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\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}

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

\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} &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}

\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)\\ \\ \\ \\ &\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)}\\ \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 ^{2r})=\sum_{0\lt n_1\lt\cdots\lt n_{2r}}\frac{(-1)^{n_1+n_2+\cdots +n_{2r}}}{(2n_1-1)(2n_2-1)\cdots (2n_{2r}-1)}=\frac{(-1)^{r}\pi^{2r}}{2^{4r}(2r)!}\\ &t(\lbrace \bar{1}\rbrace ^{2r-1})=\sum_{0\lt n_1\lt\cdots\lt n_{2r-1}}\frac{(-1)^{n_1+n_2+\cdots +n_{2r-1}}}{(2n_1-1)(2n_2-1)\cdots (2n_{2r-1}-1)}=\frac{(-1)^r \pi^{2r-1}}{2^{4r-2}(2r-1)!} \end{align}

\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} &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}