N}(-1)^n\frac{(2N-n)!(2N+n)!}=-\frac{1}{N!(3N)!}$$


 * $\phi$**

$$\sum_{n=1}^{\infty}(-1)^{n+1}\left(\frac{F_n}{F_{n+1}F_{n+2}}\right)^2=\frac{1}{\phi^3}$$

$$\sum_{n=1}^{\infty}(-1)^{n+1}\left(\frac{2F_{n+1}}{F_{n+2}F_{n+3}}\right)^2=\frac{1}{\phi^6}$$


 * Harmonic**

$$\sum_{k=2}^{n-2}(-1)^k\frac{B_{n-k}B_k}{n-k}\left(1-{n \choose k}\right )=H_nB_n$$

$$\sum_{k=2}^{n-2}(-1)^k\frac{B_{n-k}B_k}{k}\left(1-{n \choose k}\right )=H_nB_n$$

$$-\frac{1}{5}\int_{0}^{1}\int_{0}^{1}\frac{\mathrm dx \mathrm dy}{\left(\frac{1}{5}-x+x^2\right)\sqrt{1-y+\frac{y^2}{5}}}=\ln\left(\frac{1}{\phi^4}\right)\ln\left(\frac{\phi^2+1}{\phi^4}\right)$$

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$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2\phi^{j}k+(-\phi)^{-j}\right]^2}{2k-1}=\frac{L_j^2}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2\phi^{j}k-(-\phi)^{-j}\right]^2}{2k-1}=\frac{5F_j^2}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2\phi^{j} k+(-\phi)^{-j}\right]^2}{2k-1}=\color{brown}{-\frac{2L_{j}^2}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2\phi^{j} k-(-\phi)^{-j}\right]^2}{2k-1}=\color{brown}{-\frac{10F_{j}^2}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)[2\phi^{j}k+\phi^{-j}]^2}{2k-1}=\color{red}{\frac{2+L_{2j}}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)[2\phi^{j}k-\phi^{-j}]^2}{2k-1}=\color{red}{\frac{L_{2j}-2}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)[4\phi^{4j}k^2+1]}{2k-1}=\color{green}{\frac{\phi^{2j}L_{2j}}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2\phi^{j} k-\phi^{-j}\right]^2}{2k-1}=\color{orange}{\frac{2(2-L_{2j})}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2\phi^{j} k-\phi^{-j}\right]^2}{2k-1}=\color{orange}{-\frac{2(2+L_{2j})}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)[4\phi^{4j}k^2+1]}{2k-1}=\color{green}{\frac{-2\phi^{2j}L_{2j}}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)[2\phi^{2j}k^2+\phi^{-2j}k-1]}{2k-1}=\color{purple}{\frac{2-L_{2j}}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)[2\phi^{2j}k^2+\phi^{-2j}k+1]}{2k-1}=-\color{purple}{\frac{2+L_{2j}}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2\phi^{4j}k+1)k}{2k-1}=-\color{green}{\frac{\phi^{2j}L_{2j}}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2\phi^{4j}k^2-k+1)}{2k-1}=-\color{green}{\frac{\phi^{2j}L_{2j}}{\pi}}$$

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$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)[4a^2k^2+4abk+b^2]}{2k-1}=-\frac{2a^2+4ab+2b^2}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2\phi^{2j-1} k+\phi^{2j-1}\right]^2}{2k-1}=\color{purple}{-\frac{10F_{2j-1}^2}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2\phi^{2j} k+\phi^{-2j}\right]^2}{2k-1}=\color{brown}{-\frac{2L_{2j}^2}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2\phi^{j} k+\phi^{-j}\right]^2}{2k-1}=\color{green}{-\frac{2(L_{2j}+2)}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2\phi^{2j-1} k-\phi^{2j-1}\right]^2}{2k-1}=\color{orange}{-\frac{2L_{2j-1}^2}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2\phi^{2j} k-\phi^{-2j}\right]^2}{2k-1}=\color{grey}{-\frac{10F_{2j}^2}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2\phi^{j} k-\phi^{-j}\right]^2}{2k-1}=\color{pink}{-\frac{2(L_{2j}-2)}{\pi}}$$

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$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)[2a^2k^2+b^2k+ab]}{2k-1}=-\frac{(a+b)^2}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)[2\phi^{2(2j-1)}k^2+\phi^{2(-2j+1)}k+1]}{2k-1}=\color{red}{-\frac{5F_{2j-1}^2}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)[2\phi^{4j}k^2+\phi^{-4j}k+1]}{2k-1}=\color{green}{-\frac{L_{2j}^2}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)[2\phi^{2(2j-1)}k^2+\phi^{2(-2j+1)}k-1]}{2k-1}=\color{brown}{-\frac{L_{2j-1}^2}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)[2\phi^{4j}k^2+\phi^{-4j}k-1]}{2k-1}=\color{purple}{-\frac{5F_{2j}^2}{\pi}}$$

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$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)[2\phi^{2j-1}k-\phi^{-2j+1}]^2}{2k-1}=\color{red}{\frac{L_{2j-1}^2}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)[2\phi^{2j}k-\phi^{-2j}]^2}{2k-1}=\color{blue}{\frac{5F_{2j}^2}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)[2\phi^{2j-1}k+\phi^{-2j+1}]^2}{2k-1}=\color{purple}{\frac{5F_{2j-1}^2}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)[2\phi^{2j}k+\phi^{-2j}]^2}{2k-1}=\color{green}{\frac{L_{2j}^2}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2\left(\frac{\phi^n+1}{2}\right)^2k-\left(\frac{\phi^n-1}{2}\right)^2\right]^2}{2k-1}=\frac{\phi^}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2\left(\frac{\phi^n-1}{2}\right)^2k-\left(\frac{\phi^n+1}{2}\right)^2\right]^2}{2k-1}=\frac{\phi^}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2\phi^2 k-\phi^{-2}\right]^2}{2k-1}=\frac{5}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2\phi^{-2} k-\phi^{2}\right]^2}{2k-1}=\frac{5}{\pi}$$

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$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2\left(4\phi+3\right)^2k-\left(4\phi+2\right)^2\right]^2}{2k-1}=\frac{\phi^{12}}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2\left(17\phi+11\right)^2k-\left(17\phi+10\right)^2\right]^2}{2k-1}=\frac{\phi^{18}}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2\left(72\phi+45\right)^2k-\left(72\phi+44\right)^2\right]^2}{2k-1}=\frac{\phi^{24}}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2\left(\frac{\phi F_m}{2}+\frac{F_{m-1}+1}{2}\right)^2k-\left(\frac{\phi F_m}{2}+\frac{F_{m-1}-1}{2}\right)^2\right]^2}{2k-1}=\frac{\phi^}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2\left(305\phi+189\right)^2k-\left(305\phi+188\right)^2\right]^2}{2k-1}=\frac{\phi^}{\pi}$$

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$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2c^4k-(c^2-1)^2\right]^2}{2k-1}=\frac{(2c^2-1)^2}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2\phi^2 k-1\right]^2}{2k-1}=\frac{\phi^2}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2k-\phi^2\right]^2}{2k-1}=\frac{\phi^2}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2\phi^{-2}k-1\right]^2}{2k-1}=\frac{1}{\phi^2\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2k-\phi^{-2}\right]^2}{2k-1}=\frac{1}{\phi^2\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(-4\phi^2k+\phi^4+1)}{2k-1}=\frac{\phi^2}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(\phi^4k+k-\phi^2)k}{2k-1}=\frac{\phi^2}{4\pi}$$

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Pythagoras's triple:

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2a^4k-b^4)^2}{2k-1}=\frac{(a^2-b^2)^2\cdot c^4}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left(2\phi k-\frac{1}{\phi}\right)^2}{2k-1}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left(2k-9\phi^4\right)^2}{2k-1}=\frac{\phi^{12}}{\pi}\left(3+\frac{1}{\phi^2}\right)^2$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left(k-2\phi^4\right)^2}{2k-1}=\frac{\phi^{10}}{\pi}\left(1+\frac{1}{2\phi^2}\right)^2$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left(\phi^4 k-2\right)^2}{2k-1}=\frac{1}{\pi}\left(\frac{\phi}{2}+\frac{1}{\phi}\right)^2$$

$$ \frac{1}{1-\phi^8}\left[\phi^{10}\left(1+\frac{1}{2\phi^2}\right)^2-\left(\frac{\phi}{2}+\frac{1}{\phi}\right)^2\right]=-\frac{15}{4}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left(32k-\phi^6\right)^2}{2k-1}=\color{red}{\frac{(16-\phi^6)^2}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)[2k\cos^4(z)-\sin(z)^4]^2}{2k-1}=\frac{\cos^2(2z)}{\pi}$$

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$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)[(4a+2)k^2+(2b-1)k+c]}{2k-1}=\frac{a+b+c}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)[4ak^2+2bk+c]}{2k-1}=\frac{a+b+c}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2ak+c)^2}{2k-1}=\frac{(a+c)^2}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2\sin^2(\alpha)k+\cos^2(\alpha)\right]^2}{2k-1}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2\phi F_nk+F_{n-1})^2}{2k-1}=\frac{\phi^{2n}}{\pi}$$

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$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2a^2k+b^2)^2}{2k-1}=\frac{c^4}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2\phi k+1)^2}{2k-1}=\frac{\phi^4}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2k+\phi)^2}{2k-1}=\color{blue}{\frac{\phi^4}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2\phi k+1)^2}{2k-1}=\color{blue}{\frac{\phi^4}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(4k-\phi)^2}{2k-1}=\color{red}{\frac{1}{\phi^4\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2\phi k-2)^2}{2k-1}=\color{red}{\frac{1}{\phi^4\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(6k+\phi^2)^2}{2k-1}=\color{brown}{\frac{1}{\phi^4\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2\phi^2 k+3)^2}{2k-1}=\color{brown}{\frac{1}{\phi^4\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(4k-\phi^2)^2}{2k-1}=\color{brown}{\frac{1}{\phi^2\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2\phi^2 k-2)^2}{2k-1}=\color{brown}{\frac{1}{\phi^2\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2k-3\phi^2)^2}{2k-1}=\color{orange}{\frac{\phi^8}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(6\phi^2 k-1)^2}{2k-1}=\color{orange}{\frac{\phi^8}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left(\frac{2k}{\phi^2}+\frac{1}{\phi}\right)^2}{2k-1}=\color{orange}{\frac{1}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left(2\phi^2 k-\phi\right)^2}{2k-1}=\color{orange}{\frac{1}{\pi}}$$

--

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)[4a^2k^2+4ak+1]}{2k-1}=\frac{(a+1)^2}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\color{red}{(2\phi k+1)^2}}{2k-1}=\frac{\phi^4}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\color{green}{(2\phi k-1)^2}}{2k-1}=\frac{1}{\phi^2\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\color{green}{(2k+\phi)^2}}{2k-1}=\frac{\phi^4}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\color{brown}{(2k-\phi)^2}}{2k-1}=\frac{1}{\phi^2\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\color{purple}{(2k-\phi^2)^2}}{2k-1}=\frac{\phi^2}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\color{grey}{(4\phi k+1)^2}}{2k-1}=\frac{\phi^6}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\color{grey}{(4\phi k-1)^2}}{2k-1}=\frac{5}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\color{grey}{(4k-\phi)^2}}{2k-1}=\frac{1}{\phi^4\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\color{grey}{(4k+\phi)^2}}{2k-1}=\frac{5\phi^2}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\color{pink}{(4k-\phi^2)^2}}{2k-1}=\frac{1}{\phi^2\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\color{pink}{(6k-\phi^2)^2}}{2k-1}=\frac{1}{\phi^4\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\color{pink}{(6\phi^2 k-1)^2}}{2k-1}=\frac{\phi^8}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\color{orange}{(4\phi^2 k-1)^2}}{2k-1}=\frac{\phi^6}{\pi}$$

-

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)[4k^2+4ck+c^2]}{2k-1}=\frac{(c+1)^2}{\pi}$$

$$\color{red}{\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2k+\phi)^2}{2k-1}=\frac{\phi^4}{\pi}}$$

$$\color{blue}{\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2k-\phi)^2}{2k-1}=\frac{1}{\phi^2\pi}}$$

$$\color{blue}{\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left(2k+\frac{1}{\phi}\right)^2}{2k-1}=\frac{\phi^2}{\pi}}$$

$$\color{blue}{\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left(2k-\frac{1}{\phi}\right)^2}{2k-1}=\frac{1}{\phi^4\pi}}$$

$$\color{blue}{\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left(2\phi k-\frac{1}{\phi}\right)^2}{2k-1}=\frac{1}{\pi}}$$

$$\color{blue}{\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left(2k+2\phi\right)^2}{2k-1}=\frac{\phi^6}{\pi}}$$

$$\color{blue}{\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left(2k-2\phi\right)^2}{2k-1}=\frac{5}{\pi}}$$

$$\color{blue}{\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left(2k-\frac{2}{\phi}\right)^2}{2k-1}=\frac{1}{\phi^6\pi}}$$

$$\color{blue}{\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left(2k+\frac{2}{\phi}\right)^2}{2k-1}=\frac{5}{\pi}}$$

$$\color{blue}{\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left(2k-\frac{2}{\phi^2}\right)^2}{2k-1}=\frac{1}{\phi^6\pi}}$$

$$\color{blue}{\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left(2k-\frac{3}{\phi^2}\right)^2}{2k-1}=\frac{1}{\phi^8\pi}}$$

$$\color{blue}{\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left(2k-3\phi^2\right)^2}{2k-1}=\frac{1}{\phi^8\pi}}$$

$$\color{blue}{\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left(2k-2\phi^2\right)^2}{2k-1}=\frac{\phi^6}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)[(4a+2)k^2+(2b-1)k+c]}{2k-1}=\frac{a+b+c}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)[4ak^2+2bk+c]}{2k-1}=\frac{a+b+c}{\pi}$$

---

Pythagoras triples: (a,b,c)

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{[n-k)(2a^2k^2+(b^2+1)k]}{2k-1}=-\frac{c^2+1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{[n-k)(2a^2k^2+b^2k]}{2k-1}=-\frac{c^2}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{[n-k)(2k^2\sin^2(\alpha)+k\cos^2(\alpha)]}{2k-1}=-\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{[n-k)(2\phi F_nk^2+F_{n-1}k]}{2k-1}=-\frac{\phi^n}{\pi}$$

Pythagoras triples: (a,b,c)

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{[n-k)(2a^2k^2+(b^2+1)k-c]}{2k-1}=-\frac{(c-1)^2}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2\phi k^2+2k-\phi)}{2k-1}=-\frac{1}{\phi^2\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2k^2+\phi^2 k-\phi)}{2k-1}=-\frac{1}{\phi^2\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2k^2+3\phi k-\phi^2)}{2k-1}=-\frac{1}{\phi\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(4k^2+\phi k-\phi)}{2k-1}=-\frac{1}{\phi^2\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2\phi^2 k^2-3\phi k+1)}{2k-1}=\frac{1}{\phi^3\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(4\phi k^2-k-1)}{2k-1}=-\frac{1}{\phi^3\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2\phi^2 k^2-3\phi k+1)}{2k-1}=\frac{1}{\phi^3\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(6\phi k^2-\phi^2 k-1)}{2k-1}=-\frac{1}{\phi^3\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2\phi^3 k^2-4\phi k+1)}{2k-1}=\frac{1}{\phi^3\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(6\phi k^2+k-\phi^2)}{2k-1}=-\frac{1}{\phi\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2k^2-4\phi^2k+\phi^3)}{2k-1}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(10\phi^2 k^2+(1-\phi^5)k-1)}{2k-1}=-\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(6\phi^3 k^2-6\phi^2 k+1)}{2k-1}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(6\phi k^2+3k-\phi^2)}{2k-1}=-\frac{\phi^2}{\pi}$$

--

Pythagoras triples: (a,b,c)

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{[n-k)(2a^2k^2+(b^2+1)k+c]}{2k-1}=-\frac{(c+1)^2}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2\phi k^2+2k+\phi)}{2k-1}=-\frac{\phi^4}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2k^2+\phi^2 k+\phi)}{2k-1}=-\frac{\phi^4}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2\phi^3 k^2-4\phi k-1)}{2k-1}=\frac{\phi^3}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(6\phi k^2-\phi^2 k+1)}{2k-1}=-\frac{\phi^3}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2\phi^2 k^2-3\phi k-1)}{2k-1}=\frac{\phi^3}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(4\phi k^2-k+1)}{2k-1}=-\frac{\phi^3}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2\phi^2 k^2+k+\phi)}{2k-1}=-\frac{\phi^4}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(8\phi^2 k^2+k+2\phi)}{2k-1}=-\frac{\phi^6}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2\phi^2 k^2+k-\phi)}{2k-1}=-\frac{1}{\phi^2\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2\phi^4 k^2+k-\phi^2)}{2k-1}=-\frac{\phi^2}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(8\phi^4 k^2+k-2\phi^2)}{2k-1}=-\frac{\phi^6}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(18\phi^4 k^2+k-3\phi^2)}{2k-1}=-\frac{\phi^8}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(4\phi^3 k^2+3k-2\phi^2)}{2k-1}=-\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)\left[2\phi^2(\phi^2+1)k^2+2k-\phi^3\right]}{2k-1}=-\frac{3}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(8\phi^2 k^2+k-2\phi)}{2k-1}=-\frac{5}{\pi}$$

--

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(k-n)(4k^2F_{n+1}-2kF_n+\phi F_n)}{2k-1}=\frac{L_n+F_n\sqrt{5}}{\pi}$$

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$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2ak^2+bk+c)}{2k-1}=-\frac{a+b+2c}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2\phi k-1)(\phi k+1)}{2k-1}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(4k^2F_{n+1}-2kF_n+\phi F_n)}{2k-1}=-\frac{2\phi^n}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2k^2\sin^2(\alpha)+k\cos^2(\alpha)+1)}{2k-1}=\frac{1}{\pi}$$

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$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{(2n-1)(2n-3)}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)^2(1-38k)}{(2k-1)(2k-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{n-k}{2k-1}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{n-k}{2k-1}=-\frac{2}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)k}{2k-1}=-\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(k-1)}{2k-1}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(k-1)(4k-1)}{2k-1}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2k^2-1)}{2k-1}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2jk+1)k}{2k-1}=-\frac{j+1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(k-1)(2jk-1)}{2k-1}=\frac{j-1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(4k^2+2jk-1)}{2k-1}=-\frac{j}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2jk^2+2k-1)}{2k-1}=-\frac{j}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{2n-1}\right)^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2ak^2+bk+c)}{2k-1}=-\frac{a+b+2c}{\pi}$$

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$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{(2n-1)(2n-3)}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2k^2+41k+2)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{(2n-1)(2n-3)}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2k^2-103k+2)}{(2k-1)(2k+1)(2k-3)}=-\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{(2n-1)(2n-3)}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(k^2+jk+1)}{(2k-1)(2k+1)(2k-3)}=\frac{2j+31}{144\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{(2n-1)(2n-3)}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2k^2+jk+2)}{(2k-1)(2k+1)(2k-3)}=\frac{2j+62}{144\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{(2n-1)(2n-3)}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(k^2-16k+1)}{(2k-1)(2k+1)(2k-3)}=-\frac{1}{144\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{(2n-1)(2n-3)}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(k^2-15k+1)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{144\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{(2n-1)(2n-3)}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2k^2-31k+2)}{(2k-1)(2k+1)(2k-3)}=\frac{0}{144\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\left(\frac{1}{(2n-1)(2n-3)}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)F_k(A,B)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

$$ F_k(A,B)=(A-B)(2k^2-31k+2)+3(38k^2-565k+38)$$

$$ F_k(A,57)=A(2k^2-31k+2)+72k $$