User:BBBitcoin

$$\sum_{n=0}^{\infty}\left(\frac{C_{n+\frac{1}{2}}}{2^{2n+1}}\right)^2(3n+4)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\left(\frac{C_{n+\frac{1}{2}}}{2^{2n+3}}\right)^2(4n+5)=\frac{1}{\pi^2}$$

$$\sum_{n=0}^{\infty}\frac{C_{n}C_{n-\frac{1}{2}}}{16^n}n=\frac{\ln(4)}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_{n+1}C_{n-\frac{1}{2}}}{16^n}n=\frac{2}{\pi}$$

$$\sum_{n=0}^{\infty}\left(\frac{C_{n+\frac{1}{2}}}{2^{2n-1}}\right)^2\frac{1}=\pi$$; wrong

$$A_c+1\approx \frac{3}{\phi}\cdot 2^c(c+1)^2T_{1/2}(c)$$, c>=0

$$\sum_{n=0}^{\infty}\left(\frac{C_{n+1}}{2^{2n+3}}\right)^2(5n+9)=\frac{1}{\pi}$$