| $0$ | | | | $\sum\limits_{n=0}^{\infty}\frac{{({}^{2n}_{\; n}) }^2}{16^n(n+1)}=\frac{4}{\pi}$ |
| $1$ | $1$ | $0$ | $0$ | $\sum\limits_{n=0}^{\infty}\frac{{({}^{2n}_{\; n}) }^2}{16^n(n+1)}O_n=\frac{4}{\pi}(1-\ln2)$[13, (2.17)] |
| $1$ | $0$ | $0$ | $1$ | $\sum\limits_{n=0}^{\infty}\frac{{({}^{2n}_{\; n}) }^2}{16^n(n+1)}H_{n+1}=\frac{16}{\pi}(1-\ln2)$[13, (2.18)] |
| $2$ | $1$ | $0$ | $0$ | $\sum\limits_{n=0}^{\infty}\frac{{({}^{2n}_{\; n})}^2}{16^n(n+1)}(O_n^2-O_n^{(2)}) =\frac{4}{\pi}(2-2\ln2+\ln^22-\frac{\pi^2}{12})$ [14, Ex.4.2] |
| $2$ | $0$ | $0$ | $1$ | $\begin{array}{l} \sum\limits_{n=0}^{\infty}\frac{{({}^{2n}_{\; n})}^2}{16^n(n+1)}(H_{n+1}^{(2)}\!+\!H_{n+1}^2\!-\!2H_{n+1}) \!=\!\frac{4}{\pi}(16\ln^22\!-\!24\ln2\!+\!16\!-\!\frac{2\pi^2}{3}) {\rm [14, Ex.4.7]} \end{array}$ |
| $3$ | $1$ | $0$ | $0$ | $\begin{array}{l} \sum\limits_{n=0}^{\infty}\frac{{({}^{2n}_{\; n}) }^2}{16^n(n+1)}(O_{n}^3-3O_{n}O_n^{(2)}+2O_{n}^{(3)}) \\ =\frac{1}{\pi}(24-16\ln2+4\ln^22-4\ln^32-\pi^2+\pi^2\ln2-6\zeta(3))\ {\rm [14, Ex.4.12]} \end{array}$ |
| $3$ | $0$ | $0$ | $1$ | $\begin{array}{l} \sum\limits_{n=0}^{\infty}\frac{{({}^{2n}_{\; n}) }^2}{16^n(n+1)}(H_{n+1}^3 -3H_{n+1}^2+3H_{n+1}H_{n+1}^{(2)}-3H_{n+1}^{(2)}+2H_{n+1}^{(3)}) \\ =\frac{8}{\pi}(60-60\ln2+24\ln^22-32\ln^32-3\pi^2+4\pi^2\ln2-12\zeta(3))\ {\rm [14, Ex.4.13]} \end{array}$ |
| $4$ | $1$ | $0$ | $0$ | $\begin{array}{l} \sum\limits_{n=0}^{\infty}\frac{{({}^{2n}_{\; n})}^2}{16^n(n+1)}(O_{n}^{4}-6O_{n}^2O_n^{(2)} +3(O_n^{(2)})^2+8O_nO_{n}^{(3)}-6O_n^{(4)})\\ =\frac{4}{\pi}(24-24\ln2+12\ln^22-4\ln^32+\ln^42-\pi^2+\pi^2\ln2 -\frac{1}{2}\pi^2\ln^22\\ +\frac{\pi^4}{48}-6\zeta(3)+6\zeta(3)\ln2-\frac{21}{4}\zeta(4)) \end{array}$ |
| $4$ | $0$ | $0$ | $1$ | $\begin{array}{l} \sum\limits_{n=0}^{\infty}\frac{{({}^{2n}_{\; n}) }^2}{16^n(n+1)}(-4H_{n+1}^3+H_{n+1}^4-12H_{n+1}H_{n+1}^{(2)} +6H_{n+1}^2H_{n+1}^{(2)}\\ +3(H_{n+1}^{(2)})^2-8H_{n+1}^{(3)}+8H_{n+1}H_{n+1}^{(3)}+6H_{n+1}^{(4)}) \\ =\frac{4}{\pi}(1152-1920\ln2+1536\ln^22-768\ln^32+256\ln^42-64\pi^2+96\pi^2\ln2 \\ -64\pi^2\ln^22+\frac{4}{3}\pi^4-288\zeta(3)+384\ln2\cdot\zeta(3)-168\zeta(4)) \end{array}$ |