| $0$ | | | | $\sum\limits_{n=0}^{\infty}\frac{({}^{2n}_{\; n}) }{4^n(2n+1)}=\frac{\pi}{2}$ |
| $1$ | $1$ | $0$ | $0$ | $\sum\limits_{n=0}^{\infty}\frac{({}^{2n}_{\; n}) }{4^n(2n+1)}O_n=\frac{\pi}{2}\ln2$[13, (2.15)] |
| $1$ | $0$ | $0$ | $1$ | $\sum\limits_{n=0}^{\infty}\frac{({}^{2n}_{\; n})}{4^n(2n+1)}O_{n+1}=\pi\ln2$[13, (2.16)] |
| $2$ | $1$ | $0$ | $0$ | $\sum\limits_{n=0}^{\infty}\frac{({}^{2n}_{\; n})}{4^n(2n+1)}(O_n^2-O_n^{(2)})=\frac{\pi}{2}\ln^22+\frac{\pi^3}{24}$ |
| $2$ | $0$ | $0$ | $1$ | $\sum\limits_{n=0}^{\infty}\frac{({}^{2n}_{\; n}) }{4^n(2n+1)}(O_{n+1}^2-2O_{n+1}+O_{n+1}^{(2)}) =2\pi\ln^22-2\pi\ln2+\frac{\pi^3}{12}$ |
| $3$ | $1$ | $0$ | $0$ | $ \sum\limits_{n=0}^{\infty}\frac{({}^{2n}_{\; n})}{4^n(2n+1)}(O_n^3-3O_nO_n^{(2)}+2O_n^{(3)})= \frac{\pi}{8}(4\ln^32+\pi^2\ln2+6\zeta(3))$ |
| $3$ | $0$ | $0$ | $1$ | $\begin{array}{l} \sum\limits_{n=0}^{\infty}\frac{({}^{2n}_{\; n}) }{4^n(2n+1)}(3O_{n+1}^2 -O_{n+1}^3+3O_{n+1}^{(2)}-3O_{n+1}O_{n+1}^{(2)}-2O_{n+1}^{(3)})\\ =\frac{\pi}{4}(24\ln^22-16\ln^32+\pi^2-2\pi^2\ln2-6\zeta(3)) \end{array}$ |
| $4$ | $1$ | $0$ | $0$ | $\begin{array}{l} \sum\limits_{n=0}^{\infty}\frac{({}^{2n}_{\; n}) }{4^n(2n+1)}(O_{n}^{4}-6O_{n}^2O_n^{(2)}+3(O_n^{(2)})^2+8O_nO_{n}^{(3)}-6O_n^{(4)})\\ =\frac{\pi}{2}(\ln^42+\frac{\ln^22}{2}\pi^2+\frac{\pi^4}{48}+6\ln2\cdot \zeta(3)+4\zeta(4)) \end{array}$ |
| $4$ | $0$ | $0$ | $1$ | $\begin{array}{l} \sum\limits_{n=0}^{\infty}\frac{({}^{2n}_{\; n}) }{4^n(2n+1)}(-4O_{n+1}^3+O_{n+1}^4-12O_{n+1}O_{n+1}^{(2)}+6O_{n+1}^2O_{n+1}^{(2)}+3(O_{n+1}^{(2)})^2\\ -8O_{n+1}^{(3)}+8O_{n+1}O_{n+1}^{(3)}+6O_{n+1}^{(4)})=\pi(-16\ln^32+8\ln^42-2\ln2\cdot\pi^2 \\ +2\ln^22\cdot\pi^2+\frac{1}{24}\pi^4-6\zeta(3)+12\ln2\cdot\zeta(3)+\frac{21}{4}\zeta(4))\end{array}$ |