雑多

$\sum_{n=1}^\infty\frac{\sigma_\alpha (n)}{m^n}=\sum_{n=1}^\infty\frac{n^\alpha}{m^n-1}$

$\sum_{m=1}^\infty\frac{\sigma_\alpha (n)}{m^n}=\zeta(n)\zeta(n-\alpha)$

$\frac{\zeta(s)\zeta(s-a)\zeta(s-2a)}{\zeta(2s-2a)} = \sum_{n=1}^\infty \frac{\sigma_a(n^2)}{n^s}$

$\sum_{m=1}^\infty\frac{\sigma_\alpha (n)\sigma_\beta (n)}{m^n}=\frac{\zeta(n)\zeta(n-\alpha)\zeta(n-\beta)\zeta(n-\alpha-\beta)}{\zeta(2n-\alpha-\beta)}$

$\frac{\zeta^3(s)}{\zeta(2s)}=\sum_{n=1}^{\infty}\frac{d(n^2)}{n^s}$

$\frac{\zeta^4(s)}{\zeta(2s)}=\sum_{n=1}^{\infty}\frac{d(n)^2}{n^s}.$

$\sum_{n=1}^\infty \mu(n)\,\frac{q^n}{1-q^n} = q$

$\sum_{n=1}^\infty \frac {\mu (n)}{n^s} = \left( 1- \frac{1}{2^s} \right) \left( 1- \frac{1}{3^s} \right) \left( 1- \frac{1}{5^s} \right) \left( 1- \frac{1}{7^s} \right) \cdots = \frac {1}{\zeta (s)}$

$\sum_{n=1}^\infty \varphi(n)\,\frac{q^n}{1-q^n} = \frac{q}{(1-q)^2}$

$\sum_{n=1}^\infty \frac {\varphi (n)}{n^s} = \frac{\zeta (s-1)}{\zeta (s)}$

$\sum_{n=1}^\infty \lambda(n)\,\frac{q^n}{1-q^n} = \sum_{n=1}^\infty q^{n^2}$

$\sum_{n=1}^\infty \frac {\lambda (n)}{n^s} = \frac{1} {{(1+ \frac{1}{2^s})} {(1+ \frac{1}{3^s})} {(1+ \frac{1}{5^s})} {(1+ \frac{1}{7^s})} \cdots } = \frac {\zeta (2s)} {\zeta (s)}$

$\sum_{n=1}^\infty J_k(n)\,\frac{q^n}{1-q^n}=\sum_{n=1}^\infty n^kq^n$

$\sum_{n=1}^\infty \frac {J_k(n)}{n^s} = \frac{\zeta (s-k)}{\zeta (s)}$

$\sum_{k=0}^{m}a_kF_{n+k}=0;\sum_{n=0}^\infty F_nx^n=\frac{\sum_{n=0}^{m-1} \left(F_n-\sum_{k=0}^{n-1}a_kF_k\right)x^n}{\sum_{k=0}^ma_kx^k}$

$\begin{align} &\frac{1}{5^2-1}+\frac{1}{7^2-1}+\frac{1}{11^2-1}+\frac{1}{13^2-1}+\cdots \\ =&\sum_{n=1}^\infty\left(\frac{1}{3n^2-n}+\frac{1}{3n^2+n}\right) \\ =&\sum_{n=1}^\infty\left(\frac{3}{3n-1}-\frac{3}{3n+1}\right) \\ =&3-\frac{\sqrt{3}}{2}\pi \end{align}$

$\prod_{n=1}^\infty\left(\frac{a_n-b}{a_n-c}\right)=\prod_{n=1}^\infty\frac{\displaystyle 1-\frac{b}{a_n}}{\displaystyle 1-\frac{c}{a_n}}$

$\prod_{c}\left(1-\frac{1}{c^k}\right)$

ここで、少なくともcは合成数、奇数の合成数、3で割り切れない合成数、2でも3でも割り切れない合成数、5で割り切れない合成数、などをとることができる。また、kがとることができるのは2,4,6,8,10,12,16,20,24,30…である。

ここから和として展開した時にまた様々な意味を成す。

$\begin{align} &1+\frac{1}{a+a^2}=1+\frac{1}{a^2}\frac{1}{1+\frac{1}{a}}=1+\frac{1}{a^2}-\frac{1}{a^3}+\frac{1}{a^4}-\frac{1}{a^5}\cdots \end{align}$

$\begin{align} &1+\frac{1}{a+a^2+a^3+a^4}=1+\frac{1}{a^4}\frac{1}{1+\frac{1}{a}}\frac{1}{1+\frac{1}{a^2}} \\ =&1+\frac{1}{a^4}\left(\sum_{n=0}^\infty\left(-\frac{1}{a}\right)^n\right)\left(\sum_{n=0}^\infty\left(-\frac{1}{a^2}\right)^n\right) \end{align}$

$\begin{align} \prod_{n=1}^\infty\left(1+\frac{1}{a_n^k}\right)&=\sum_{n=1}^\infty\frac{p_d(n)}{n^k} \\ \prod_{n=1}^\infty\frac{1}{(1+\frac{1}{a_n^k}}&=\sum_{n=1}^\infty\frac{p_e(n)-p_o(n)}{n^k} \\ \prod_{n=1}^\infty\left(1-\frac{1}{a_n^k}\right)&=\sum_{n=1}^\infty\frac{p_{ed}(n)-p_{od}(n)}{n^k} \\ \prod_{n=1}^\infty\frac{1}{1-\frac{1}{a_n^k}}&=\sum_{n=1}^\infty\frac{p(n)}{n^k} \\ \prod_{n=1}^\infty\frac{1-\frac{1}{a_n^mk}}{1-\frac{1}{a_n^k}}&=\sum_{n=1}^\infty\frac{p_{d(m)}(n)}{n^k} \\ \prod_{n=1}^\infty\frac{1}{1-\frac{1}{a_{ln}^k}}-\prod_{n=1}^\infty\frac{1-\frac{1}{a_n^mk}}{1-\frac{1}{a_n^k}}&=\sum_{n=1}^\infty\frac{p_{ed(m+1,l)}(n)}{n^k} \\ \end{align}$

p(n)はnを自然数の積に分割する方法の数、p_d(n)はnを異なる自然数の積に分割する方法の数、p_e(n)はnを偶数個の自然数の積に分割する方法の数、p_o(n)はnを奇数個の自然数の積に分割する方法の数

$\zeta^{m+1}(n)=\sum_{k=0}^\infty\frac{D_m(k)}{k^n}$

$D_m(k)=\sum_{p|k,p:prime}{}_{a_p+m}C_m$

$\sum_{n=0}^\infty\frac{d(k)^m}{k^n}$の一般解を求めることは困難である。m=2の場合ですらゼータ関数で表すことは難しい。

素因数分解の冪p_kにそれぞれ1を足したものではなく冪そのものをかけたものをd_a(n)とすると、

$\begin{align} \sum_{k=0}^\infty\frac{d_a(k)}{k^n}&=\prod_p\left(1+\sum_{k=1}^\infty \frac{k}{p^{nk}}\right)=\prod_p\left(\sum_{k=0}^\infty \frac{k+1}{p^{nk}}-\sum_{k=1}^\infty \frac{1}{p^{nk}}\right) \\ &=\prod_p\left(\frac{p^{2n}}{(p^n-1)^2}-\frac{1}{p^n-1}\right)=\prod_p\left(\frac{p^{2n}-p^n+1}{(p^n-1)^2}\right) \\ &=\prod_p\left(\frac{p^{3n}+1}{(p^n-1)^2(p^n+1)}\right)=\frac{\zeta(6n)}{\zeta(3n)\zeta(2n)\zeta(n)} \end{align}$

$\zeta(nk)\frac{\zeta(mk)}{\zeta(2mk)}=\sum_{s}\frac{1}{s^k}$

$\zeta(nk)\frac{1}{\zeta(mk)}=\sum_{s}\frac{(-1)^{\Omega(n)}}{s^k}$

sとはすべての素因数の冪に対してnで割った余りが0あるいはmであるような数

Ω(n)は素因数のうち冪をnで割った余りがmであるものの数。

$\begin{align} \frac{\zeta(s(2m+1))\zeta(2s)}{\zeta(s(4m+2))\zeta(s)}&=\prod_{p:prime}\frac{1+\frac{1}{p^{s(2m+1)}}}{1+\frac{1}{p^s}}=\prod_{p:prime}\left(\sum_{n=0}^{2m}\frac{(-1)^n}{p^{sn}}\right)=\sum_{n=1}^\infty\frac{\mu_{2m+1}(n)}{n^s} \\ \frac{\zeta(2s)}{\zeta(2ms)\zeta(s)}&=\prod_{p:prime}\frac{1-\frac{1}{p^{2ms}}}{1+\frac{1}{p^s}}=\prod_{p:prime}\left(\sum_{n=0}^{2m-1}\frac{(-1)^n}{p^{sn}}\right)=\sum_{n=1}^\infty\frac{\mu_{2m}(n)}{n^s} \\ \frac{\zeta(s)}{\zeta(ms)}&=\prod_{p:prime}\frac{1-\frac{1}{p^{ms}}}{1-\frac{1}{p^s}}=\prod_{p:prime}\left(\sum_{n=0}^{m-1}\frac{1}{p^{sn}}\right)=\sum_{n=1}^\infty\frac{|\mu_{m}(n)|}{n^s} \\ \end{align}$

$\mu_m(n)= \begin{cases} 0 & when\ n\ have\ a\ \text{m}th\ power\ factor\\ (-1)^n & otherwise \end{cases}$

${\pi^2 \over 6} {\pi^4 \over 90}{\pi^6 \over 945}\frac{\pi^8}{9450}\frac{\pi^{10}}{93555}\frac{691\pi^{12}}{638512875}\frac{2\pi^{14}}{18243225}\frac{3617\pi^{16}}{325641566250}\frac{43867\pi^{18}}{38979295480125}$

$\frac{174611\pi^{20}}{1531329465290625}\frac{155366\pi^{22}}{13447856940643125}\frac{236364091\pi^{24}}{201919571963756521875}$

$\zeta (2n)=(-1)^{n+1} \, \frac{\, B_{2n}(2\pi)^{2n}}{2(2n)!}$

$\begin{align} \sum_{k=0}^\infty\frac{d_a(k)}{k^2}&=\frac{2764}{5005} \\ \sum_{k=0}^\infty\frac{d_a(k)}{k^4}&=\frac{4727281820}{5138572439} \\ \end{align}$

$\begin{align} \sum_{n=1}^\infty\frac{\mu_2(n)}{n^2}&=\frac{6}{\pi^2} \\ \sum_{n=1}^\infty\frac{\mu_2(n)}{n^4}&=\frac{90}{\pi^4} \\ \sum_{n=1}^\infty\frac{\mu_2(n)}{n^6}&=\frac{945}{\pi^6} \\ \sum_{n=1}^\infty\frac{\mu_2(n)}{n^8}&=\frac{9450}{\pi^8} \\ \sum_{n=1}^\infty\frac{\mu_2(n)}{n^{10}}&=\frac{93555}{\pi^{10}} \\ \sum_{n=1}^\infty\frac{\mu_2(n)}{n^{12}}&=\frac{638512875}{691\pi^{12}} \\ \sum_{n=1}^\infty\frac{\mu_2(n)}{n^{14}}&=\frac{18243225}{2\pi^{14}} \\ \sum_{n=1}^\infty\frac{\mu_2(n)}{n^{16}}&=\frac{325641566250}{3617\pi^{16}} \\ \sum_{n=1}^\infty\frac{\mu_2(n)}{n^{18}}&=\frac{38979295480125}{42867\pi^{18}} \\ \sum_{n=1}^\infty\frac{\mu_2(n)}{n^{20}}&=\frac{1531329465290625}{174611\pi^{20}} \\ \sum_{n=1}^\infty\frac{\mu_2(n)}{n^{22}}&=\frac{13447856940643125}{155366\pi^{22}} \\ \sum_{n=1}^\infty\frac{\mu_2(n)}{n^{24}}&=\frac{201919571963756521875}{236364091\pi^{24}} \\ \sum_{n=1}^\infty\frac{\mu_3(n)}{n^2}&=\frac{45045}{691\pi^4} \\ \sum_{n=1}^\infty\frac{\mu_3(n)}{n^4}&=\frac{2081121837795}{236364091\pi^8} \\ \sum_{n=1}^\infty\frac{\mu_4(n)}{n^2}&=\frac{630}{\pi^6} \\ \sum_{n=1}^\infty\frac{\mu_4(n)}{n^4}&=\frac{3101348250}{3617\pi^{12}} \\ \sum_{n=1}^\infty\frac{\mu_4(n)}{n^6}&=\frac{206499314355208875}{236364091\pi^{18}} \\ \sum_{n=1}^\infty\frac{\mu_5(n)}{n^2}&=\frac{1091215125}{174611\pi^8} \\ \sum_{n=1}^\infty\frac{\mu_6(n)}{n^2}&=\frac{42567525}{691\pi^{10}} \\ \sum_{n=1}^\infty\frac{\mu_6(n)}{n^4}&=\frac{1923043542511966875}{236364091\pi^{20}} \\ \sum_{n=1}^\infty\frac{\mu_8(n)}{n^2}&=\frac{21709437750}{3617\pi^{14}} \\ \sum_{n=1}^\infty\frac{\mu_{10}(n)}{n^2}&=\frac{102088631019375}{174611\pi^{18}} \\ \sum_{n=1}^\infty\frac{\mu_{12}(n)}{n^2}&=\frac{13641304797583768125}{236364091\pi^{22}} \\ \sum_{n=1}^\infty\frac{\mu_\infty(n)}{n^2}&=\frac{\pi^2}{15} \\ \sum_{n=1}^\infty\frac{\mu_\infty(n)}{n^4}&=\frac{\pi^4}{105} \\ \sum_{n=1}^\infty\frac{\mu_\infty(n)}{n^6}&=\frac{691\pi^6}{675675} \\ \sum_{n=1}^\infty\frac{\mu_\infty(n)}{n^8}&=\frac{3617\pi^8}{34459425} \\ \sum_{n=1}^\infty\frac{\mu_\infty(n)}{n^{10}}&=\frac{174611\pi^{10}}{16368226875} \\ \sum_{n=1}^\infty\frac{\mu_\infty(n)}{n^{12}}&=\frac{236364091\pi^{12}}{218517792968475} \\ \sum_{n=1}^\infty\frac{|\mu_2(n)|}{n^2}&=\frac{15}{\pi^2} \\ \sum_{n=1}^\infty\frac{|\mu_2(n)|}{n^4}&=\frac{105}{\pi^4} \\ \sum_{n=1}^\infty\frac{|\mu_2(n)|}{n^6}&=\frac{675675}{691\pi^6} \\ \sum_{n=1}^\infty\frac{|\mu_2(n)|}{n^8}&=\frac{34459425}{3617\pi^8} \\ \sum_{n=1}^\infty\frac{|\mu_2(n)|}{n^{10}}&=\frac{16368226875}{174611\pi^{10}} \\ \sum_{n=1}^\infty\frac{|\mu_2(n)|}{n^{12}}&=\frac{218517792968475}{236364091\pi^{12}} \\ \sum_{n=1}^\infty\frac{|\mu_3(n)|}{n^2}&=\frac{315}{2\pi^4} \\ \sum_{n=1}^\infty\frac{|\mu_3(n)|}{n^4}&=\frac{14189175}{1362\pi^8} \\ \sum_{n=1}^\infty\frac{|\mu_3(n)|}{n^6}&=\frac{41247931725}{43867\pi^{12}} \\ \sum_{n=1}^\infty\frac{|\mu_3(n)|}{n^8}&=\frac{42374300944710375}{472728182\pi^{16}} \\ \sum_{n=1}^\infty\frac{|\mu_4(n)|}{n^2}&=\frac{1575}{\pi^6} \\ \sum_{n=1}^\infty\frac{|\mu_4(n)|}{n^4}&=\frac{3618239625}{3617\pi^{12}} \\ \sum_{n=1}^\infty\frac{|\mu_4(n)|}{n^6}&=\frac{213671504723551875}{236364091\pi^{18}} \\ \sum_{n=1}^\infty\frac{|\mu_5(n)|}{n^2}&=\frac{31185}{2\pi^8} \\ \sum_{n=1}^\infty\frac{|\mu_5(n)|}{n^4}&=\frac{34029543673125}{349222\pi^{16}} \\ \sum_{n=1}^\infty\frac{|\mu_6(n)|}{n^2}&=\frac{212837625}{1382\pi^{10}} \\ \sum_{n=1}^\infty\frac{|\mu_6(n)|}{n^4}&=\frac{4487101599194589375}{472728182\pi^{20}} \\ \sum_{n=1}^\infty\frac{|\mu_7(n)|}{n^2}&=\frac{6081075}{4\pi^{12}} \\ \sum_{n=1}^\infty\frac{|\mu_8(n)|}{n^2}&=\frac{54273594375}{3617\pi^{14}} \\ \sum_{n=1}^\infty\frac{|\mu_9(n)|}{n^2}&=\frac{12993098493375}{87734\pi^{16}} \\ \sum_{n=1}^\infty\frac{|\mu_{10}(n)|}{n^2}&=\frac{510443155096875}{349222\pi^{18}} \\ \sum_{n=1}^\infty\frac{|\mu_{11}(n)|}{n^2}&=\frac{4482618980214375}{310732\pi^{20}} \\ \sum_{n=1}^\infty\frac{|\mu_{12}(n)|}{n^2}&=\frac{67306523987918840625}{472728182\pi^{22}} \\ \end{align}$

$\begin{align} \sum_{n=1}^\infty\frac{\mu_{2(o)}(n)}{n^2}&=\frac{9}{2\pi^2} \\ \sum_{n=1}^\infty\frac{\mu_{2(o)}(n)}{n^4}&=\frac{15}{2\pi^4} \\ \sum_{n=1}^\infty\frac{\mu_{2(o)}(n)}{n^6}&=\frac{11340}{691\pi^6} \\ \sum_{n=1}^\infty\frac{\mu_{2(o)}(n)}{n^8}&=\frac{278775}{7234\pi^8} \\ \sum_{n=1}^\infty\frac{\mu_{2(o)}(n)}{n^{10}}&=\frac{16247385}{174611\pi^{10}} \\ \sum_{n=1}^\infty\frac{\mu_{2(o)}(n)}{n^{12}}&=\frac{37139825022304}{163327586881\pi^{12}} \\ \sum_{n=1}^\infty\frac{\mu_{3(o)}(n)}{n^2}&=\frac{127575}{2744\pi^4} \\ \sum_{n=1}^\infty\frac{\mu_{3(o)}(n)}{n^4}&=\frac{173107835946045}{214818594628\pi^8} \\ \sum_{n=1}^\infty\frac{\mu_{4(o)}(n)}{n^2}&=\frac{945}{2\pi^6} \\ \sum_{n=1}^\infty\frac{\mu_{4(o)}(n)}{n^4}&=\frac{516891375}{7234\pi^{12}} \\ \sum_{n=1}^\infty\frac{\mu_{4(o)}(n)}{n^6}&=\frac{2586095184171500}{236364091\pi^{18}} \\ \sum_{n=1}^\infty\frac{\mu_{5(o)}(n)}{n^2}&=\frac{3262813785}{698444\pi^8} \\ \sum_{n=1}^\infty\frac{\mu_{6(o)}(n)}{n^2}&=\frac{127702575}{2764\pi^{10}} \\ \sum_{n=1}^\infty\frac{\mu_{6(o)}(n)}{n^4}&=\frac{641014514170655625}{945456364\pi^{20}} \\ \sum_{n=1}^\infty\frac{\mu_{8(o)}(n)}{n^2}&=\frac{32564156625}{7234\pi^{14}} \\ \sum_{n=1}^\infty\frac{\mu_{10(o)}(n)}{n^2}&=\frac{306265893058125}{698444\pi^{18}} \\ \sum_{n=1}^\infty\frac{\mu_{12(o)}(n)}{n^2}&=\frac{40023914392751304375}{945456364\pi^{22}} \\ \sum_{n=1}^\infty\frac{\mu_{\infty(o)}(n)}{n^2}&=\frac{\pi^2}{20} \\ \sum_{n=1}^\infty\frac{\mu_{\infty(o)}(n)}{n^4}&=\frac{\pi^4}{1260} \\ \sum_{n=1}^\infty\frac{\mu_{\infty(o)}(n)}{n^6}&=\frac{4\pi^6}{225225} \\ \sum_{n=1}^\infty\frac{\mu_{\infty(o)}(n)}{n^8}&=\frac{59\pi^8}{137837700} \\ \sum_{n=1}^\infty\frac{\mu_{\infty(o)}(n)}{n^{10}}&=\frac{521\pi^{10}}{49104680625} \\ \sum_{n=1}^\infty\frac{\mu_{\infty(o)}(n)}{n^{12}}&=\frac{872492\pi^{12}}{3277766894527125} \\ \sum_{n=1}^\infty\frac{\mu_{2(e)}(n)}{n^2}&=\frac{21}{2\pi^2} \\ \sum_{n=1}^\infty\frac{\mu_{2(e)}(n)}{n^4}&=\frac{195}{2\pi^4} \\ \sum_{n=1}^\infty\frac{\mu_{2(e)}(n)}{n^6}&=\frac{664335}{691\pi^6} \\ \sum_{n=1}^\infty\frac{\mu_{2(e)}(n)}{n^8}&=\frac{68640075}{7234\pi^8} \\ \sum_{n=1}^\infty\frac{\mu_{2(e)}(n)}{n^{10}}&=\frac{16351979490}{174611\pi^{10}} \\ \sum_{n=1}^\infty\frac{\mu_{2(e)}(n)}{n^{12}}&=\frac{150958655116193925}{163327586881\pi^{12}} \\ \sum_{n=1}^\infty\frac{\mu_{3(e)}(n)}{n^2}&=\frac{307755}{2764\pi^4} \\ \sum_{n=1}^\infty\frac{\mu_{3(e)}(n)}{n^4}&=\frac{2081121837795}{236364091\pi^8} \\ \sum_{n=1}^\infty\frac{\mu_{4(e)}(n)}{n^2}&=\frac{2205}{2\pi^6} \\ \sum_{n=1}^\infty\frac{\mu_{4(e)}(n)}{n^4}&=\frac{3101348250}{3617\pi^{12}} \\ \sum_{n=1}^\infty\frac{\mu_{4(e)}(n)}{n^6}&=\frac{206499314355208875}{236364091\pi^{18}} \\ \sum_{n=1}^\infty\frac{\mu_{5(e)}(n)}{n^2}&=\frac{7627674285}{698444\pi^8} \\ \sum_{n=1}^\infty\frac{\mu_{6(e)}(n)}{n^2}&=\frac{297972675}{2764\pi^{10}} \\ \sum_{n=1}^\infty\frac{\mu_{6(e)}(n)}{n^4}&=\frac{8333188684218523125}{945456364\pi^{20}} \\ \sum_{n=1}^\infty\frac{\mu_{8(e)}(n)}{n^2}&=\frac{75983032125}{7234\pi^{14}} \\ \sum_{n=1}^\infty\frac{\mu_{{10}(e)}(n)}{n^2}&=\frac{714620417135625}{698444\pi^{18}} \\ \sum_{n=1}^\infty\frac{\mu_{{12}(e)}(n)}{n^2}&=\frac{94589133583086376875}{945456364\pi^{22}} \\ \sum_{n=1}^\infty\frac{\mu_{\infty(e)}(n)}{n^2}&=\frac{7\pi^2}{60} \\ \sum_{n=1}^\infty\frac{\mu_{\infty(e)}(n)}{n^4}&=\frac{13\pi^4}{1260} \\ \sum_{n=1}^\infty\frac{\mu_{\infty(e)}(n)}{n^6}&=\frac{703\pi^6}{675675} \\ \sum_{n=1}^\infty\frac{\mu_{\infty(e)}(n)}{n^8}&=\frac{14527\pi^8}{137837700} \\ \sum_{n=1}^\infty\frac{\mu_{\infty(e)}(n)}{n^{10}}&=\frac{524354\pi^{10}}{49104680625} \\ \sum_{n=1}^\infty\frac{\mu_{\infty(e)}(n)}{n^{12}}&=\frac{322393987\pi^{12}}{3277766894527125} \\ \end{align}$

ただし、

$d_a(m)=\prod\nu(p);m=\prod_{p;prime} p^{\nu(p)}$

$\mu_m(n)= \begin{cases} 0 & when\ n\ have\ a\ \text{m}th\ power\ factor\\ (-1)^n & otherwise \end{cases}$

$\mu_{m(o)}(n)$はnがm個以上の素因数の重なりを持たず、かつ奇数個の素因数を持つとき1を、それ以外のとき0を返す関数、

$\mu_{m(o)}(n)$はnがm個以上の素因数の重なりを持たず、かつ偶数個の素因数を持つとき1を、それ以外のとき0を返す関数である。

また、$|\mu_\infty(n)|=1$なので割愛した。

$\prod_{p:prime}\left(1+\frac{1}{p^2+p^4}\right)=\prod_{p:prime}\left(1+\frac{1}{[p^2]_3-1}\right)=\frac{21}{2\pi^2}$

$\prod_{p:prime}\left(1+\frac{1}{p^2+p^4+p^6}\right)=\prod_{p:prime}\left(1+\frac{1}{[p^2]_4-1}\right)=\frac{10}{\pi^2}$

$\prod_{p:prime}\left(1+\frac{1}{p^2+p^4+p^6+p^8}\right)=\prod_{p:prime}\left(1+\frac{1}{[p^2]_5-1}\right)=\frac{99}{10\pi^2}$

$\prod_{p:prime}\left(1-\frac{1}{[p^2]_3}\right)=\frac{2}{21}\pi^2$

$\prod_{p:prime}\left(1-\frac{1}{[p^2]_4}\right)=\frac{1}{10}\pi^2$

$\prod_{p:prime}\left(1-\frac{1}{[p^2]_5}\right)=\frac{10}{99}\pi^2$

$\begin{align} &\coth x -\cot x -\frac{2x}{3} =2\sum_{k=1}^\infty\frac{2^{4k+2}B_{4k+2}}{(4k+2)!}x^{4k+1}=2\sum_{k=1}^\infty\frac{2^{4k+2}}{(4k+1)!}x^{4k+1}\sum_{n=1}^\infty\frac{n^{4k+1}}{e^{2\pi n}-1} \\ =&4\sum_{n=1}^\infty\frac{1}{e^{2\pi n}-1}\sum_{k=1}^\infty\frac{(2nx)^{4k+1}}{(4k+1)!}=2\sum_{n=1}^\infty\frac{\sinh 2nx +\sin 2nx-4nx}{e^{2\pi n}-1} \end{align}$

$\sum_{n=1}^\infty\frac{\sinh 2nx +\sin 2nx}{e^{2\pi n}-1}=\frac{\coth x -\cot x}{2} -\frac{x}{6}-\frac{x}{2\pi}$

$\sum_{m=2}^\infty\sum_{n=2}^\infty\frac{1}{m^n}=\sum_{p:prime}\sum_{k=2}^\infty\frac{|\mu(k)|d(k)}{k^p}$