メモ2　(随時更新予定)

\begin{align*}     \zeta(3)&=\frac{1}{16}\sum_{0< n}(-1)^{n+1}\frac{56n^2-32n+5}{n^3(n-\frac{1}{2})^2}\frac{1}{\binom{2n}{n}\binom{3n}{n}}\\     10-\pi^2&=\frac{1}{24}     \sum_{0< n}(-1)^{n+1}\frac{56n^2+24n+3}{n^2(n+\frac{1}{2})(n+\frac{1}{6})}\frac{\binom{2n}{n}^2}{\binom{3n}{n}\binom{6n}{3n}}   \end{align*}

\begin{align*}     -2\zeta(3)-{10\sqrt{3}}\pi-80\log(2)+90\log(3)     &\overset{?}{=}\sum_{0\leq n< m}(-1)^{n}\binom{2n}{n}\binom{3n}{n}\frac{(-1)^{m}(56m^2-32m+5)}{m^3(m-\frac{1}{2})^2}\frac{1}{\binom{2m}{m}\binom{3m}{m}}\\     &+\frac{1}{162}\sum_{0\leq n< m}(-1)^{n}\binom{2n}{n}\binom{3n}{n}\frac{(-1)^{m}(56m^2+80m+29)}{(m+1)(m+\frac{1}{2})^2(m+\frac{1}{3})(m+\frac{2}{3})}\frac{1}{\binom{2m}{m}\binom{3m}{m}}\\     &+\frac{1}{6}\sum_{0< n\leq m}\frac{(-1)^n\binom{3n}{n}\binom{6n}{3n}}{n\binom{2n}{n}^2}\frac{(-1)^{m}(56m^2+24m+3)}{m^2(m+\frac{1}{2})(m+\frac{1}{6})}\frac{\binom{2m}{m}^2}{\binom{3m}{m}\binom{6m}{3m}}\\     &-\frac{1}{27}\sum_{0< n\leq m}\frac{(-1)^{n}\binom{3n}{n}\binom{6n}{3n}}{(n-\frac{1}{6})\binom{2n}{n}^2}\frac{(-1)^{m}(56m^2+24m+3)}{m^2(m+\frac{1}{2})(m+\frac{1}{6})}\frac{\binom{2m}{m}^2}{\binom{3m}{m}\binom{6m}{3m}}   \end{align*}

$\boldsymbol{k}=(k_1,\ldots,k_r)$:許容インデックス
$\boldsymbol{\epsilon}=(\epsilon_1,\ldots,\epsilon_r)\in\lbrace 0,1\rbrace^r$
$$\varsigma\begin{pmatrix}\boldsymbol{\epsilon}\\\boldsymbol{k}\end{pmatrix} :=\sum_{0=n_0< n_1<\cdots< n_r}\frac{n_r}{(n_1-\frac{1}{2})^{k_1}(n_2-\frac{1}{2})^{k_2}\cdots(n_r-\frac{1}{2})^{k_r}}\frac{\binom{2n_r}{n_r}}{2^{2n_r}}\Bigg(\frac{2^{2n_{i-1}}}{\binom{2n_{i-1}}{n_{i-1}}}\frac{\binom{2n_i}{n_i}}{2^{2n_i}}\Bigg)^{\epsilon_i}$$
このとき,正整数$a_1,\ldots,a_s,b_1,\ldots,b_s$と$(\epsilon_1,\ldots,\epsilon_r)\in\lbrace 0,1\rbrace^r$について次が成立する。
$$\varsigma\begin{pmatrix}   \lbrace0\rbrace^{a_1-1},&\epsilon_1,&\ldots,&\lbrace0\rbrace^{a_s-1},&\epsilon_s\\    \lbrace1\rbrace^{a_1-1},&b_1+1,&\ldots,&\lbrace1\rbrace^{a_s-1},&b_s+1\end{pmatrix}    =    \varsigma\begin{pmatrix}   \lbrace0\rbrace^{b_s-1},&\epsilon_s,&\ldots,&\lbrace0\rbrace^{b_1-1},&\epsilon_1\\   \lbrace1\rbrace^{b_s-1},&a_s+1,&\ldots,&\lbrace1\rbrace^{b_1-1},&a_1+1\end{pmatrix}$$

上の関係式の一般化は次のようになる。
$$\varsigma_{(x)}\begin{pmatrix}\boldsymbol{\epsilon}\\\boldsymbol{k}\end{pmatrix} :=\sum_{0=n_0< n_1<\cdots< n_r}\prod_{i=1}^{r}\frac{1}{((n_i-\frac{1}{2})-x)(n_i-\frac{1}{2})^{k_i-1}}\frac{(\frac{1}{2}-x)_{n_r}}{(1-x)_{n_r-1}}\Bigg(\frac{2^{2n_{i-1}}}{\binom{2n_{i-1}}{n_{i-1}}}\frac{\binom{2n_i}{n_i}}{2^{2n_i}}\Bigg)^{\epsilon_i}$$
このとき,正整数$a_1,\ldots,a_s,b_1,\ldots,b_s$と$(\epsilon_1,\ldots,\epsilon_r)\in\lbrace 0,1\rbrace^r$について次が成立する。
$$\varsigma_{(x)}\begin{pmatrix}   \lbrace0\rbrace^{a_1-1},&\epsilon_1,&\ldots,&\lbrace0\rbrace^{a_s-1},&\epsilon_s\\    \lbrace1\rbrace^{a_1-1},&b_1+1,&\ldots,&\lbrace1\rbrace^{a_s-1},&b_s+1\end{pmatrix}    =    \varsigma_{(x)}\begin{pmatrix}   \lbrace0\rbrace^{b_s-1},&\epsilon_s,&\ldots,&\lbrace0\rbrace^{b_1-1},&\epsilon_1\\   \lbrace1\rbrace^{b_s-1},&a_s+1,&\ldots,&\lbrace1\rbrace^{b_1-1},&a_1+1\end{pmatrix}$$

$\boldsymbol{k}=(k_1,\ldots,k_r)=(\lbrace1\rbrace^{a_1-1},b_1+1,\ldots,\lbrace1\rbrace^{a_s-1},b_s+1)$:許容インデックス
$\boldsymbol{l}=(l_1,\ldots,l_s)$:インデックス
$\boldsymbol{m}=(m_1,\ldots,m_t)$:インデックス
$\boldsymbol{\epsilon(\boldsymbol{k})}=(\lbrace0\rbrace^{a_1-1},\epsilon_1,\ldots,\lbrace0\rbrace^{a_s-1},\epsilon_s)\quad(\epsilon_1,\ldots,\epsilon_s)\in\lbrace 0,1\rbrace^s$
\begin{align*}   \varsigma\begin{pmatrix}\boldsymbol{\epsilon}(\boldsymbol{k})\\\boldsymbol{k}|\boldsymbol{l}|\boldsymbol{m}\end{pmatrix} :=&\sum_{0=n_0< n_1<\cdots< n_r}\frac{1}{(n_1-\frac{1}{2})^{k_1}(n_2-\frac{1}{2})^{k_2}\cdots(n_r-\frac{1}{2})^{k_r}}\frac{\binom{2n_r}{n_r}}{2^{2n_r}}\Bigg(\frac{2^{2n_{i-1}}}{\binom{2n_{i-1}}{n_{i-1}}}\frac{\binom{2n_i}{n_i}}{2^{2n_i}}\Bigg)^{\epsilon_i}\\ &\times \sum_{0=a_0< a_1<\cdots< a_s\leq n_r}\frac{1}{(a_1-\frac{1}{2})^{l_1}(a_2-\frac{1}{2})^{l_2}\cdots(a_s-\frac{1}{2})^{l_s}}\sum_{0=b_0< b_1\leq \cdots\leq b_t\leq n_r}\frac{1}{b^{m_1}_1b^{m_2}_2\cdots b^{m_t}_t} \end{align*}

このとき、任意の非負整数$N$に対して次が成立する。
\begin{align*}   &\sum_{\substack{|\boldsymbol{e}|=n_1\\n_1+n_2+n_3=N}}(-1)^{n_2}\varsigma\begin{pmatrix}\boldsymbol{\epsilon(\boldsymbol{k})}\\\boldsymbol{k}_{\downarrow}\oplus\boldsymbol{e}|\lbrace 1\rbrace^{n_2}|\lbrace 1\rbrace^{n_3}\end{pmatrix} +\frac{1}{2}\sum_{\substack{|\boldsymbol{e}|=n_1\\n_1+n_2+n_3=N}}(-1)^{n_2}\varsigma\begin{pmatrix}\boldsymbol{\epsilon(\boldsymbol{k})}\\\boldsymbol{k}\oplus\boldsymbol{e}|\lbrace 1\rbrace^{n_2}|\lbrace 1\rbrace^{n_3}\end{pmatrix} -\sum_{\substack{|\boldsymbol{e}|=n_1\\n_1+n_2+n_3=N-1}}(-1)^{n_2}\varsigma\begin{pmatrix}\boldsymbol{\epsilon(\boldsymbol{k})}\\\boldsymbol{k}\oplus\boldsymbol{e}|\lbrace 1\rbrace^{n_2}|\lbrace 1\rbrace^{n_3}\end{pmatrix}\\ =&\sum_{\substack{|\boldsymbol{e}|=n_1\\n_1+n_2+n_3=N}}(-1)^{n_2}\varsigma\begin{pmatrix}\boldsymbol{\epsilon(\boldsymbol{k^\dagger})}\\(\boldsymbol{k}^{\dagger})_{\downarrow}\oplus\boldsymbol{e}|\lbrace 1\rbrace^{n_2}|\lbrace 1\rbrace^{n_3}\end{pmatrix} +\frac{1}{2}\sum_{\substack{|\boldsymbol{e}|=n_1\\n_1+n_2+n_3=N}}(-1)^{n_2}\varsigma\begin{pmatrix}\boldsymbol{\epsilon(\boldsymbol{k^\dagger})}\\\boldsymbol{k^{\dagger}}\oplus\boldsymbol{e}|\lbrace 1\rbrace^{n_2}|\lbrace 1\rbrace^{n_3}\end{pmatrix} -\sum_{\substack{|\boldsymbol{e}|=n_1\\n_1+n_2+n_3=N-1}}(-1)^{n_2}\varsigma\begin{pmatrix}\boldsymbol{\epsilon(\boldsymbol{k^\dagger})}\\\boldsymbol{k^\dagger}\oplus\boldsymbol{e}|\lbrace 1\rbrace^{n_2}|\lbrace 1\rbrace^{n_3}\end{pmatrix} \end{align*}

$$\int_{0}^{1}\ (x^{-x})^{{(x^{-x})}^{(x^{-x})\cdots}}dx =\zeta(2)$$
$$\int_{0}^{1}\ (x^{-yx})^{{(x^{-yx})}^{(x^{-yx})\cdots}}dx =\frac{{\rm{{Li}_2}}(y)}{y}$$

$$\int_{0}^{1}(\log x)^k\ (x^{-x})^{{(x^{-x})}^{(x^{-x})\cdots}}dx =(-1)^k\sum_{i=0}^{k}{k\brack i}\zeta(k-i+2)\quad\left({k\brack i}\text{は第一種スターリング数}\right)$$
$$\int_{0}^{1}(\log x)^k\ (x^{-yx})^{{(x^{-yx})}^{(x^{-yx})\cdots}}dx =(-1)^k\sum_{i=0}^{k}{k\brack i}\frac{\mathrm{Li}_{k-i+2}(y)}{y}$$

$$\int_{0}^{1}\Bigg(\log(\log \frac{1}{x})\Bigg)^k\ (x^{-x})^{{(x^{-x})}^{(x^{-x})\cdots}}dx =\frac{1}{\Gamma(n)}\sum_{i=0}^{k}(-1)^{k-i}\binom{k}{i}\sum_{n=1}^{\infty}\frac{(\log n)^{k-i}}{n^2}\frac{d^i}{ds^i}\Gamma(s)\Bigg|_{n}$$
$$\int_{0}^{1}\Bigg(\log(\log \frac{1}{x})\Bigg)^k\ (x^{-yx})^{{(x^{-yx})}^{(x^{-yx})\cdots}}dx =\frac{1}{\Gamma(n)}\sum_{i=0}^{k}(-1)^{k-i}\binom{k}{i}\sum_{n=1}^{\infty}\frac{y^{n-1}(\log n)^{k-i}}{n^2}\frac{d^i}{ds^i}\Gamma(s)\Bigg|_{n}$$

$$\int_{0}^{1}\frac{K(t)\log(1-t^2)}{\sqrt{1-t^2}}dt=-\frac{\pi^4}{6\Gamma(\frac{3}{4})^4}$$

$$\sum_{0< m\leq n}\frac{\binom{2n}{n}^3}{m^22^{6n}}-\sum_{0\leq m< n}\frac{\binom{2n}{n}^3}{(m+\frac{1}{2})^22^{6n}}=\frac{\pi}{\Gamma(\frac{3}{4})^4}\Big(8\beta(2)-5\zeta(2)\Big)$$

$$\sum_{n=0}^{\infty}\frac{(\frac{1}{2})_n(\frac{1}{4})_n(\frac{3}{4})_n}{n!^3}\Bigg(2\sum_{k=1}^{n}\frac{1}{k}-\sum_{k=1}^{n}\frac{1}{k-\frac{1}{2}}\Bigg) =\frac{\pi}{\Gamma(\frac{5}{8})^2\Gamma(\frac{7}{8})^2}\Bigg(\sqrt{2}\pi-6\log(2)\Bigg)$$

$$\zeta(3)=\frac{1}{2^{11}\cdot 7}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(7168n^5-12672n^4+8592n^3-2804n^2+441n-27)}{n^5(n-\frac{1}{2})^3}\frac{2^{12n}}{\binom{3n}{n}\binom{2n}{n}\binom{4n}{2n}^3}$$

$$-\frac{1}{2}\log2 =\sum_{0< n,m}\frac{(-1)^m}{2^nm}\Bigg(\frac{n!m!}{(n+m)!}\Bigg)^2+2\sum_{0< n,m}\frac{(-1)^m}{n2^n}\Bigg(\frac{n!m!}{(n+m)!}\Bigg)^2$$
$$-\frac{1}{4}(\log2)^2=\sum_{0< n,m}\frac{(-1)^n}{n}\frac{1}{m2^m}\Bigg(\frac{n!m!}{(n+m)!}\Bigg)^2$$
$$\sum_{0< n}\frac{z^{n}}{n}-\frac{1}{z-1}\sum_{0< n}\frac{(\frac{z}{z-1})^{n}}{n} =2\sum_{0< n,m}z^{n}\frac{(\frac{z}{z-1})^m}{m}\Bigg(\frac{n!m!}{(n+m)!}\Bigg)^2-\frac{2}{z-1}\sum_{0< n,m}\frac{z^n}{n}\left(\frac{z}{z-1}\right)^{m}\Bigg(\frac{n!m!}{(n+m)!}\Bigg)^2$$
$$\sum_{0< n}\frac{z^n}{n^2}+\sum_{0< n}\frac{(\frac{z}{z-1})^n}{n^2} =2\sum_{0< n,m}\frac{z^n}{n}\frac{(\frac{z}{z-1})^m}{m}\Bigg(\frac{n!m!}{(n+m)!}\Bigg)^2$$

$$\sum_{\substack{0< n_1< n_2\\0< m}}\frac{1}{m2^m}\frac{(-1)^{n_1+n_2}}{n_1 n_2}\Bigg(\frac{n_2!m!}{(n_2+m)!}\Bigg)^2 =\frac{31}{64}\zeta(3)-\frac{\pi^2}{16}\log(2)+\frac{\log^3(2)}{24}$$

$$\sum_{0< n}\frac{2^{6n}}{n^3\binom{2n}{n}^3}=\frac{1}{32}\sum_{0< n}\frac{(-1)^{n+1}(112n^2+8n+1)}{(n-\frac{3}{4})^2(n-\frac{1}{4})^3}\frac{(\frac{3}{4})^2_n}{27^n(\frac{1}{12})_n(\frac{5}{12})_n}$$

$$\sum_{0< n,m,k}\frac{1}{nmk^2}\frac{n!m!}{(n+m)!}\frac{m!k!}{(m+k)!}\frac{k!n!}{(k+n)!}=\frac{\pi^4}{320}$$
$$\sum_{0< n,m,k}\frac{1}{n^2m^2k^2}\frac{n!m!}{(n+m)!}\frac{m!k!}{(m+k)!}\frac{k!n!}{(k+n)!}=\frac{\pi^6}{5040}$$


インデックス$(k_1,k_2,\ldots,k_r),(l_1,l_2,\ldots,l_s)$(ただし、$(l_1,l_2,\ldots,l_s)=(1)_{\rightarrow^{a_1-1}\uparrow^{b_1}\cdots\rightarrow^{a_t}\uparrow^{b_t}}$という表示を持つとする。)
に対して次が成立する。
\begin{align*}   \sum_{\substack{0< n_1<\cdots< n_r\\0< m_1<\cdots< m_s\\0< k}}\frac{1}{n^{k_1}_1\cdots n^{k_r}_r}\frac{1}{m^{l_1}\cdots m^{l_s}_s}\frac{1}{k^2}\frac{n_r!m_s!}{(n_r+m_s)!}\frac{m_s!k!}{(m_s+k)!}\frac{k!n_r!}{(k+n_r)!} =&\zeta((k_1,\ldots,k_r,2)_{\rightarrow^{b_t}\uparrow^{a_t}\cdots\rightarrow^{b_1}\uparrow^{a_1}}) \end{align*}
特に$\boldsymbol{l}=(l_1,l_2,\ldots,l_s)$が許容インデックスのときは
$$\sum_{\substack{0< n_1<\cdots< n_r\\0< m_1<\cdots< m_s\\0< k}}\frac{1}{n^{k_1}_1\cdots n^{k_r}_r}\frac{1}{m^{l_1}\cdots m^{l_s}_s}\frac{1}{k^2}\frac{n_r!m_s!}{(n_r+m_s)!}\frac{m_s!k!}{(m_s+k)!}\frac{k!n_r!}{(k+n_r)!}\\ =\zeta(\boldsymbol{k},2,\boldsymbol{l}^{\dagger})$$

$\boldsymbol{k}=(k_1,k_2,\ldots,k_r),\boldsymbol{l}=(l_1,l_2,\ldots,l_s)$を許容インデックスとする。このとき次が成立する。

$$\sum_{\substack{0< n_1<\cdots< n_r\\0< m_1<\cdots< m_s\\0< k}}\frac{1}{n^{k_1}_1\cdots n^{k_r}_r}\frac{1}{m^{l_1}_1\cdots m^{l_s}_s}\frac{1}{k^2} \frac{n_r!m_s!k!(n_r+m_s+k)!}{(n_r+m_s)!(m_s+k)!(k+n_r)!}\\ =\zeta(\boldsymbol{k},2)\zeta(\boldsymbol{l})+\zeta(\boldsymbol{l},2)\zeta(\boldsymbol{k})-\zeta(\boldsymbol{k},2,\boldsymbol{l}^{\dagger})$$

$\boldsymbol{k}=(k_1,k_2,\ldots,k_r),\boldsymbol{l}=(l_1,l_2,\ldots,l_s)$を許容インデックスとする。このとき次が成立する。
$$\sum_{\substack{0< n_1<\cdots< n_r\\0< m_1<\cdots< m_s\\0< a_1,a_2}}\frac{1}{n^{k_1}_1\cdots n^{k_r}_r}\frac{1}{m^{l_1}_1\cdots m^{l_s}_s}\frac{1}{a^2_1a^2_2} \frac{n_r!^2m_s!^2a_1!^2a_2!^2(n_r+m_s+a_1+a_2)!}{(n_r+m_s)!(n_r+a_1)!(n_r+a_2)!(m_s+a_1)!(m_s+a_2)!(a_1+a_2)!}\\ =\zeta(\boldsymbol{k},2,2)\zeta(\boldsymbol{l})+\zeta(\boldsymbol{l},2,2)\zeta(\boldsymbol{k})+\zeta(\boldsymbol{l},2)\zeta(\boldsymbol{k},2)-2\zeta(\boldsymbol{k},2,2,\boldsymbol{l}^{\dagger})$$

許容インデックス$\boldsymbol{k}=(k_1,k_2,\ldots,k_a),\boldsymbol{l}=(l_1,l_2,\ldots,l_b),\boldsymbol{j}=(j_1,j_2,\ldots,j_c)$に対して
$$Z(\boldsymbol{k}|\boldsymbol{l}|\boldsymbol{j}):=\sum_{\substack{0< n_1<\cdots< n_a\\0< m_1<\cdots< m_b\\0< r_1<\cdots< r_c}} \left(\prod_{i=1}^{a}\frac{1}{n^{k_i}_i}\right)\left(\prod_{i=1}^{b}\frac{1}{m^{l_i}_i}\right)\left(\prod_{i=1}^{c}\frac{1}{r^{j_i}_i}\right)\frac{n_a!m_b!r_c!(n_a+m_b+r_c)!}{(n_a+m_b)!(m_b+r_c)!(r_c+n_a)!}$$
とする。このとき、任意の許容インデックスの組$\boldsymbol{k},\boldsymbol{l},\boldsymbol{j}$に対して次が成立する。
$$Z(\boldsymbol{k},2|\boldsymbol{l}|\boldsymbol{j})-Z(\boldsymbol{k}|\boldsymbol{l}|\boldsymbol{j},2)=\zeta(\boldsymbol{k},2,\boldsymbol{l}^{\dagger})\zeta(\boldsymbol{j})-\zeta(\boldsymbol{j},2,\boldsymbol{l}^{\dagger})\zeta(\boldsymbol{k})$$

$$H(a,b):=\sum_{0\leq n_1<\cdots< n_a< n< m_1<\cdots< m_b}\left(\prod_{i=1}^{a}\frac{1}{(n_i+\frac{1}{2})^2}\right)\frac{\binom{2n}{n}^3}{2^{6n}}\left(\prod_{j=1}^{b}\frac{1}{m^2_j}\right)$$
このとき、任意の非負整数$r$に対して次が成立する。
$$\sum_{k=0}^{r}H(k,r-k)=\frac{\pi}{\Gamma(\frac{3}{4})^2}\frac{(-1)^r}{(2r)!}\frac{d^{2r}}{dx^{2r}}\frac{1}{\Gamma(\frac{3}{4}-x)\Gamma(\frac{3}{4}+x)}\Bigg|_{x=0}$$

$$A(\lbrace 2\rbrace^a,-1,\lbrace 2\rbrace^b):= \sum_{0\leq n_1<\cdots< n_a< n< m_1<\cdots< m_b}\frac{1}{(n_1+\frac{1}{2})^2\cdots(n_a+\frac{1}{2})^2} \frac{(-1)^n(4n+1)\binom{2n}{n}^5}{2^{10n}}\frac{1}{m_1^2\cdots m_b^2}$$
とする。このとき任意の非負整数$r$に対して、次の等式が成立する。
$$\sum_{k=0}^{r}A(\lbrace2\rbrace^k,-1,\lbrace2\rbrace^{r-k}) =\frac{(-1)^r}{(2r)!}\frac{d^{2r}}{dz^{2r}}\frac{2}{\Gamma(\frac{3}{4}-\frac{z}{2})^2\Gamma(\frac{3}{4}+\frac{z}{2})^2}\Bigg|_{z=0}$$

$$A(\lbrace2\rbrace^a,-1,\lbrace2\rbrace^b):=\sum_{0\leq n_1<\cdots< n_a< n\leq m_1\leq\cdots\leq m_b}\frac{1}{(n_1+\frac{1}{2})^2\cdots(n_a+\frac{1}{2})^2}\frac{(6n+1)\binom{2n}{n}^3}{2^{8n}}\frac{1}{(m_1+\frac{1}{2})^2\cdots(m_b+\frac{1}{2})^2}$$
$$B(\lbrace2\rbrace^a,1,\lbrace2\rbrace^b):=\sum_{0\leq n_1<\cdots< n_a< n\leq m_1\leq\cdots\leq m_b}\frac{1}{(n_1+\frac{1}{2})^2\cdots(n_a+\frac{1}{2})^2}\frac{\binom{2n}{n}^3}{(n+\frac{1}{2})2^{8n}}\frac{1}{(m_1+\frac{1}{2})^2\cdots(m_b+\frac{1}{2})^2}$$
このとき、任意の正整数$n$に対して次が成立する。
$$\sum_{\substack{a+b=n\\(a,b)\in\mathbb{Z}^2_{\ge 0}}}(-1)^aA(\lbrace2\rbrace^a,-1,\lbrace2\rbrace^b)=2\sum_{\substack{a+b=n-1\\(a,b)\in\mathbb{Z}^2_{\ge 0}}}(-1)^aB(\lbrace2\rbrace^a,1,\lbrace2\rbrace^b)$$

$$\sum_{n=0}^{\infty}\frac{(-1)^n(80n^2-8n-1)}{4n-1}\frac{\binom{2n}{n}\binom{4n}{2n}^3}{2^{20n}}\frac{n!^2}{(\frac{3}{4})^2_n}=\frac{4\sqrt{2\pi}}{\Gamma(\frac{1}{4})^2}$$

\begin{align*}   &\sum_{0< n,m}\frac{(-1)^{n+m}}{n^4m^4}\left(\frac{n!m!}{(n+m)!}\right)^2   +2\sum_{\substack{0< n_1< n_2\\0< m}}\frac{(-1)^{n_1+n_2}}{n^4_1n_2}\frac{(-1)^m}{m^3}\left(\frac{n_2!m!}{(m+m_2)!}\right)^2   +2\sum_{\substack{0< n_1< n_2\\0< m}}\frac{(-1)^{n_1+n_2}}{n^4_1}\frac{(-1)^m}{m^4}\left(\frac{n_2!m!}{(n_2+m)!}\right)^2\\   &-2\sum_{\substack{0< n_1< n_2\\0< m}}\frac{(-1)^{n_1+n_2}}{n^4_1n^3_2}\frac{(-1)^m}{m}\left(\frac{n_2!m!}{(n_2+m)!}\right)^2 -2\sum_{\substack{0< n_1< n_2\\0< m}}\frac{(-1)^{n_1+n_2}}{n^4_1n^4_2}(-1)^m\left(\frac{n_2!m!}{(n_2+m)!}\right)^2 =-\frac{41\pi^8}{7257600} \end{align*}

\begin{align*}   \sum_{n=0}^{\infty}\frac{\binom{2n}{n}^3}{2^{8n}}&=\frac{2}{3}\frac{2^{\frac{1}{3}}\sqrt{\pi}}{\Gamma(\frac{5}{6})^3}\\   \sum_{n=0}^{\infty}n\frac{\binom{2n}{n}^3}{2^{8n}}&=\frac{2}{3\pi}-\frac{1}{9}\frac{2^{\frac{1}{3}}\sqrt{\pi}}{\Gamma(\frac{5}{6})^3}\\   \sum_{n=0}^{\infty}n^2\frac{\binom{2n}{n}^3}{2^{8n}}&=\frac{1}{3}\frac{\Gamma(\frac{5}{6})^3}{2^{\frac{1}{3}}\pi^{\frac{5}{2}}}+\frac{1}{54}\frac{2^{\frac{1}{3}}\sqrt{\pi}}{\Gamma(\frac{5}{6})^3}\\ \end{align*}

$$\sum_{n=0}^{\infty}(6n+1)\frac{\binom{2n}{n}^3}{2^{8n}}=\frac{4}{\pi}$$
$$\sum_{n=0}^{\infty}(36n^2-1)\frac{\binom{2n}{n}^3}{2^{8n}}=\frac{2^{\frac{5}{3}}\cdot3\Gamma(\frac{5}{6})^3}{\pi^{\frac{5}{2}}}$$

\begin{align*}   \frac{2^{\frac{5}{3}}}{9}\frac{\pi}{\Gamma(\frac{5}{6})^6}   =&\frac{\Gamma(\frac{5}{6})^3}{2^{\frac{1}{3}}\pi^{\frac{5}{2}}}\sum_{0\leq m< n}\frac{1}{(m+\frac{1}{2})^2}\frac{\binom{2n}{n}^3}{2^{8n}}\\   &-\frac{1}{3\pi}\sum_{0\leq m< n}\frac{1}{(m+\frac{1}{2})^2}\frac{(6n+1)\binom{2n}{n}^3}{2^{8n}}\\   &+\frac{1}{18}\frac{2^{\frac{1}{3}}\sqrt{\pi}}{\Gamma(\frac{5}{6})^3}\sum_{0\leq m< n}\frac{1}{(m+\frac{1}{2})^2}\frac{(36n^2-1)\binom{2n}{n}^3}{2^{8n}} \end{align*}

\begin{align*}   \int_{0}^{1}{}_3F_2\left[\begin{matrix}\frac{1}{2},\frac{1}{2},\frac{1}{2}\\1,1\end{matrix};\frac{x}{4}\right]^2dx   =&\frac{\Gamma(\frac{5}{6})^3}{2^{\frac{1}{3}}\pi^{\frac{5}{2}}}\sum_{0\leq m< n}\frac{1}{(m+\frac{1}{2})^3}\frac{\binom{2n}{n}^3}{2^{8n}}\\   &-\frac{1}{3\pi}\sum_{0\leq m< n}\frac{1}{(m+\frac{1}{2})^3}\frac{(6n+1)\binom{2n}{n}^3}{2^{8n}}\\   &+\frac{1}{18}\frac{2^{\frac{1}{3}}\sqrt{\pi}}{\Gamma(\frac{5}{6})^3}\sum_{0\leq m< n}\frac{1}{(m+\frac{1}{2})^3}\frac{(36n^2-1)\binom{2n}{n}^3}{2^{8n}} \end{align*}

\begin{align*}   S:=\sum_{n=0}^{\infty}\frac{\binom{2n}{n}^3}{2^{12n}}&=\frac{2}{7\pi}\frac{\Gamma(\frac{1}{7})\Gamma(\frac{2}{7})\Gamma(\frac{4}{7})}{\Gamma(\frac{3}{7})\Gamma(\frac{5}{7})\Gamma(\frac{6}{7})}=\frac{16}{7\sqrt{7}}\frac{\pi^2}{\Gamma(\frac{3}{7})^2\Gamma(\frac{5}{7})^2\Gamma(\frac{6}{7})^2}\\   \sum_{n=0}^{\infty}n\frac{\binom{2n}{n}^3}{2^{12n}}&=\frac{8}{21\pi}-\frac{5}{42}S\\   \sum_{n=0}^{\infty}n^2\frac{\binom{2n}{n}^3}{2^{12n}}&=\frac{2^5}{3^2\cdot7^2\pi^2S}-\frac{2^3}{3^3\cdot7\pi}+\frac{43}{2^2\cdot3^3\cdot7^2}S \end{align*}

$$\sum_{n=0}^{\infty}(42n+5)\frac{\binom{2n}{n}^3}{2^{12n}} =\frac{16}{\pi}$$
$$\sum_{n=0}^{\infty}(1764n^2+196n+9)\frac{\binom{2n}{n}^3}{2^{12n}} =\frac{3584\pi^2}{\sqrt{7}\Gamma(\frac{1}{7})^2\Gamma(\frac{2}{7})^2\Gamma(\frac{4}{7})^2}$$

\begin{align*}   \frac{2^7}{7^3}\frac{\pi^4}{\Gamma(\frac{3}{7})^4\Gamma(\frac{5}{7})^4\Gamma(\frac{6}{7})^4}   =&\frac{2^5}{7\pi^2S}\sum_{0\leq m< n}\frac{1}{(m+\frac{1}{2})^2}\frac{\binom{2n}{n}^3}{2^{12n}}\\   &-\frac{4}{7\pi}\sum_{0\leq m< n}\frac{1}{(m+\frac{1}{2})^2}\frac{(42n+5)\binom{2n}{n}^3}{2^{12n}}\\   &+\frac{S}{28}\sum_{0\leq m< n}\frac{1}{(m+\frac{1}{2})^2}\frac{(1764n^2+196n+9)\binom{2n}{n}^3}{2^{12n}} \end{align*}

\begin{align*}   \int_{0}^{1}{}_3F_2\left[\begin{matrix}\frac{1}{2},\frac{1}{2},\frac{1}{2}\\1,1\end{matrix};\frac{x}{64}\right]^2dx   =&\frac{32}{7\pi^2S}\sum_{0\leq m< n}\frac{1}{(m+\frac{1}{2})^3}\frac{\binom{2n}{n}^3}{2^{12n}}\\   &-\frac{4}{7\pi}\sum_{0\leq m< n}\frac{1}{(m+\frac{1}{2})^3}\frac{(42n+5)\binom{2n}{n}^3}{2^{12n}}\\   &+\frac{S}{28}\sum_{0\leq m< n}\frac{1}{(m+\frac{1}{2})^3}\frac{(1764n^2+196n+9)\binom{2n}{n}^3}{2^{12n}} \end{align*}

\begin{align*}   R:=&\sum_{n=0}^{\infty}\frac{(4n)!}{(n!)^4}\frac{1}{396^{4n}}\\   &\sum_{n=0}^{\infty}\frac{(4n)!}{(n!)^4}\frac{n}{396^{4n}}=\frac{9801}{2\sqrt{2}\cdot26390\pi}-\frac{1103}{26390}R\\   &\sum_{n=0}^{\infty}\frac{(4n)!}{(n!)^4}\frac{n^2}{396^{4n}}=\frac{96059601}{2\cdot58\cdot96059600}\frac{1}{\pi^2R}-\frac{2713808691}{2\sqrt{2}\cdot1820\cdot96059600}\frac{1}{\pi}+\frac{35427684841}{2\cdot211120\cdot96059600}R \end{align*}

$$\sum_{n=0}^{\infty}\frac{(4n)!}{(n!)^4}\frac{89143308800n^2+3725845296n+77862889}{396^{4n}}   =\frac{768476808}{\pi^2R}$$

\begin{align} C:=&\sum_{n=0}^{\infty}\frac{(-1)^n(6n)!}{(3n)!(n!)^3}\frac{1}{640320^{3n}}\\ &\sum_{n=0}^{\infty}\frac{(-1)^n(6n)!}{(3n)!(n!)^3}\frac{n}{640320^{3n}}=\frac{213440\sqrt{10005}}{272570067}\frac{1}{\pi}-\frac{13591409}{545140134}C\\ &\sum_{n=0}^{\infty}\frac{(-1)^n(6n)!}{(3n)!(n!)^3}\frac{n^2}{640320^{3n}}=\frac{75965686528000}{24764813808128163}\frac{1}{C π^2}-\frac{4960117854824320\sqrt{10005}}{254061009406602376209}\frac{1}{\pi}+\frac{5720391597869733347}{18405308681456083254252}C \end{align}