メモ (随時更新予定)

○○進法と無限積
\begin{align} &\sum_{n=0}^{∞}\frac{2^n}{1+2^{2^n}}=1\\ &\sum_{n=0}^{∞}\frac{2^{2^{n}+2n}}{(1+2^{2^n})^2}=2\\ &\sum_{n=0}^{\infty}\frac{2^{2^n+3n}(2^{2^n}-1)}{(2^{2^n}+1)^3}=6\\ &\sum_{n=0}^{\infty}\frac{2^{2^n+4n}(2^{2^{n+1}}-4\cdot2^{2^n}+1)}{(2^{2^n}+1)^4}=26\\ \end{align}

\begin{align} \quad&\boldsymbol{k}=(k_1,k_2,\cdots ,k_r)\\ &\zeta(\boldsymbol{k})=(-1)^r\sum_{\substack{0\lt m_1,m_2,\cdots,m_r\\0\leq n_1,n_2,\cdots,n_r}}\frac{(-1)^{m_1+m_2+\cdots+m_r}2^{n_1+n_2+\cdots+n_r}}{(m_1 2^{n_1})^{k_1}(m_1 2^{n_1}+m_2 2^{n_2})^{k_2}\cdots(m_1 2^{n_1}+m_2 2^{n_2}+\cdots+m_r 2^{n_r})^{k_r}} \end{align}

$$ \begin{align} &\frac{e^\pi-e^{-\pi}}{2\pi}=\exp \left( \frac{\zeta(2)}{1}-\frac{\zeta(4)}{2}+\frac{\zeta(6)}{3}-\frac{\zeta(8)}{4}+\cdots \right)\\ &\frac{e^\frac{\pi}{2}+e^{-\frac{\pi}{2}}}{\pi}=\exp \left( \frac{\eta(2)}{1}-\frac{\zeta(4)}{2}+\frac{\eta(6)}{3}-\frac{\zeta(8)}{4}+\cdots \right)\\ &\frac{e^\frac{\pi}{2}+e^{-\frac{\pi}{2}}}{2}=\exp \left( \frac{t(2)}{1}-\frac{t(4)}{2}+\frac{t(6)}{3}-\frac{t(8)}{4}+\cdots \right)\\ \end{align} $$

級数botⅡの問題
\begin{align} &\sum_{n=0}^{\infty}\frac{1}{n+\frac{1}{2} }\frac{\binom{2n}{n}}{2^{4n}}=\frac{2\pi}{3} \\ &\sum_{n=0}^{\infty}\frac{1}{(n+\frac{1}{2})^3}\frac{\binom{2n}{n}}{2^{4n}}=\frac{7\pi^3}{27} \end{align}

\begin{align} &\sin\Big(\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\zeta(4n-2)}{2n-1}z^{4n-2}\Big)=\frac{\sin(\frac{\pi}{\sqrt2}z)\cosh(\frac{\pi}{\sqrt2}z)-\cos(\frac{\pi}{\sqrt2}z)\sinh(\frac{\pi}{\sqrt2}z)}{\sqrt{\cosh(\sqrt2\pi z)-\cos(\sqrt2\pi z)}}\\ &\cos\Big(\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\zeta(4n-2)}{2n-1}z^{4n-2}\Big)=\frac{\sin(\frac{\pi}{\sqrt2}z)\cosh(\frac{\pi}{\sqrt2}z)+\cos(\frac{\pi}{\sqrt2}z)\sinh(\frac{\pi}{\sqrt2}z)}{\sqrt{\cosh(\sqrt2\pi z)-\cos(\sqrt2\pi z)}} \\ \\ \\ &\sin\Big(\sum_{n=1}^{\infty}\frac{(-1)^{n-1}t(4n-2)}{2n-1}z^{4n-2}\Big)=\frac{\sqrt2 \sin(\frac{\sqrt2}{4}\pi z)\sinh(\frac{\sqrt2}{4}\pi z)}{\sqrt{\cos(\frac{\pi}{\sqrt2}z)+\cosh(\frac{\pi}{\sqrt2}z)}}\\ &\cos\Big(\sum_{n=1}^{\infty}\frac{(-1)^{n-1}t(4n-2)}{2n-1}z^{4n-2}\Big)=\frac{\sqrt2 \cos(\frac{\sqrt2}{4}\pi z)\cosh(\frac{\sqrt2}{4}\pi z)}{\sqrt{\cos(\frac{\pi}{\sqrt2}z)+\cosh(\frac{\pi}{\sqrt2}z)}} \\ \\ \\ &\sin\Big(\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\eta(4n-2)}{2n-1}z^{4n-2}\Big)=2x \frac{\sinh(\frac{\sqrt2}{4}\pi z)\cosh(\frac{\sqrt2}{4}\pi z)-\sin(\frac{\sqrt2}{4}\pi z)\cos(\frac{\sqrt2}{4}\pi z)}{\sqrt{\cosh(\sqrt2 \pi z)-\cos(\sqrt2 \pi z)}}\\ &\cos\Big(\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\eta(4n-2)}{2n-1}z^{4n-2}\Big)=2x \frac{\sinh(\frac{\sqrt2}{4}\pi z)\cosh(\frac{\sqrt2}{4}\pi z)+\sin(\frac{\sqrt2}{4}\pi z)\cos(\frac{\sqrt2}{4}\pi z)}{\sqrt{\cosh(\sqrt2 \pi z)-\cos(\sqrt2 \pi z)}} \\ \\ \end{align}

\begin{align} &\zeta(\lbrace 2 \rbrace^r)=\sum_{0\lt n_1\lt\cdots\lt n_r}\frac{1}{n^2_1n^2_2\cdots n^2_r}=\frac{\pi^{2r}}{(2r+1)!}\\ &\zeta(\lbrace 4 \rbrace^r)=\sum_{0\lt n_1\lt\cdots\lt n_r}\frac{1}{n^4_1n^4_2\cdots n^4_r}=\frac{2^{2r+1}\pi^{4r}}{(4r+2)!}\\ &\zeta(\lbrace 6 \rbrace^r)=\sum_{0\lt n_1\lt\cdots\lt n_r}\frac{1}{n^6_1n^6_2\cdots n^6_r}=\frac{6\cdot2^{6r}\pi^{6r}}{(6r+3)!}\\ &\zeta(\lbrace 8 \rbrace^r)=\sum_{0\lt n_1\lt\cdots\lt n_r}\frac{1}{n^8_1n^8_2\cdots n^8_r}=\frac{2^{4r+1}\pi^{8r}}{(8r+4)!}\lbrace (2+\sqrt2)^{4r+2}+(2-\sqrt2)^{4r+2} \rbrace\\ &\zeta(\lbrace 10 \rbrace^r)=\sum_{0\lt n_1\lt\cdots\lt n_r}\frac{1}{n^{10}_1n^{10}_2\cdots n^{10}_r}=\frac{10\cdot2^{10r}\pi^{10r}}{(10r+5)!}(1+L_{10r+5}) \quad (L_nはn番目のリュカ数) \\ \\ \\ \end{align}

\begin{align} \zeta(\lbrace 2m \rbrace^r)&=\sum_{0\lt n_1 \lt \cdots \lt n_r}\frac{1}{{n^{2m}_1}{n^{2m}_2}\cdots{n^{2m}_r}}\\ &=\frac{(-i)^{m+1}}{2^{m-1}}\frac{(-1)^{(m+1)r}\pi^{2mr}}{(2mr+m)!}\overline{ \sum_{}^{} }(e^{\frac{i \pi }{m}}\pm e^{\frac{2i \pi }{m}}\pm \cdots \pm e^{\frac{mi \pi }{m}})^{2mr+m}\\ \end{align}

\begin{align} &t(\lbrace 2 \rbrace^r)=\sum_{0\lt n_1\lt\cdots\lt n_r}\frac{1}{(2n_1-1)^2(2n_2-1)^2\cdots (2n_r-1)^2}=\frac{\pi^{2r}}{2^{2r}(2r)!}\\ &t(\lbrace 4 \rbrace^r)=\sum_{0\lt n_1\lt\cdots\lt n_r}\frac{1}{(2n_1-1)^4(2n_2-1)^4\cdots (2n_r-1)^4}=\frac{\pi^{4r}}{2^{2r}(4r)!}\\ &t(\lbrace 6 \rbrace^r)=\sum_{0\lt n_1\lt\cdots\lt n_r}\frac{1}{(2n_1-1)^6(2n_2-1)^6\cdots (2n_r-1)^6}=\frac{3\pi^{6r}}{4(6r)!}\\ \end{align}

気に入っている積分1
\begin{align} \int_0^1\frac{{\rm{Li}}_2(\frac{1+x}{2})-\frac{\pi^2}{12}+\frac{1}{2}(\log2)^2}{x} dx=\frac{13}{8}\zeta(3)-\frac{\pi^2}{6}\log2 \end{align}

\begin{align} &\Omega(n):nの重複を含めた素因数の個数\\ &\omega(n):nの重複を含めない素因数の個数\\ &A(n):nの重複を含めた素因数の和\\ &t(s):=\sum_{n=1}^{\infty}\frac{1}{(2n-1)^s}\\ &正の整数nとその素因数分解n=p^{\alpha_1}_1p^{\alpha_2}_2 \cdots p^{\alpha_{\omega(n)}}_{\omega(n)}に対し、\\ &Y(n):=\prod_{k=1}^{\omega(n)}\alpha_k \quad (ただしY(1)=1)\\ &\sigma_a(n):nの約数のa乗和(約数関数) \\ \\ \\ &\sum_{n=1}^{\infty}\frac{(-1)^{\Omega(n)}}{n^s}=\frac{\zeta(2s)}{\zeta(s)}\\ &\sum_{n=1}^{\infty}\frac{2^{\omega(n)}}{n^s}=\frac{\zeta(s)^2}{\zeta(2s)}\\ &\sum_{n=1}^{\infty}\frac{(-1)^{A(n)}}{n^s}=\frac{2^s}{2^s-1}\frac{t(2s)}{t(s)}\\ &\sum_{n=1}^{\infty}\frac{\sigma_a(n)(-1)^{\Omega(n)}}{n^s}=\frac{\zeta(2s)\zeta(2s-2a)}{\zeta(s)\zeta(s-a)}\\ &\sum_{n=1}^{\infty}\frac{Y(n^2)}{n^s}=\frac{\zeta(s)^2\zeta(2s)}{\zeta(4s)}\\ &\sum_{n=1}^{\infty}\frac{Y(n^4)}{n^s}=\frac{\zeta(s)^4}{\zeta(2s)^2}\\ &\Big(\sum_{n=1}^{\infty}\frac{Y(n)}{n^s}\Big)\Big(\sum_{n=1}^{\infty}\frac{Y(n^3)}{n^s}\Big)=\frac{\zeta(s)^4\zeta(2s)}{\zeta(6s)}\\ \\ \\ &\sum_{n=1}^{\infty}\frac{\sigma_a(n)}{n^s}=\zeta(s)\zeta(s-a)\\ &\sum_{n=1}^{\infty}\frac{\sigma_a(n)\sigma_b(n)}{n^s}=\frac{\zeta(s)\zeta(s-a)\zeta(s-b)\zeta(s-a-b)}{\zeta(2s-a-b)}\\ \\ &\sum_{n=0}^{\infty}\frac{(-1)^n\sigma_a(2n+1)}{(2n+1)^s}=\beta(s)\beta(s-a)\\ &\sum_{n=0}^{\infty}\frac{(-1)^n\sigma_a(2n+1)\sigma_b(2n+1)}{(2n+1)^s}=\frac{\beta(s)\beta(s-a)\beta(s-b)\beta(s-a-b)}{t(2s-a-b)}\\ &\sum_{0\lt n_1 \lt n_2}\frac{\sigma_a(n_1) \sigma_a(n_2)}{n^s_1 n^s_2}=\frac{1}{2}\Big \lbrace (\zeta(s)\zeta(s-a))^2-\frac{\zeta(2s)\zeta(2s-a)^2\zeta(2s-2a)}{\zeta(4s-2a)} \Big \rbrace\\ \end{align}

\begin{align} &\sum_{0\lt n_1 \lt n_2 \lt n_3}\frac{1}{(2n_1-1)(2n_2-1)(2n_3-1)5^{n_3-n_2}}=\frac{1}{20}\zeta(3) \end{align}

\begin{align} &t(\lbrace \bar{1} \rbrace ^{r})=\sum_{0\lt n_1\lt\cdots\lt n_{r}}\frac{(-1)^{n_1+n_2+\cdots +n_{r}}}{(2n_1-1)(2n_2-1)\cdots (2n_{r}-1)}=\frac{(-1)^{ \lbrack \frac{1+r}{2} \rbrack}\pi^{r}}{2^{2r}r!}\\ &t(\lbrace \bar{3} \rbrace ^{r})=\sum_{0\lt n_1\lt\cdots\lt n_{r}}\frac{(-1)^{n_1+n_2+\cdots +n_{r}}}{(2n_1-1)^3(2n_2-1)^3\cdots (2n_{r}-1)^3}=\frac{3\pi^{3r}}{2^{3r+1}(3r)!}(-1)^{ \lbrack \frac{1+r}{2} \rbrack}\\ \end{align}

\begin{align} \\ &\zeta(\{\bar{2}\}^r)=\sum_{0\lt n_1\lt\cdots\lt n_r}\frac{(-1)^{n_1+n_2+\cdots +n_r}}{n_1^{2}n_2^{2}\cdots n_r^{2}}=\frac{(-1)^{ \lbrack \frac{1+r}{2} \rbrack}\pi^{2r}}{2^r(2r+1)!} \\ \\ &\zeta(\{\bar{4}\}^r)=\sum_{0\lt n_1\lt\cdots\lt n_r}\frac{(-1)^{n_1+n_2+\cdots +n_r}}{n_1^{4}n_2^{4}\cdots n_r^{4}}=\frac{(-1)^{ \lbrack \frac{1+r}{2} \rbrack}\pi^{4r}}{2^{r}(4r+2)!} \Big( (\sqrt2+1)^{2r+1}-(\sqrt2-1)^{2r+1}\Big)\\ \\ \end{align}

コネクターと級数
許容インデックス$\boldsymbol{ k }=(k_1,k_2,\cdots ,k_r)$に対して
$$U(\boldsymbol{k}):=\sum_{0\lt n_1 \lt n_2 \lt \cdots \lt n_r}\frac{n_r}{(n_1-\frac{1}{2})^{k_1}(n_2-\frac{1}{2})^{k_2} \cdots (n_r-\frac{1}{2})^{k_r}}\frac{\binom{2n_r}{n_r}}{2^{2n_r}}$$
とすると$U(\boldsymbol{ k })$は双対性を持つ。すなわち、
$\boldsymbol{ k }$の双対インデックス$\boldsymbol{ k^{\dagger}}$に対し
$$U(\boldsymbol{ k })=U(\boldsymbol{ k^{\dagger} })$$
が成り立つ。

多重ゼータ値の双対性2
コネクターと級数
\begin{align} &\boldsymbol{k}=(k_1,k_2,\cdots,k_r)\\ &(a)_n:ポッホハマー記号,\boldsymbol{\alpha}=(\lbrace \alpha \rbrace^{dep(\boldsymbol{k^\dagger})})\\ &\sum_{0\lt n_1\lt\cdots\lt n_r}\frac{2^{2n_r}}{n_1^{k_1}n_2^{k_2}\cdots n_r^{k_r}{ 2n_r \choose n_r }}=2^{wt(\boldsymbol{k})}t(\boldsymbol{k^\dagger})\\ &\sum_{0\lt n_1\lt\cdots\lt n_r}\frac{n_r!}{n_1^{k_1}n_2^{k_2}\cdots n_r^{k_r}(\alpha)_{n_r}}=\zeta(\boldsymbol{k^\dagger};\boldsymbol{\alpha})\\ \end{align}

\begin{align} &\boldsymbol{ k }=(k_1,k_2,\cdots ,k_r)に対して\\ & \varsigma(\boldsymbol{k}):=\sum_{0\leq n_1 \lt n_2 \lt \cdots \lt 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}}\\ &\boldsymbol{k'}:=((\boldsymbol{k}_{ \uparrow})^{\dagger})_{ \downarrow}とすると\\ &\varsigma(\boldsymbol{k})=\varsigma(\boldsymbol{k'})が成り立つ\\ \end{align}

コネクターと級数
\begin{align} \sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{(n-\frac{1}{2})(m-\frac{1}{2})}\frac{n!m!}{(n+m-1)!}\frac{\binom{2n}{n}\binom{2m}{m}}{2^{2n+2m}}=\pi\\ \sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{(n-\frac{1}{2})^2(m-\frac{1}{2})^2}\frac{n!m!}{(n+m-1)!}\frac{\binom{2n}{n}\binom{2m}{m}}{2^{2n+2m}}=\frac{\pi^3}{6} \end{align}

$$\sum_{0\lt n_1 \lt n_2 \lt \cdots \lt n_r}\frac{n_r}{(n_1-\frac{1}{2})^2(n_2-\frac{1}{2})^2\cdots (n_r-\frac{1}{2})^2}\frac{\binom{2n}{n}}{2^{2n}}=\frac{\pi^{2r-1}}{(2r-1)!}$$

\begin{align} &M:=M(n_1,n_2,\cdots,n_{2r})=\sum_{k=1}^{2r}n_k(-1)^k=n_{2r}-n_{2r-1}+\cdots+n_2-n_1\\ \\ &\sum_{0\lt n_1\lt\cdots\lt n_{2r}}\frac{(-1)^M}{(n_1-\frac{1}{2})(n_2-\frac{1}{2})\cdots (n_{2r}-\frac{1}{2})\cdot3^M}=\frac{(-1)^r\pi^{2r}}{3^{2r}(2r)!}\\ \\ \\ &\sum_{0\lt n_1 \lt \cdots \lt n_{2r}}\frac{1}{(n_1-\frac{1}{2})(n_2-\frac{1}{2})\cdots(n_{2r}-\frac{1}{2})2^{M}}=\frac{2^{2r}}{(2r)!}\Big(\log(1+\sqrt2)\Big)^{2r}\\ &\sum_{0\lt n_1 \lt \cdots \lt n_{2r}}\frac{1}{(n_1-\frac{1}{2})(n_2-\frac{1}{2})\cdots(n_{2r}-\frac{1}{2})3^{M}}=\frac{2^{2r}}{(2r)!}\Big(\log(\frac{\sqrt2+\sqrt6}{2})\Big)^{2r}\\ &\sum_{0\lt n_1 \lt \cdots \lt n_{2r}}\frac{1}{(n_1-\frac{1}{2})(n_2-\frac{1}{2})\cdots(n_{2r}-\frac{1}{2})5^{M}}=\frac{2^{2r}}{(2r)!}\Big(\log(\phi)\Big)^{2r}\\ \end{align}

\begin{align} &M':=M'(n_1,n_2,\cdots,n_{2r-1})=\sum_{k=1}^{2r-1}n_k(-1)^{k+1}=n_{2r-1}-n_{2r-2}+\cdots +n_1\\ \\ &\sum_{0\lt n_1 \lt \cdots \lt n_{2r-1}}\frac{(-1)^{M'}}{(n_1-\frac{1}{2})(n_2-\frac{1}{2})\cdots(n_{2r-1}-\frac{1}{2})3^{M'}}=\frac{1}{\sqrt3}\frac{(-1)^{r} \pi^{2r-1}}{3^{2r-1}(2r-1)!} \\ \\ &\sum_{0\lt n_1 \lt \cdots \lt n_{2r-1}}\frac{1}{(n_1-\frac{1}{2})(n_2-\frac{1}{2})\cdots(n_{2r-1}-\frac{1}{2})2^{M'}}=\frac{2^{2r-1}}{(2r-1)!}\frac{\Big(\log(1+\sqrt2)\Big)^{2r-1}}{\sqrt2}\\ &\sum_{0\lt n_1 \lt \cdots \lt n_{2r-1}}\frac{1}{(n_1-\frac{1}{2})(n_2-\frac{1}{2})\cdots(n_{2r-1}-\frac{1}{2})3^{M'}}=\frac{2^{2r-1}}{(2r-1)!}\frac{\Big(\log(\frac{\sqrt2+\sqrt6}{2})\Big)^{2r-1}}{\sqrt3}\\ &\sum_{0\lt n_1 \lt \cdots \lt n_{2r-1}}\frac{1}{(n_1-\frac{1}{2})(n_2-\frac{1}{2})\cdots(n_{2r-1}-\frac{1}{2})5^{M'}}=\frac{2^{2r-1}}{(2r-1)!}\frac{\Big(\log(\phi)\Big)^{2r-1}}{\sqrt5}\\ \end{align}

\begin{align} &\sum_{0\leq n\lt m}\frac{1}{m(n+m)}\frac{\binom{2m+4n}{m+2n}\binom{2m}{m}}{2^{6n}\binom{2m}{m}}=\frac{\pi^3}{6\Gamma(\frac{3}{4})^4}\\ &\sum_{0\leq n\lt m \lt l}\frac{2^{2l}}{m(n+m)(l+2n)(l+3n)}\frac{\binom{2m+4n}{m+2n}}{\binom{2l+4n}{l+2n}\binom{2m}{m}}=\frac{\pi^5}{24\Gamma(\frac{3}{4})^4} \end{align}

\begin{align} \quad&\sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty}\sum_{n_3=1}^{\infty}\frac{(n_1-1)!(n_2-1)!(n_3-1)!}{(n_1+n_2+n_3)!}=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{n+m}\frac{(n-1)!(m-1)!}{(n+m)!} \\ \\ &\sum_{0\lt n_1,n_2,\cdots ,n_r}\frac{(n_1-1)!(n_2-1)!\cdots(n_r-1)!}{(n_1+n_2+\cdots +n_r)!}=\sum_{0\lt n_1,n_2,\cdots ,n_{r-1}}\frac{1}{n_1+n_2+\cdots +n_{r-1}}\frac{(n_1-1)!(n_2-1)!\cdots(n_{r-1}-1)!}{(n_1+n_2+\cdots +n_{r-1})!} \\ \\ &\sum_{0\lt n_1.n_2.n_3}\frac{n_1!n_2!n_3!}{(n_1+n_2+n_3)!}\frac{\binom{2n_1}{n_1}\binom{2n_2}{n_2}\binom{2n_3}{n_3}}{\binom{2n_1+2n_2+2n_3}{n_1+n_2+n_3}}\frac{n_1+n_2+n_3-\frac{1}{2}}{(n_1-\frac{1}{2})(n_2-\frac{1}{2})(n_3-\frac{1}{2})}=\sum_{0\lt n,m}\frac{n!m!}{(n+m)!}\frac{\binom{2n}{n}\binom{2m}{m}}{\binom{2n+2m}{n+m}}\frac{n+m-\frac{1}{2}}{(n-\frac{1}{2})(m-\frac{1}{2})(n+m-1)} \end{align}

AMZVメモ
\begin{align} \quad\zeta(\bar{1})&=-\ln2\\ \zeta(\bar{1},\bar{1})&=\frac{1}{2}\ln^{2}2-\frac{1}{2}\zeta(2)\\ \zeta(\bar{1},\bar{1},\bar{1})&=-\frac{1}{6}\ln^ {3}2+\frac{\pi^2}{12}\ln2-\frac{1}{4}\zeta(3)\\ \zeta(\bar{1},\bar{1},\bar{1},\bar{1})&=\frac{1}{24}\ln^{4}2-\frac{\pi^2}{24}\ln^{2}2+\frac{5}{16}\zeta(3)\ln2+\frac{\pi^4}{1440}\\ \\ \zeta(1,\bar{1},1,\bar{1})&=\frac{1}{24}\ln^{4}2+\frac{\pi^4}{720}-\frac{5}{32}\zeta(3)\ln2 \end{align}

\begin{align} &\sum_{\substack{n \lt m \\(n,m)\in( \mathbb{Z}\backslash \lbrace 0 \rbrace )^2}}\frac{1}{n^3 m^3}=-\zeta(6) \\ &\sum_{\substack{n \lt m \\(n,m)\in( \mathbb{Z}\backslash \lbrace 0 \rbrace )^2}}\frac{1}{n^2 m^4}=\frac{\pi^6}{378} \end{align}

\begin{align} &\sum_{\substack{n_1 \lt n_2 \lt \cdots \lt n_r \\(n_1,n_2,\cdots,n_r)\in( \mathbb{Z}\backslash \lbrace 0 \rbrace )^r}}\frac{1}{n^2_1 n^2_2\cdots n^2_r}=\frac{2^{2r+1}}{(2r+2)!}\pi^{2r}\\ \\ &\sum_{\substack{n_1 \lt n_2 \lt \cdots \lt n_r \\(n_1,n_2,\cdots,n_r)\in( \mathbb{Z}\backslash \lbrace 0 \rbrace )^r}}\frac{1}{n^4_1 n^4_2\cdots n^4_r}=\frac{2^{6r+4}+2^{4r+3}(-1)^r}{(4r+4)!}\pi^{4r}\\ \end{align}

\begin{align} &\sum_{\substack{n_1 \lt n_2 \lt \cdots \lt n_{2r-1} \\(n_1,n_2,\cdots,n_{2r-1})\in( \mathbb{Z}\backslash \lbrace 0 \rbrace )^{2r-1}}}\frac{(-1)^{n_1+n_2+\cdots+n_{2r-1}}}{n^2_1 n^2_2\cdots n^2_{2r-1}}=\frac{(-1)^r2^{2r}\pi^{4r-2}}{(4r)!}\\ \\ &\sum_{\substack{n_1 \lt n_2 \lt \cdots \lt n_{2r} \\(n_1,n_2,\cdots,n_{2r})\in( \mathbb{Z}\backslash \lbrace 0 \rbrace )^{2r}}}\frac{(-1)^{n_1+n_2+\cdots+n_{2r}}}{n^2_1 n^2_2\cdots n^2_{2r}}=\frac{2\pi^{4r}}{(4r+2)!}\\ \end{align}

\begin{align} &\sum_{\substack{n_1 \lt n_2 \lt \cdots \lt n_r \\(n_1,n_2,\cdots,n_r)\in \mathbb{Z}^r}}\frac{1}{(n_1+\frac{1}{2})^2 (n_2+\frac{1}{2})^2\cdots (n_r+\frac{1}{2})^2}=\frac{2^{2r-1}}{(2r)!}\pi^{2r}\\ \\ &\sum_{\substack{n_1 \lt n_2 \lt \cdots \lt n_r \\(n_1,n_2,\cdots,n_r)\in \mathbb{Z}^r}}\frac{1}{(n_1+\frac{1}{2})^4 (n_2+\frac{1}{2})^4\cdots (n_r+\frac{1}{2})^4}=\frac{{2^{6r-2}+2^{4r-1}(-1)^r}}{(4r)!}\pi^{4r}\\ \end{align}

\begin{align} &\frac{1}{\sqrt{2}\cos\frac{\pi}{4}(1-x)}=\frac{1}{\sin \frac{\pi x}{4}+\cos \frac{\pi x}{4}}=1+a_1x+a_2x^2+a_3x^3+\cdots\\ \\ &a_n=\sum_{ \substack{wt(\boldsymbol{k})=n}} t(\boldsymbol{k})\\ &\boldsymbol{k}\text{は}\bar{1},2,\bar{3},4,\cdots\text{で構成されるインデックス}\\ \end{align}

\begin{align} \quad \sum_{0\leq n_1}\frac{1}{(2n_1+1)}\frac{\binom{2n_1}{n_1}^2}{2^{4n_1}}&=\frac{4}{\pi}\beta(2)\\ \sum_{0\leq n_1 \lt n_2}\frac{1}{(2n_1+1)^2(2n_2+1)}\frac{\binom{2n_2}{n_2}^2}{2^{4n_2}}&=\frac{\pi}{2}\beta(2)-\frac{4}{\pi}\beta(4)\\ \sum_{0\leq n_1 \lt n_2 \lt n_3}\frac{1}{(2n_1+1)^2(2n_2+1)^2(2n_3+1)}\frac{\binom{2n_3}{n_3}^2}{2^{4n_3}}&=\frac{\pi^3}{96}\beta(2)-\frac{\pi}{2}\beta(4)+\frac{4}{\pi}\beta(6)\\ \\ \end{align}
\begin{align} &\sum_{0\leq n_1 \lt \cdots \lt n_{r+1}}\frac{1}{(2n_1+1)^2\cdots(2n_r+1)^2(2n_{r+1}+1) }\frac{{\binom{2n_{r+1}}{n_{r+1}}}^2}{2^{4n_{r+1}}}=\frac{4}{\pi}\sum_{k=0}^{r}\beta(2k+2)\frac{(-1)^k\pi^{2r-2k}}{2^{2r-2k}(2r-2k)!}\\ \end{align}

$$\quad\sum_{n=0}^{\infty}(-1)^n\frac{{\binom{2n}{n}}^5(4n+1)}{2^{10n}}=\frac{2}{\Gamma(\frac{3}{4})^4} $$

$$ \quad\frac{1}{4}\sum_{0\lt n}\frac{2^{2n}}{n^4\binom{2n}{n}}+2\sum_{0\leq n\lt m}\frac{1}{(2n+1)^3(2m)(2m+1)}=\frac{\pi^2}{4}\ln^22$$

$$\quad \sum_{0\lt n}\frac{2^{2n}}{n^3\binom{2n}{n}}=3\sum_{0\lt n}\frac{2^{2n}}{n^3}\frac{(n!)^3}{(3n)!}+2\sum_{0\lt n\lt m}\frac{2^{2n}}{nm^2\binom{2n}{n}}\frac{(2m)!n!}{(2m+n)!}=\pi^2\ln2-\frac{7}{2}\zeta(3)$$

(n!m!/(n+m)!)²が含まれる級数
$$\quad \zeta(2)=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{n^2}\Bigg(\frac{n!m!}{(n+m)!}\Bigg)^2+2\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{nm}\Bigg(\frac{n!m!}{(n+m)!}\Bigg)^2$$

\begin{align} \\ \quad \zeta(3)&=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{nm^2}\frac{n!m!}{(n+m)!}\\ &=3\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{nm^2}\Bigg(\frac{n!m!}{(n+m)!}\Bigg)^2 \\ \\ \zeta(3)&=6\sum_{\substack{0\lt n_1 \lt n_2\\0\lt m}}\frac{1}{n_1n_2m}\Bigg(\frac{n_2!m!}{(n_2+m)!}\Bigg)^2+3\sum_{\substack{0\lt n_1 \lt n_2\\0\lt m}}\frac{1}{n_1n_2^2}\Bigg(\frac{n_2!m!}{(n_2+m)!}\Bigg)^2 \\ \\ \zeta(3)&=2\sum_{\substack{0\lt n_1 \lt n_2\\0\lt m}}\frac{1}{n_1n_2m}\Bigg(\frac{n_2!m!}{(n_2+m)!}\Bigg)^2+\sum_{\substack{0\lt n_1 \lt n_2\\0\lt m}}\frac{1}{n_1m^2}\Bigg(\frac{n_2!m!}{(n_2+m)!}\Bigg)^2 \\ \\ \end{align}

$$\quad \zeta(3)+\sum_{\substack{0\lt n\\0\lt m}}\frac{1}{n^3}\Bigg(\frac{n!m!}{(n+m)!}\Bigg)^2=2\sum_{\substack{0\lt n\\0\lt m}}\frac{1}{n^2}\Bigg(\frac{n!m!}{(n+m)!}\Bigg)^2+\sum_{\substack{0\lt n\\0\lt m}}\frac{1}{nm}\Bigg(\frac{n!m!}{(n+m)!}\Bigg)^2$$

\begin{align} &\frac{1}{1^{2k}}-\frac{1}{3^{2k}}-\frac{1}{5^{2k}}+\frac{1}{7^{2k}}+\frac{1}{9^{2k}}-\frac{1}{11^{2k}}-\frac{1}{13^{2k}}+\cdots=\frac{\pi}{2\sqrt2}t^{\star}(\lbrace \bar{1} \rbrace^{2k-1}) \\ \\ &\frac{1}{1^{2k+1}}+\frac{1}{3^{2k+1}}-\frac{1}{5^{2k+1}}-\frac{1}{7^{2k+1}}+\cdots=\frac{\pi}{2\sqrt2}t^{\star}(\lbrace \bar{1} \rbrace^{2k}) \end{align}

$$\zeta^{\star}(\lbrace 3 \rbrace^{k})=\frac{3\pi}{4}\sum_{n=0}^{\infty}\frac{1}{(2n+1)^{3k-1}}\frac{\binom{2n}{n}}{2^{4n}}\prod_{m=n+1}^{\infty}\frac{1}{\Big(1+\frac{3(2n+1)^2}{(2m+1)^2}\Big)}-\frac{3}{8}\sum_{n=1}^{\infty}\frac{1}{n^{3k}\binom{2n}{n}}\prod_{m=n+1}^{\infty}\frac{1}{\Big(1+\frac{3n^2}{m^2}\Big)}$$

$$\frac{\pi}{2}=\sum_{n=0}^{\infty}\frac{1}{2n+1}\frac{\binom{2n}{n}\binom{4n+4}{2n+2}}{\binom{3n+2}{n}2^{6n+4}}+\sum_{n=0}^{\infty}\frac{1}{4n+1}\frac{\binom{2n}{n}\binom{4n}{2n}}{\binom{3n}{n}2^{6n}}+\sum_{n=0}^{\infty}\frac{1}{4n+3}\frac{\binom{2n}{n}\binom{4n+2}{2n+1}}{\binom{3n+1}{n}2^{6n+2}}$$

$$\sum_{0\leq n}\Bigg(\frac{1}{(2n+1)^3\binom{3n+1}{n}}-\frac{H_{2n}-H_n}{(2n+1)^2\binom{3n+1}{n}}\Bigg)=\sum_{0\lt n}\Bigg(\frac{2}{n^3\binom{3n}{n}}+\frac{7}{4}\frac{H_{2n}-H_n}{n^2\binom{3n}{n}}\Bigg)$$

$$\sum_{n=0}^{\infty}\frac{\tanh((n+\frac{1}{2})\pi)}{(n+\frac{1}{2})^3}=\frac{\pi^3}{4}$$
$$\sum_{n=0}^{\infty}\frac{\tanh((n+\frac{1}{2})\pi)}{(n+\frac{1}{2})^7}=\frac{7\pi^7}{180}$$

$$2\pi^2\log^22=\sum_{0< n_1< n_2< n_3}\frac{\binom{2n_1}{n_1}}{n_12^{2n_1}}\frac{1}{n_2}\frac{2^{2n_3}}{n_3^2\binom{2n_3}{n_3}}+2\sum_{0< n_1< n_2< n_3}\frac{1}{n_1}\frac{\binom{2n_2}{n_2}}{n_22^{2n_2}}\frac{2^{2n_3}}{n_3^2\binom{2n_3}{n_3}}$$
$$14\zeta(1,3)-16t(1,3)=\sum_{0< n_1< n_2< n_3}\frac{1}{n_1}\frac{2^{2n_2}}{n_2^2\binom{2n_2}{n_2}}\frac{\binom{2n_3}{n_3}}{n_32^{2n_3}}+2\sum_{0< n_1< n_2\leq n_3}\frac{1}{n_1}\frac{2^{2n_2}}{n_2^2\binom{2n_2}{n_2}}\frac{(-1)^{n_2+n_3}}{n_2+n_3}$$
$$14\zeta(3)=\sum_{0< n\leq m}\frac{1}{(n-\frac{1}{2})}\frac{2^{2m}}{m^2\binom{2m}{m}}$$
$$16t(1,3)=\sum_{0< n_1< n_2\leq n_3}\frac{\binom{2n_1}{n_1}}{n_1 2^{2n_1}}\frac{1}{n_2-\frac{1}{2}}\frac{2^{2n_3}}{n^2_3\binom{2n_3}{n_3}}-\sum_{0< n_1< n_2< n_3}\frac{\binom{2n_1}{n_1}}{n_1 2^{2n_1}}\frac{1}{n_2}\frac{2^{2n_3}}{n^2_3\binom{2n_3}{n_3}}$$

$$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{2^{2n}}{n^2\binom{2n}{n}}\frac{2^{2m}}{m^2\binom{2m}{m}}\frac{n!m!}{(n+m)!}=\frac{\pi^4}{24}$$
$$\sum_{0< n< m}\frac{\binom{2n}{n}}{n^22^{2n}}\frac{2^{2m}}{m^2\binom{2m}{m}}=\sum_{0< n< m}\frac{2^{2n}}{n^2\binom{2n}{n}}\frac{1}{m^2}$$

$$\Xi(k_1,k_2,\ldots,k_r):=\sum_{0< n_1<\cdots< n_r}\frac{1}{n^{k_1}_1\cdots n^{k_{r-1}}_{r-1}{n_{r}}^{k_{r}}}\frac{(a)_{n_{r-1}}}{n_{r-1}!}\frac{n_{r}!}{(a)_{n_{r}}}$$
$$\text{特に}r=1\text{のとき},\Xi(k):=\sum_{n=1}^{\infty}\frac{n!}{n^k(a)_n}$$
重さ$k$深さ$r$の許容インデックス全体がなす集合を$I_{r}(k,r)$とする。このとき次が成立する。(多重ゼータ値の和公式の一般化)
$$\Xi(k)=\sum_{\boldsymbol{k}\in I_{r}(k,r)}\Xi(\boldsymbol{k})$$

log2に収束する級数
$$\log2=\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}\frac{\binom{2n}{n}}{\binom{4n}{2n}}+\sum_{n=1}^{\infty}\frac{1}{n(2n-1)}\frac{\binom{2n}{n}}{\binom{4n}{2n}}$$

$$\log2=\sum_{n=1}^{\infty}\frac{1}{(2n)^2}\frac{2^{4n}}{\binom{4n}{2n}\binom{2n}{n}}+\sum_{n=1}^{\infty}\frac{1}{n(2n-1)}\frac{2^{4n}}{\binom{4n}{2n}\binom{2n}{n}}$$
$\downarrow\text{一般化}\\$

$$\sum_{n=1}^{\infty}\frac{1}{2n(2n-2x-1)} =\sum_{n=1}^{\infty}\frac{1}{(2n-2x-1)^2}\frac{(1-2x)^2_{2n}(1-x)_{2n}}{(1-2x)_{4n}(1-x)^2_{n}}+\sum_{n=1}^{\infty}\frac{1}{n(2n-2x-1)}\frac{(1-2x)^2_{2n}(1-x)_{2n}}{(1-2x)_{4n}(1-x)^2_{n}}$$

$$\sum_{n=1}^{\infty}\frac{1}{2n(2n-2x-1)} =\sum_{n=1}^{\infty}\frac{1}{(2n)^2}\frac{(1-2x)^2_{2n}(1-x)_{2n}n!^2}{(1-2x)_{4n}(1-x)^2_{n}(\frac{1}{2}-x)^2_{n}}+\sum_{n=1}^{\infty}\frac{1}{n(2n-2x-1)}\frac{(1-2x)^2_{2n}(1-x)_{2n}n!^2}{(1-2x)_{4n}(1-x)^2_{n}(\frac{1}{2}-x)^2_{n}}$$

任意に正整数の組$(n,m)$について次が成立する。
$$\sum_{n< k}\frac{1}{k(k-\frac{1}{2})}\frac{\binom{2k}{k}\binom{2m}{m}}{\binom{2k+2m}{k+m}}-\sum_{n< k}\frac{1}{k(k-\frac{1}{2})}\frac{\binom{2k}{k}}{2^{2k}} =\sum_{m< k}\frac{1}{k(k-\frac{1}{2})}\frac{\binom{2k}{k}\binom{2n}{n}}{\binom{2k+2n}{k+n}}-\sum_{m< k}\frac{1}{k(k-\frac{1}{2})}\frac{\binom{2k}{k}}{2^{2k}}$$

\begin{align*}   &\frac{2^{2a}}{\binom{2a}{a}}\sum_{a\leq n_1\leq\cdots\leq n_r}\frac{1}{n_1(n_1+\frac{1}{2})\cdots n_r(n_r+\frac{1}{2})}\frac{\binom{2n_r}{n_r}}{2^{2n_r}}\\ &=\sum_{0< n_1\leq n_2<\cdots< n_{2r-1}\leq n_{2r}}\frac{1}{n_1}\frac{2^{2n_1}}{\binom{2n_r}{n_r}}\frac{1}{n_2+\frac{1}{2}}\frac{\binom{2n_2}{n_2}}{2^{2n_2}}\cdots\frac{1}{n_{2r-1}}\frac{2^{2n_{2r-1}}}{\binom{2n_{2r-1}}{n_{2r-1}}}\frac{1}{n_{2r}+\frac{1}{2}}\frac{\binom{2n_{2r}}{n_{2r}}}{2^{2n_{2r}}}\frac{n_{2r}!a!}{(n_{2r}+a)!} \end{align*}

$\beta_n:=\frac{\binom{2n}{n}}{2^{2n}}$
\begin{align*}   &\sum_{0\leq n< n_1<\cdots< n_r}\beta^3_n\frac{1}{n^3_1\cdots n^3_r\beta^3_{n_r}}=\sum_{0\leq n< n_1<\cdots< n_{2r-1}}\beta^3_n\frac{1}{n^2_1\beta_{n_1}}\frac{\beta_{n_2}}{n_2}\cdots\frac{1}{n^2_{2r-3}\beta_{n_{2r-3}}}\frac{\beta_{n_{2r-2}}}{n_{2r-2}}\frac{1}{n^3_{2r-1}\beta^3_{n_{2r-1}}} \end{align*}

$\boldsymbol{k}^{\lor}$:Hoffman双対インデックス
\begin{align*}   &H^{\star}_N(\boldsymbol{k};x):=\sum_{1\leq n_1\leq \cdots\leq n_a\leq N}\frac{1}{(n_1-x)^{k_1}\cdots(n_a-x)^{k_a}}\frac{(-1)^{n_a-1}(1-x)_N}{(1-x)_{n_a}(N-n_a)!}\\   &G^{\star}_N(\boldsymbol{k};x):=\sum_{1\leq n_1\leq \cdots\leq n_a\leq N}\frac{1}{(n_1-x)^{k_1}\cdots(n_a-x)^{k_a}}\frac{(1-x)_{n_1-1}}{(n_1-1)!} \end{align*}
このとき,次が成立する.
$$H^{\star}_N(\boldsymbol{k};x)=G^{\star}_N(\boldsymbol{k}^{\lor};x)$$

$\boldsymbol{k}=(k_1,\ldots,k_r)$:許容インデックス
$\boldsymbol{\epsilon}=(\epsilon_1,\ldots,\epsilon_r)\in\lbrace 0,1\rbrace^r$
$$\nu_{a,b}\begin{pmatrix}\boldsymbol{\epsilon}\\\boldsymbol{k}\end{pmatrix} :=\sum_{0=n_0< n_1<\cdots< n_r}\prod_{i=1}^{r}\frac{1}{n^{k_i}_i}\Bigg(\frac{n^2_i}{n^2_i-a^2}\frac{(1-b)_{n_{i-1}}}{n_{i-1}!}\frac{n_{i-1}!^2}{(1-a)_{n_{i-1}}(1+a)_{n_{i-1}}}\frac{n_i!}{(1-b)_{n_i}}\frac{(1-a)_{n_{i}}(1+a)_{n_{i}}}{n_{i}!^2}\Bigg)^{\epsilon_i}$$
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。
$$\nu_{a,b}\begin{pmatrix}   \lbrace0\rbrace^{a_1-1},&\epsilon_1,&\ldots,&\lbrace0\rbrace^{a_s-1},&\epsilon_s\\    \lbrace1\rbrace^{a_1-1},&b_1+1,&\ldots,&\lbrace1\rbrace^{a_s-1},&b_s+1\end{pmatrix}    =    \nu_{a,b}\begin{pmatrix}   \lbrace0\rbrace^{b_s-1},&\epsilon_s,&\ldots,&\lbrace0\rbrace^{b_1-1},&\epsilon_1\\   \lbrace1\rbrace^{b_s-1},&a_s+1,&\ldots,&\lbrace1\rbrace^{b_1-1},&a_1+1\end{pmatrix}$$

$\boldsymbol{k}':=((\boldsymbol{k}_{\uparrow})^{\dagger})_{\downarrow}$
$$\zeta_{\alpha,\beta,\gamma,\delta}(\boldsymbol{k}) :=\frac{\Gamma(\beta)\Gamma(\alpha+\gamma)}{\Gamma(\alpha+\beta+\gamma)} \sum_{0\leq n_1<\cdots< n_r}\frac{(1-\delta)_{n_1}}{n_1!}\frac{1}{(n_1+\gamma)^{k_1}\cdots(n_r+\gamma)^{k_r}}\frac{(\alpha+\gamma)_{n_r}}{(\alpha+\beta+\gamma)_{n_r}}$$
このとき
$$\zeta_{\alpha,\beta,\gamma,\delta}(\boldsymbol{k}) =\zeta_{\delta,\gamma,\beta,\alpha}(\boldsymbol{k}')$$

$$\pi=\sum_{0< k}\frac{3k-1}{k(k-\frac{1}{2})}\frac{2^{2k}}{\binom{4k}{2k}}$$

$$\sum_{0\leq m\leq n}\frac{\binom{2m}{m}^2}{2^{4m}}\frac{1}{(2n+1)^2}=\frac{7}{2\pi}\zeta(3)$$
$$\sum_{0\leq m\leq n}\frac{\binom{2m}{m}^2}{2^{4m}}\frac{1}{(2n+1)^3}=\frac{4}{\pi}\beta(2)^2$$

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

$$\frac{\pi^3}{16}\sum_{n=0}^{\infty}(4n+1)\frac{\binom{2n}{n}^6}{2^{12n}} =\sum_{n=0}^{\infty}\frac{\binom{2n}{n}^4}{2^{8n}}\Bigg(2\beta(2)+\sum_{m=1}^{n}\frac{2^{4m}}{(2m)^2\binom{2m}{m}^2}\Bigg)$$

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

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

$$\sum_{n=1}^{\infty}\frac{2^{6n}}{n^3\binom{2n}{n}^3}\Bigg(\sum_{m=0}^{n-1}\frac{\binom{2m}{m}^2}{2^{4m}}\Bigg)^2-\pi\sum_{n=1}^{\infty}\frac{\binom{2n}{n}^3}{2^{6n}}\sum_{m=1}^{n}\frac{2^{2m}}{m^2\binom{2m}{m}} =\frac{8\pi^2}{\Gamma(\frac{3}{4})^4}\beta(2)$$

$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}2^{6n}}{(2n)^3\binom{2n}{n}^3}\sum_{m=0}^{n-1}\frac{\binom{2m}{m}^2}{2^{4m}} =\sum_{n=1}^{\infty}\frac{(-1)^{n}2^{2n}}{(2n)^2\binom{2n}{n}}\sum_{m=0}^{n-1}(-1)^{m}\frac{\binom{2n}{n}^3}{2^{6n}} -\sqrt{2}\log\Big(1+\sqrt{2}\Big)\sum_{n=1}^{\infty}\frac{(-1)^{n}2^{2n}}{(2n)\binom{2n}{n}}\sum_{m=0}^{n-1}(-1)^{m}\frac{\binom{2n}{n}^3}{2^{6n}}$$

$$7\zeta(3)=\sum_{0\leq n_1< n_2}\frac{1}{n_1+\frac{1}{2}}\frac{3n_2-1}{n^3_2}\frac{2^{4n_2}}{\binom{2n_2}{n_2}^3}-\sum_{0< n}\frac{1}{n^3}\frac{2^{4n}}{\binom{2n}{n}^3}$$

$$\frac{\pi^4}{6}=\sum_{0\leq n_1< n_2}\frac{1}{(n_1+\frac{1}{2})^2}\frac{3n_2-1}{n^3_2}\frac{2^{4n_2}}{\binom{2n_2}{n_2}^3}-4\sum_{0< n_1< n_2}\frac{1}{n^2_1}\frac{3n_2-1}{n^3_2}\frac{2^{4n_2}}{\binom{2n_2}{n_2}^3}$$

$A(n):=\sin\Big(\frac{\pi}{4}+\frac{n\pi}{2}\Big)$
$B(n):=\cos\Big(\frac{\pi}{4}+\frac{n\pi}{2}\Big)$
$$t^{\star}(k_1,\ldots,\bar{k_i},\ldots,k_r):=\sum_{0\leq n_1\leq \dots\leq n_r}\frac{(-1)^{n_i}}{(2n_1+1)^{k_1}\dots(2n_r+1)^{k_r}}$$
$$\zeta^{\star}(k_1,\ldots,\bar{k_i},\ldots,k_r):=\sum_{0< n_1\leq \dots\leq n_r}\frac{(-1)^{n_i-1}}{n^{k_1}_1\dots n^{k_r}_r}$$

\begin{align} \\ \end{align}

$$\frac{\pi}{4}t^{\star}(\lbrace\bar{1}\rbrace^{2k-1})=\sum_{n=0}^{\infty}\frac{B(n)}{(2n+1)^{2k}}$$
$$\frac{\pi}{4}t^{\star}(\lbrace\bar{1}\rbrace^{2k})=\sum_{n=0}^{\infty}\frac{A(n)}{(2n+1)^{2k+1}}$$
\begin{align} \\ \end{align}

$$\frac{\pi}{12}t^{\star}(\lbrace\bar{3}\rbrace^{2k-1})=\sum_{n=0}^{\infty}\frac{B(n)}{(2n+1)^{6k-2}}\frac{1}{\cosh\Big(\frac{\sqrt{3}\pi}{4}(2n+1)\Big)+B(n)}$$
$$\frac{\pi}{12}t^{\star}(\lbrace\bar{3}\rbrace^{2k})=\sum_{n=0}^{\infty}\frac{A(n)}{(2n+1)^{6k+1}}\frac{1}{\cosh\Big(\frac{\sqrt{3}\pi}{4}(2n+1)\Big)+B(n)}$$
\begin{align} \end{align}
$$\beta_{n}:=\frac{\binom{2n}{n}}{2^{2n}}$$
$$\zeta^{\star}(\lbrace\bar{1}\rbrace^{2k-1})=\sum_{n=1}^{\infty}\frac{\sqrt{2}B(n-1)}{n^{2k-1}}\beta_{\lbrack\frac{n}{2}\rbrack}$$
$$\zeta^{\star}(\lbrace\bar{1}\rbrace^{2k})=\sum_{n=1}^{\infty}\frac{\sqrt{2}A(n-1)}{n^{2k}}\beta_{\lbrack\frac{n}{2}\rbrack}$$

$$\sum_{n=0}^{\infty}\frac{48n^2+32n+3}{n+\frac{1}{2}}\frac{\binom{2n}{n}\binom{4n}{2n}^2}{2^{12n}}=\frac{16\sqrt{2}}{\pi}$$
$$\sum_{0\leq n}\frac{1701n^4+2754n^3+1566n^2+351n+22}{(n+\frac{1}{2})^3}\frac{{\binom{3n}{n}}^3}{3^{9n}}=\frac{324\sqrt{3}}{\pi}$$
$$\sum_{0\leq k}\frac{5376k^4+8704k^3+4896k^2+1056k+57}{(k+\frac{1}{2})^3}\frac{\binom{4k}{2k}^3}{2^{18k}}=\frac{1024\sqrt{2}}{\pi}$$

$$\zeta(3)=\frac{9}{2}\sum_{0< n}\frac{1}{n^3\binom{2n}{n}^2}+\frac{3}{2}\sum_{0< m< n}\frac{\binom{2m}{m}}{m}\frac{21n-8}{n^3\binom{2n}{n}^3}$$

\begin{align*} \sum_{0< n}\Bigg(\frac{1}{n}-3n\binom{2n}{n}\sum_{n< m}\frac{1}{m^2\binom{2m}{m}}\Bigg)&=\frac{8}{3\sqrt{3}}\sum_{0< n}\frac{\sin(\frac{\pi n}{3})}{n^2}\\ \sum_{0< n}\Bigg(\frac{1}{n^2}-3\binom{2n}{n}\sum_{n< m}\frac{1}{m^2\binom{2m}{m}}\Bigg)&=\frac{\pi^2}{3}-\frac{12}{3\sqrt{3}}\sum_{0< n}\frac{\sin(\frac{\pi n}{3})}{n^2} \end{align*}

$$\sum_{0\leq m< n}\frac{2^{4m}}{(m+\frac{1}{2})^2\binom{2m}{m}}\frac{\binom{2n}{n}}{(n+\frac{1}{2})2^{4n}}=\frac{16}{9}\pi\beta(2)-\frac{35}{9}\zeta(3)$$

$$\frac{64}{3\pi}\beta(2)=\sum_{0< m\leq n}\frac{1}{m(m-\frac{1}{2})}\frac{2^{4m}}{\binom{2m}{m}}\frac{(6n+1)\binom{2n}{n}^3}{2^{8n}}$$
$$\frac{4096}{3\pi}\beta(2)=24\sum_{0\leq n}\frac{1}{n+\frac{1}{2}}\frac{\binom{2n}{n}^2}{2^{8n}}+\sum_{0< m\leq n}\frac{1}{m(m-\frac{1}{2})}\frac{2^{4m}}{\binom{2m}{m}}\frac{(42n+5)\binom{2n}{n}^3}{2^{12n}}$$

\begin{align*} \pi&=\sum_{0\leq n}\frac{1}{n+\frac{1}{2}}\frac{\binom{2n}{n}^3}{2^{8n}}+\sum_{0\leq m< n}\frac{1}{(m+\frac{1}{2})^2}\frac{(6n+1)\binom{2n}{n}^3}{2^{8n}}\\ \frac{8}{3}\pi&=4\sum_{0\leq n}\frac{1}{n+\frac{1}{2}}\frac{\binom{2n}{n}^3}{2^{12n}}+\sum_{0\leq m< n}\frac{1}{(m+\frac{1}{2})^2}\frac{(42n+5)\binom{2n}{n}^3}{2^{12n}} \end{align*}

$$\frac{\pi}{\Gamma(\frac{3}{4})^4}=\frac{1}{2^{10}\cdot3^3}\sum_{0\leq n}(-1)^{n}\frac{28672n^5+56832n^4+43584n^3+16000n^2+2760n+171}{(n+\frac{1}{2})^3(n+\frac{1}{12})(n+\frac{5}{12})}\frac{1}{3^{3n}}\frac{(\frac{3}{4})^2_{n}}{(\frac{1}{12})_{n}(\frac{5}{12})_{n}}\frac{\binom{4n}{2n}^3}{2^{12n}}$$