FANDOM


オイラーの方法でもたらされる級数は積の形に直すことが可能であるが、符号を反転させられるのは周期が4,6,8,12,24の場合だけである。

4の時は4n+1,4n+3

6の時は6n+1,6n+5

8の時は8n+1,8n+3,8n+5,8n+7(2種類の級数の和)

12の時は12n+1,12n+5,12n+7,12n+11(2種類の級数の和)

24の時は24n+1,24n+5,24n+7,24n+11,24n+13,24n+17,24n+19,24n+23(4種類の級数の和)

が分母となる。

また、オイラーの方法でもたらされる関数をarctanのように多項式に治すことはできても意味のある特殊値をもたらすことは実質不可能。

$ \sum_{\begin{smallmatrix} (l,m,n)\in\mathbb{N}^3 & \\ l\neq m\neq n\neq l& \end{smallmatrix}} $

$ \frac{\displaystyle \sum_{n=1}^{\infty}(n+2)^2}{\displaystyle n!} $

$ \begin{align} &\sum_{n=0}^\infty a_nx^n \\ =&\sqrt{\displaystyle \sum_{n=0}^\infty\left(\sum_{k=0}^na_ka_{n-k}\right)x^n} \\ =&\sqrt{\displaystyle 1+2a_1x\left(1+\sum_{n=2}^\infty\frac{\displaystyle \sum_{k=0}^na_ka_{n-k}}{\displaystyle 2a_1}x^{n-1}\right)} \end{align} $

$ \begin{align} &\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}\cdots \\ =&x\sqrt{1-\frac{2}{3}x^2\sqrt{1-\frac{23}{15}x^2\sqrt{1-\frac{11623}{4830}x^2\sqrt{\cdots}}}} \end{align} $

$ \sqrt{1-\frac{2}{3}\sqrt{1-\frac{23}{15}\sqrt{1-\frac{11623}{4830}\sqrt{\cdots}}}}=\frac{\pi}{4} $

$ \sqrt{1-\frac{2}{3}\cdot\frac{1}{3}\sqrt{1-\frac{23}{15}\cdot\frac{1}{3}\sqrt{1-\frac{11623}{4830}\cdot\frac{1}{3}\sqrt{\cdots}}}}=\frac{\pi}{2\sqrt{3}} $

$ \begin{align} &\sqrt[3]{\displaystyle 3\left(\left(\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}\right)\left(\sum_{\begin{smallmatrix} (m,n)\in\mathbb{O}^2 & \\ m\neq n& \end{smallmatrix}}\frac{(-1)^{\frac{m+n}{2}}}{mn}\right)+\sum_{\begin{smallmatrix} (l,m,n)\in\mathbb{O}^3 & \\ l\neq m\neq n\neq l& \end{smallmatrix}}\frac{(-1)^{\frac{l+m+n-1}{2}}}{lmn}\right)} \\ &\overline{+\sqrt[3]{\displaystyle 3\left(\left(\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3}\right)\left(\sum_{\begin{smallmatrix} (m,n)\in\mathbb{O}^2 & \\ m\neq n& \end{smallmatrix}}\frac{(-1)^{\frac{m+n}{2}}}{m^3n^3}\right)+\sum_{\begin{smallmatrix} (l,m,n)\in\mathbb{O}^3 & \\ l\neq m\neq n\neq l& \end{smallmatrix}}\frac{(-1)^{\frac{l+m+n-1}{2}}}{l^3m^3n^3}\right)}} \\ &\overline{\overline{+\sqrt[3]{\displaystyle 3\left(\left(\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^9}\right)\left(\sum_{\begin{smallmatrix} (m,n)\in\mathbb{O}^2 & \\ m\neq n& \end{smallmatrix}}\frac{(-1)^{\frac{m+n}{2}}}{m^9n^9}\right)+\sum_{\begin{smallmatrix} (l,m,n)\in\mathbb{O}^3 & \\ l\neq m\neq n\neq l& \end{smallmatrix}}\frac{(-1)^{\frac{l+m+n-1}{2}}}{l^9m^9n^9}\right)+\cdots}}} \\ =&\frac{\pi}{4}\ (\mathbb{O}=\{n=1(\text{mod}2)|n\in\mathbb{N}\}) \end{align} $

$ x^{\log (1+y)}=x^{y-\frac{y^2}{2}+\frac{y^3}{3}-\cdots}=x^{y(1-\frac{y}{2}(1-\frac{2}{3}y\cdots))}=x\sqrt[\frac{1}{y}]{x\sqrt[-\frac{2}{y}]{x\sqrt[-\frac{3}{y}]{x^2\sqrt[-\frac{4}{y}]{\cdots}}}} $

$ \frac{\pi^2}{6}=\left(\sqrt[4]{e\sqrt[8]{e\sqrt[12]{e^2\sqrt[16]{e^3\cdots}}}}\right)\left(\sqrt[9]{e\sqrt[18]{e\sqrt[27]{e^2\sqrt[36]{e^3\cdots}}}}\right)\left(\sqrt[25]{e\sqrt[50]{e\sqrt[75]{e^2\sqrt[100]{e^3\cdots}}}}\right)\cdots $

$ \frac{\pi}{2}=\left(\sqrt[4]{e\sqrt[8]{e\sqrt[12]{e^2\sqrt[16]{e^3\cdots}}}}\right)\left(\sqrt[16]{e\sqrt[32]{e\sqrt[48]{e^2\sqrt[64]{e^3\cdots}}}}\right)\left(\sqrt[36]{e\sqrt[72]{e\sqrt[108]{e^2\sqrt[144]{e^3\cdots}}}}\right)\cdots $

$ \sum_{n=1}^\infty p(n)e^{-n\pi}=\prod_{n=1}^\infty\frac{1}{(1-e^{-n\pi})}=\phi(e^{-\pi})=\frac{e^{\frac{\pi}{24}}\Gamma\left(\frac{1}{4}\right)}{2^{\frac{7}{8}}\pi^{\frac{3}{4}}} $

$ \sum_{n=1}^\infty p(n)e^{-2n\pi}=\prod_{n=1}^\infty\frac{1}{(1-e^{-2n\pi})}=\phi(e^{-2\pi})=\frac{e^{\frac{\pi}{12}}\Gamma\left(\frac{1}{4}\right)}{2\pi^{\frac{3}{4}}} $

$ \sum_{n=1}^\infty p(n)e^{-4n\pi}=\prod_{n=1}^\infty\frac{1}{(1-e^{-4n\pi})}=\phi(e^{-4\pi})=\frac{e^{\frac{\pi}{6}}\Gamma\left(\frac{1}{4}\right)}{2^{\frac{11}{8}}\pi^{\frac{3}{4}}} $

$ \sum_{n=1}^\infty p(n)e^{-8n\pi}=\prod_{n=1}^\infty\frac{1}{(1-e^{-8n\pi})}=\phi(e^{-8\pi})=\frac{e^{\frac{\pi}{3}}\Gamma\left(\frac{1}{4}\right)}{2^{\frac{29}{16}}\pi^{\frac{3}{4}}}\sqrt[4]{\sqrt{2}-1} $

$ \sum_{n=1}^\infty p_d(n)e^{-n\pi}=\prod_{n=1}^\infty(1+e^{-n\pi})=\frac{\phi(e^{-\pi})}{\phi(e^{-2\pi})}=\frac{2^{\frac{1}{8}}}{e^{\frac{\pi}{24}}} $

$ \sum_{n=1}^\infty p_d(n)e^{-2n\pi}=\prod_{n=1}^\infty(1+e^{-2n\pi})=\frac{\phi(e^{-2\pi})}{\phi(e^{-4\pi})}=\frac{2^{\frac{3}{8}}}{e^{\frac{\pi}{12}}} $

$ \sum_{n=1}^\infty p_d(n)e^{-4n\pi}=\prod_{n=1}^\infty(1+e^{-4n\pi})=\frac{\phi(e^{-4\pi})}{\phi(e^{-8\pi})}=\frac{2^{\frac{7}{16}}}{e^{\frac{\pi}{6}}}\sqrt[4]{\sqrt{2}+1} $

$ \begin{align} \prod_{n=1}^\infty(1+x^{2n-1})^8&=\prod_{n=1}^\infty(1-x^{2n-1})^8+16x\prod_{n=1}^\infty(1+x^{2n})^8 \\ \prod_{n=1}^\infty\left(\frac{(1+x^{n})}{(1+x^{2n})}\right)^8&=\prod_{n=1}^\infty\left(\frac{(1-x^{n})}{(1-x^{2n})}\right)^8+16x\prod_{n=1}^\infty\left(\frac{(1-x^{4n})}{(1-x^{2n})}\right)^8 \\ \prod_{n=1}^\infty\left(\frac{(1-x^{2n})^2}{(1-x^{n})(1-x^{4n})}\right)^8&=\prod_{n=1}^\infty\left(\frac{(1-x^{n})}{(1-x^{2n})}\right)^8+16x\prod_{n=1}^\infty\left(\frac{(1-x^{4n})}{(1-x^{2n})}\right)^8 \\ \frac{\phi^{16}(x^2)}{\phi^8(x)\phi^8(x^4)}&=\frac{\phi^8(x)}{\phi^8(x^2)}+16x\frac{\phi^8(x^4)}{\phi^8(x^2)} \\ \phi^{24}(x^2)&=\phi^16(x)\phi^8(x^4)+16x\phi^{16}(x^4)\phi^8(x) \\ \phi^8(x^4)&=\frac{-\phi^16(x)+\sqrt{\phi^{32}(x)+64x\phi^{24}(x^2)\phi^{8}(x)}}{32x\phi^8(x)} \\ a_{n+2}&=\frac{-a_n^2+\sqrt{a_n(a_n^3+64xa_{n+1}^3)}}{32xa_{n}} \\ \frac{b_n}{b_{n+1}}&=\frac{1}{b_n}+16xb_{n+1} \\ b_n^2&=b_{n+1}+16xb_nb_{n+1}^2 \\ b_{n+1}&=\frac{-1+\sqrt{1+64xb_n^3}}{32xb_n} \end{align} $

$ \begin{align} b_0&=\left(\frac{\phi(e^{-2\pi})}{\phi(e^{-\pi})}\right)^8=\frac{e^{\frac{\pi}{3}}}{2} \\ b_1&=\frac{e^{\frac{2}{3}\pi}}{8} \\ \end{align} $

$ x=e^{-2^n\pi},b_n=e^{\frac{2^n}{3}\pi}c_n $

$ \begin{align} b_{n+1}&=\frac{-1+\sqrt{1+64xb_n^3}}{32xb_n} \\ e^{\frac{2^{n+1}}{3}\pi}c_{n+1}&=\frac{-1+\sqrt{1+64e^{-2^n\pi}e^{2^n\pi}c_n^3}}{32e^{-2^n\pi}e^{\frac{2^n}{3}\pi}c_n} \\ e^{\frac{2^{n+1}}{3}\pi}c_{n+1}&=\frac{-1+\sqrt{1+64c_n^3}}{32e^{-\frac{2^{n+1}}{3}\pi}c_n} \\ c_{n+1}&=\frac{-1+\sqrt{1+64c_n^3}}{32c_n} \\ \end{align} $

$ c_0=\frac{1}{2},c_1=\frac{1}{8},c_2=\frac{1}{16}(3\sqrt{2}-4) $

$ d_n=4c_n,d_{n+1}=\frac{-1+\sqrt{1+d_n^3}}{2d_n} $

$ d_0=2,d_1=\frac{1}{2},d_2=\frac{1}{4}(3\sqrt{2}-4),d_3=\frac{3}{4}\sqrt{60+43\sqrt{2}}-4-3\sqrt{2},d_{-1}=4+6\sqrt{2} $

$ \begin{align} \prod_{n=1}^\infty (1-x^n)&=\sum_{n=1}^\infty (p_e(n)-p_o(n))x^n \\ \prod_{n=1}^\infty (1+x^n)&=\sum_{n=1}^\infty p_d(n)x^n \\ \prod_{n=1}^\infty (1-x^n)^{-1}&=\sum_{n=1}^\infty p(n)x^n \\ \prod_{n=1}^\infty (1+x^n)^{-1}&=\sum_{n=1}^\infty (p_{ed}(n)-p_{od}(n))x^n \\ \frac{1}{2}\left(\prod_{n=1}^\infty (1+x^n)^{-1}+\prod_{n=1}^\infty (1-x^n)\right)&=\sum_{n=1}^\infty (p_e(n))x^n \\ \frac{1}{2}\left(\prod_{n=1}^\infty (1+x^n)^{-1}-\prod_{n=1}^\infty (1-x^n)\right)&=\sum_{n=1}^\infty (p_o(n))x^n \\ \frac{1}{2}\left(\prod_{n=1}^\infty (1+x^n)+\prod_{n=1}^\infty (1+x^n)^{-1}\right)&=\sum_{n=1}^\infty (p_{ed}(n))x^n \\ \frac{1}{2}\left(\prod_{n=1}^\infty (1+x^n)-\prod_{n=1}^\infty (1+x^n)^{-1}\right)&=\sum_{n=1}^\infty (p_{od}(n))x^n \\ \prod_{n=1}^\infty \frac{1-x^{2^kn}}{1-x^n}&=\sum_{n=1}^\infty p_{d2^k-1}(n)x^n \\ \prod_{n=1}^\infty (1-x^n)^{-1}-\prod_{n=1}^\infty \frac{1-x^{2^kn}}{1-x^n}&=\sum_{n=1}^\infty p_{e2^k}(n)x^n \\ \prod_{n=1}^\infty \frac{1-x^{2^mn}}{1-x^n}-\prod_{n=1}^\infty \frac{1-x^{2^ln}}{1-x^n}&=\sum_{n=1}^\infty p_{e2^l,d2^m-1}(n)x^n \\ \prod_{n=1}^\infty \frac{1+x^{2^ln}}{1-x^{2^mn}}&=\sum_{n=1}^\infty p^{(2)}_{2^l,2^m}(n)x^n \\ \prod_{n=1}^\infty \frac{1-x^{2^ln}}{(1-x^{n})(1-x^{2^mn})}&=\sum_{n=1}^\infty p^{(3)}_{2^l,2^m}(n)x^n \\ \prod_{n=1}^\infty \frac{1-x^{2^ln}}{(1-x^{n})(1-x^{2^mn})}-\prod_{n=1}^\infty \frac{1-x^{2^sn}}{(1-x^{2n})(1-x^{2^mn})}&=\sum_{n=1}^\infty p^{(4)}_{2^l,2^s,2^m}(n)x^n \\ \prod_{n=1}^\infty \frac{(1-x^n)(1-x^{2^ln})}{(1-x^{2n})(1-x^{2^mn})}&=\sum_{n=1}^\infty (p^{(3e)}_{2^l,2^m}-p^{(3o)}_{2^l,2^m}(n))x^n \\ \prod_{n=1}^\infty \frac{(1-x^n)(1-x^{2^ln})}{(1-x^{2n})(1-x^{2^mn})}-\prod_{n=1}^\infty \frac{(1-x^n)(1-x^{2^sn})}{(1-x^{2n})(1-x^{2^mn})}&=\sum_{n=1}^\infty (p^{(4e)}_{2^l,2^s,2^m}(n)-p^{(4o)}_{2^l,2^s,2^m}(n))x^n \\ \frac{1}{2}\left(\prod_{n=1}^\infty \frac{1-x^{2^ln}}{(1-x^{n})(1-x^{2^mn})}+\prod_{n=1}^\infty \frac{(1-x^n)(1-x^{2^ln})}{(1-x^{2n})(1-x^{2^mn})}\right)&=\sum_{n=1}^\infty (p^{(3e)}_{2^l,2^m})x^n \\ \frac{1}{2}\left(\prod_{n=1}^\infty \frac{1-x^{2^ln}}{(1-x^{n})(1-x^{2^mn})}-\prod_{n=1}^\infty \frac{(1-x^n)(1-x^{2^ln})}{(1-x^{2n})(1-x^{2^mn})}\right)&=\sum_{n=1}^\infty (p^{(3o)}_{2^l,2^m})x^n \\ \frac{1}{2}\left(\prod_{n=1}^\infty \frac{1-x^{2^ln}}{(1-x^{n})(1-x^{2^mn})}+\prod_{n=1}^\infty \frac{(1-x^n)(1-x^{2^ln})}{(1-x^{2n})(1-x^{2^mn})}\right.-& \\ \left.\prod_{n=1}^\infty \frac{1-x^{2^sn}}{(1-x^{n})(1-x^{2^mn})}-\prod_{n=1}^\infty \frac{(1-x^n)(1-x^{2^sn})}{(1-x^{2n})(1-x^{2^mn})}\right)&=\sum_{n=1}^\infty (p^{(4e)}_{2^l,2^s,2^m}(n))x^n \\ \frac{1}{2}\left(\prod_{n=1}^\infty \frac{1-x^{2^ln}}{(1-x^{n})(1-x^{2^mn})}- \prod_{n=1}^\infty \frac{(1-x^n)(1-x^{2^ln})}{(1-x^{2n})(1-x^{2^mn})}-\right.& \\ \left.\prod_{n=1}^\infty \frac{1-x^{2^sn}}{(1-x^{n})(1-x^{2^mn})}+\prod_{n=1}^\infty \frac{(1-x^n)(1-x^{2^sn})}{(1-x^{2n})(1-x^{2^mn})}\right)&=\sum_{n=1}^\infty (p^{(4o)}_{2^l,2^s,2^m}(n))x^n \\ \end{align} $

p(n)はnを自然数に分割する方法の数、p_d(n)はnを異なる自然数に分割する方法の数、p_e(n)はnを偶数個の自然数に分割する方法の数、p_o(n)はnを奇数個の自然数に分割する方法の数,p_dm(n)はnをm回まで同じ自然数を使っていいという条件付きで自然数に分割する方法の数(mで割った余りが0でない数のみを用いてnを分割する方法の数でもある。)、p_em(n)は同じ自然数をm個以上含むように自然数を分割する方法の数、p_el,dm(n)は同じ自然数をl個以上含み、かつm+1個以上は含まないように自然数を分割する方法の数、p^(2)_{l,m}(n)はmで割った余りが0かlの数を使ってnを分割する方法の数、p^(3)_{l,m}(n)はmで割った余りが0以上l未満の数を使ってnを分割する方法の数、p^(4)_{l,s,m}(n)はmで割った余りが0以上l未満の数を使い、かつ余りがs以上の数を1つ以上使ってnを分割する方法の数、p^(3e)_{l,m}(n)はmで割った余りが0以上l未満の数を使ってnを偶数個に分割する方法の数、p^(3d)_{l,m}(n)はmで割った余りが0以上l未満の数を使ってnを奇数個に分割する方法の数。

$ x\prod_{n=1}^\infty\left(\frac{1-x^{2n}}{1-x^{2n-1}}\right)^8=\sum_{n=0}^\infty\sum_{m=0}^\infty 2^{3m}\sigma_3(2n+1)x^{2^m(2n+1)}=\sum_{n=1}^\infty2^{3b(n)}\sigma_3(Od(n))x^n $

$ \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) $

$ \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)} $

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

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

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

$ \prod_{n=0}^\infty(1-x^{2^n})=\sum_{n=0}^\infty p(n)x^n $

p(n)はA106400(Thue–Morse_sequence)

$ \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} $

$ 1-\frac{1}{p^n}=\frac{1}{1+\frac{1}{p^n-1}} $

$ \sum_{n=-\infty}^{\infty}{q^{n(n+1)}z^{n}} =\prod_{m=1}^{\infty}{\left(1-q^{2m}\right)\left(1+q^{2m}z\right)\left(1+q^{2m-2}z^{-1}\right)} $

$ \begin{align} f(a,b) &=\sum_{n=-\infty}^\infty a^{\frac{n(n+1)}{2}} b^{\frac{n(n-1)}{2}} \\ &= \prod_{n=0}^\infty (1+a^{n+1}b^n)(1+a^nb^{n+1})(1-a^{n+1}b^{n+1}) \end{align} $

$ \prod_{n=1}^{\infin}(1-x^n)=\sum_{n=-\infin}^{\infin}(-1)^nx^{n(3n-1)/2} $

$ \begin{align} &\sum_{n=-\infty}^{\infty}q^{n(3n+1)}(z^{3n}-z^{-3n-1}) \\ =&\prod_{m=1}^{\infty}(1-q^{2m})(1-q^{2m}z)(1-q^{2m-2}z^{-1})(1-q^{4m-2}z^{2})(1-q^{4m-2}z^{-2}) \\ \end{align} $

$ \begin{align} &\prod_{n=1}^\infty(1-q^{2n})(1-xq^{2n-2})(1-\frac{q^{2n}}{x})(1-x^2q^{4n-2})(1-\frac{q^{4n-2}}{x^2})(1-x^2q^{4n-4})(1-\frac{q^{4n}}{x^2}) \\ =&\sum_{n=-\infty}^\infty(-1)^nq^{5n^2+n}\left(\sum_{n=-\infty}^\infty(-1)^n(q^{5n^2+n}x^{5n+3}-q^{5n^2-3n}x^{5n})\right)- \\ &\sum_{n=-\infty}^\infty(-1)^nq^{5n^2+3n}\left(\sum_{n=-\infty}^\infty(-1)^n(q^{5n^2+n}x^{5n+2}-q^{5n^2-n}x^{5n+1})\right) \\ \end{align} $

$ \begin{align} \prod_{n=2}^\infty\left(1-\frac{1}{n^2}\right)&=\frac{1}{2} \\ \prod_{n=2}^\infty\left(1-\frac{1}{n^3}\right)&=\frac{\cosh\left(\frac{\sqrt{3}}{2}\pi\right)}{3\pi} \\ \prod_{n=2}^\infty\left(1-\frac{1}{n^4}\right)&=\frac{\sinh\pi}{4\pi} \\ \prod_{n=2}^\infty\left(1-\frac{1}{n^6}\right)&=\frac{1+\cosh(\sqrt{3}\pi)}{12\pi^2} \\ \prod_{n=1}^\infty\left(1+\frac{1}{n^2}\right)&=\frac{\sinh\pi}{\pi} \\ \prod_{n=1}^\infty\left(1+\frac{1}{n^3}\right)&=\frac{\cosh\left(\frac{\sqrt{3}}{2}\pi\right)}{\pi} \\ \prod_{n=1}^\infty\left(1+\frac{1}{n^4}\right)&=\frac{\cosh(\pi\sqrt{2})-\cos(\pi\sqrt{2})}{2\pi^2} \\ \prod_{n=1}^\infty\left(1+\frac{1}{n^6}\right)&=\frac{\sinh\pi(\cosh\pi-\cos(\sqrt{3}\pi))}{2\pi^3} \\ \prod_{n=3}^\infty\left(1-\frac{4}{n^2}\right)&=\frac{1}{6} \\ \prod_{n=3}^\infty\left(1-\frac{8}{n^3}\right)&=\frac{\sinh(\pi\sqrt{3})}{42\pi\sqrt{3}} \\ \prod_{n=3}^\infty\left(1-\frac{16}{n^4}\right)&=\frac{\sinh(2\pi)}{120\pi} \\ \end{align} $

$ \begin{align} \prod_{p\equiv 0 (mod k)}(1-x^p)\prod_{p\not\equiv 0 (mod k)}(1+x^p)&=\sum_{n=1}^\infty p_{1ek}(n)-p_{1ok}x^n \\ \prod_{p\equiv 0 (mod k)}(1+x^p)\prod_{p\not\equiv 0 (mod k)}(1-x^p)&=\sum_{n=1}^\infty p_{2ek}(n)-p_{2ok}x^n \\ \prod_{p\equiv 0 (mod k)}\frac{1}{(1-x^p)}\prod_{p\not\equiv 0 (mod k)}\frac{1}{(1+x^p)}&=\sum_{n=1}^\infty p_{3ek}(n)-p_{3ok}x^n \\ \prod_{p\equiv 0 (mod k)}\frac{1}{(1+x^p)}\prod_{p\not\equiv 0 (mod k)}\frac{1}{(1-x^p)}&=\sum_{n=1}^\infty p_{4ek}(n)-p_{4ok}x^n \\ \end{align} $

p_{1ek}(n)はkで割った余りが0でない数を偶数回用いてnを相異なる自然数の和に分割する方法、p_{2ek}はkで割った余りが0になるを偶数回用いてnを相異なる自然数の和に分割する方法、p_{3ek}はkで割った余りが0になるを偶数回用いてnを自然数の和に分割する方法、p_{4ek}(n)はkで割った余りが0でない数を偶数回用いてnを自然数の和に分割する方法

$ \prod_{p:prime}^\infty(1+\frac{1}{\sum_{k=1}^tp^{mk}})=\sum_{n=1}^\infty\frac{\sigma_{(m,k)}''(n)}{n^m} $

nを自然数の積に分割するある分け方をd,d全体の集合をD、与えられた自然数の組を約数函数のようにm乗してk回まで使って足し合わせる方法の数をσ'_(m,k)(n)とすると、

$ \sigma_m''(n)=\sum_{d\in D}\frac{1}{\sigma_{(m,n)}} $

これらの記号を統一する方法

kで割った余りがs以上t以下である数を(ない数を)奇数個(偶数個)用い、同じものをa個以上使わず、かつ一種類はb個以上用いてnを自然数の和に分割する方法

積はバリエーションが多く、各々でまとめたほうがいいかも

メビウスはかなりよくまとまっている

{a_n}の中からk回まで同じものを用いて、mを自然数の積に分割する方法をδ、その時の自然数の個数をj(δ)、mにおけるδ全体の集合をΔ、このときδの要素をn個まで重複を許して用いて、その積として表せる数のt乗の和を$ \sigma_{n,t}(\delta) $とする。

このとき$ \sigma_{n,t}(\delta)=\prod_{a\in\delta}\frac{a^{nt}-1}{a^t-1} $

さらにp(δ)をcで割った余りが{d_0,d_1,…}であるようなδのみを抽出するという場合もある。

さらに要素となる自然数自体も±判定に使われる。

これより、自然数の和の分け方を表す記号を仮に

$ p_{(s,t,k),(a,b),(\{a,e,o\})}(n) $

とする。

個人的には積のほうも

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

という形で表したい。

このときはとりあえず

$ \begin{align} q_{\{a_n\},f,\mathbb{Y}}(m)&=\sum_{\delta\in\Delta}f(j(\delta))\\ \end{align} $

と表し、詳しい部分は適宜解説を入れるものとする。

積について、素数、合成数、自然数と分けなくてはならない。自然数と素数がないと合成数は作れないが、素体となる積は自然数の側で余りを判別することが難しく、素数の側で分子を1以外にできないので、合成数は分母の冪を変化させるのみにとどまると思われる。

自然数の積はコンパスと定規で作図可能な偶数と3のみで、素数の積は偶数のみ。

これより合成数はコンパスと定規で作図可能な偶数、すなわち2,4,6,8,10,12,16,20までにとどまる。

ここで、

$ \begin{align} \prod_{n=1}^\infty(1+\frac{x^k}{n^k})&=\frac{1}{x^k}\prod_{n=0}^{k-1}\frac{1}{\Gamma(xe^{\frac{2\pi in}{k}})} \\ \prod_{n=1}^\infty(1+\frac{k^3}{n^3})&=\frac{1}{k^3}\frac{1}{\Gamma(k)\Gamma(\frac{-1+\sqrt{3}i}{2}k)\Gamma(\frac{-1-\sqrt{3}i}{2}k)} \\ &=\frac{1}{k^3}\frac{\prod_{t=0}^{k-1}(\frac{-1+\sqrt{3}i}{2}k+t)}{\Gamma(k)\Gamma(\frac{1+\sqrt{3}i}{2}k)\Gamma(\frac{-1-\sqrt{3}i}{2}k)} \\ &=\frac{1}{k^3}\frac{\prod_{t=0}^{k-1}(\frac{-1+\sqrt{3}i}{2}k+t)\frac{1+\sqrt{3}i}{2}k\sin\frac{1+\sqrt{3}i}{2}k\pi}{\Gamma(k)\pi} \\ &=\frac{\prod_{t=0}^{k}(\frac{-1+\sqrt{3}i}{2}k+t)\sin\frac{1+\sqrt{3}i}{2}k\pi}{k!k^2\pi} \\ &=\begin{cases}\frac{\prod_{t=0}^{k}(\frac{-1+\sqrt{3}i}{2}k+t)(-1)^{\frac{k+1}{2}}\cosh\frac{\sqrt{3}}{2}k\pi}{k!k^2\pi}&k\ is\ odd \\ \frac{\prod_{t=0}^{k}(\frac{-1+\sqrt{3}i}{2}k+t)(-1)^{\frac{k}{2}}i\sinh\frac{\sqrt{3}}{2}k\pi}{k!k^2\pi}&k\ is\ even \\ \end{cases} \\ &=\begin{cases}\frac{\prod_{t=0}^{\frac{k-1}{2}}(-(\frac{k}{2}-t)^2-\frac{3}{4}k^2)(-1)^{\frac{k+1}{2}}\cosh\frac{\sqrt{3}}{2}k\pi}{k!k^2\pi}&k\ is\ odd \\ \frac{\frac{\sqrt{3}}{2}ki\prod_{t=0}^{\frac{k}{2}-1}(-(\frac{k}{2}-t)^2-\frac{3}{4}k^2)(-1)^{\frac{k}{2}}i\sinh\frac{\sqrt{3}}{2}k\pi}{k!k^2\pi}&k\ is\ even \\ \end{cases} \\ &=\begin{cases}\frac{\prod_{t=0}^{\frac{k-1}{2}}(k^2-tk+t^2)\cosh\frac{\sqrt{3}}{2}k\pi}{k!k^2\pi}&k\ is\ odd \\ \frac{\frac{\sqrt{3}}{2}k\prod_{t=0}^{\frac{k}{2}-1}(k^2-tk+t^2)\sinh\frac{\sqrt{3}}{2}k\pi}{k!k^2\pi}&k\ is\ even \\ \end{cases} \\ &=\begin{cases}\frac{\prod_{t=1}^{\frac{k-1}{2}}(k^2-tk+t^2)\cosh\frac{\sqrt{3}}{2}k\pi}{k!\pi}&k\ is\ odd \\ \frac{\frac{\sqrt{3}}{2}k\prod_{t=1}^{\frac{k}{2}-1}(k^2-tk+t^2)\sinh\frac{\sqrt{3}}{2}k\pi}{k!\pi}&k\ is\ even \\ \end{cases} \\ \end{align} $

$ \begin{align} \prod_{n=1}^\infty\left(1-\frac{x}{n^{2k}}\right)&=\prod_{n=1}^\infty\prod_{m=0}^{k-1}\left(1-\frac{e^{\frac{2\pi ik}{n}}\sqrt[k]{x}}{n^2}\right) \\ &=\prod_{m=0}^{k-1}\frac{\sin\pi e^{\frac{\pi ik}{n}}\sqrt[2k]{x}}{\pi e^{\frac{\pi ik}{n}}\sqrt[2k]{x}} \\ &=\frac{\sin(\pi \sqrt[2k]{x})}{\pi^k e^{\frac{\pi i(k-1)}{2}}\sqrt{x}}\prod_{m=0}^{k-1}\sin(\pi \sqrt[2k]{x}(\cos\frac{\pi m}{k}+i\sin\frac{\pi m}{k})) \\ &=\frac{\sin(\pi \sqrt[2k]{x})}{\pi^k i^{k-1}\sqrt{x}}\prod_{m=1}^{k-1}\sin(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\cos(\pi \sqrt[2k]{x}i\sin\frac{\pi m}{k}) \\ &+\cos(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\sin(\pi \sqrt[2k]{x}i\sin\frac{\pi m}{k}) \\ &=\frac{\sin(\pi \sqrt[2k]{x})}{\pi^k i^{k-1}\sqrt{x}}\prod_{m=1}^{k-1}\sin(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\cosh(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k}) \\ &+i\cos(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\sinh(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k}) \\ &=\begin{cases} \frac{\sin(\pi \sqrt[2k]{x})}{\pi^k i^{k-1}\sqrt{x}}\prod_{m=1}^{\frac{k-1}{2}}&\\ \left(\sin(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\cosh(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right.& \\ \left.+i\cos(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\sinh(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right)& \\ \left(\sin(\pi \sqrt[2k]{x}\cos\frac{\pi (k-m)}{k})\cosh(\pi \sqrt[2k]{x}\sin\frac{\pi (k-m)}{k})\right.& \\ \left.+i\cos(\pi \sqrt[2k]{x}\cos\frac{\pi (k-m)}{k})\sinh(\pi \sqrt[2k]{x}\sin\frac{\pi (k-m)}{k})\right)&k\ is\ odd \\ \frac{\sin(\pi \sqrt[2k]{x})\sin(\pi\sqrt[2k]{x}i)}{\pi^k i^{k-1}\sqrt{x}}\prod_{m=1}^{\frac{k}{2}-1}&\\ \left(\sin(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\cosh(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right.& \\ \left.+i\cos(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\sinh(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right)&\\ \left(\sin(\pi \sqrt[2k]{x}\cos\frac{\pi (k-m)}{k})\cosh(\pi \sqrt[2k]{x}\sin\frac{\pi (k-m)}{k})\right.& \\ \left.+i\cos(\pi \sqrt[2k]{x}\cos\frac{\pi (k-m)}{k})\sinh(\pi \sqrt[2k]{x}\sin\frac{\pi (k-m)}{k})\right)&k\ is\ even \\ \end{cases} \\ &=\begin{cases} \frac{\sin(\pi \sqrt[2k]{x})}{\pi^k i^{k-1}\sqrt{x}}\prod_{m=1}^{\frac{k-1}{2}}&\\ \left(\sin(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\cosh(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right.& \\ \left.+i\cos(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\sinh(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right)& \\ \left(\sin(-\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\cosh(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right.& \\ \left.+i\cos(-\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\sinh(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right)&k\ is\ odd \\ \frac{\sin(\pi \sqrt[2k]{x})i\sinh(\pi\sqrt[2k]{x})}{\pi^k i^{k-1}\sqrt{x}}\prod_{m=1}^{\frac{k}{2}-1}&\\ \left(\sin(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\cosh(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right.& \\ \left.+i\cos(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\sinh(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right)&\\ \left(\sin(-\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\cosh(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right.& \\ \left.+i\cos(-\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\sinh(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right)&k\ is\ even \\ \end{cases} \\ &=\begin{cases} \frac{\sin(\pi \sqrt[2k]{x})}{\pi^k i^{k-1}\sqrt{x}}\prod_{m=1}^{\frac{k-1}{2}}&\\ \left(\sin(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\cosh(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right.& \\ \left.+i\cos(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\sinh(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right)& \\ \left(-\sin(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\cosh(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right.& \\ \left.+i\cos(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\sinh(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right)&k\ is\ odd \\ \frac{\sin(\pi \sqrt[2k]{x})\sinh(\pi\sqrt[2k]{x})}{\pi^k i^{k-2}\sqrt{x}}\prod_{m=1}^{\frac{k}{2}-1}&\\ \left(\sin(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\cosh(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right.& \\ \left.+i\cos(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\sinh(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right)&\\ \left(-\sin(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\cosh(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right.& \\ \left.+i\cos(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\sinh(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right)&k\ is\ even \\ \end{cases} \\ &=\begin{cases} \frac{\sin(\pi \sqrt[2k]{x})}{\pi^k i^{k-1}\sqrt{x}}\prod_{m=1}^{\frac{k-1}{2}}&\\ -\left(\sin^2(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\cosh^2(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right.& \\ \left.+\cos^2(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\sinh^2(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right)&k\ is\ odd \\ \frac{\sin(\pi \sqrt[2k]{x})\sinh(\pi\sqrt[2k]{x})}{\pi^k i^{k-2}\sqrt{x}}\prod_{m=1}^{\frac{k}{2}-1}&\\ -\left(\sin^2(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\cosh^2(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right.& \\ \left.+i\cos^2(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\sinh^2(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right)&k\ is\ even \\ \end{cases} \\ &=\begin{cases} \frac{\sin(\pi \sqrt[2k]{x})}{\pi^k \sqrt{x}}\prod_{m=1}^{\frac{k-1}{2}}&\\ \left(\sin^2(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\cosh^2(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right.& \\ \left.+\cos^2(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\sinh^2(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right)&k\ is\ odd \\ \frac{\sin(\pi \sqrt[2k]{x})\sinh(\pi\sqrt[2k]{x})}{\pi^k \sqrt{x}}\prod_{m=1}^{\frac{k}{2}-1}&\\ \left(\sin^2(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\cosh^2(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right.& \\ \left.+\cos^2(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\sinh^2(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right)&k\ is\ even \\ \end{cases} \\ &=\begin{cases} \frac{\sin(\pi \sqrt[2k]{x})}{\pi^k \sqrt{x}}\prod_{m=1}^{\frac{k-1}{2}}&\\ \left(\sin^2(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})(\sinh^2(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})+1)\right.& \\ \left.+(1-\sin^2(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k}))\sinh^2(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right)&k\ is\ odd \\ \frac{\sin(\pi \sqrt[2k]{x})\sinh(\pi\sqrt[2k]{x})}{\pi^k \sqrt{x}}\prod_{m=1}^{\frac{k}{2}-1}&\\ \left(\sin^2(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})(\sinh^2(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})+1)\right.& \\ \left.+(1-\sin^2(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k}))\sinh^2(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right)&k\ is\ even \\ \end{cases} \\ &=\begin{cases} \frac{\sin(\pi \sqrt[2k]{x})}{\pi^k \sqrt{x}}\prod_{m=1}^{\frac{k-1}{2}}\left(\sin^2(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\right.\left.+\sinh^2(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right)&k\ is\ odd \\ \frac{\sin(\pi \sqrt[2k]{x})\sinh(\pi\sqrt[2k]{x})}{\pi^k \sqrt{x}}& \\ \prod_{m=1}^{\frac{k}{2}-1}\left(\sin^2(\pi \sqrt[2k]{x}\cos\frac{\pi m}{k})\right.\left.+\sinh^2(\pi \sqrt[2k]{x}\sin\frac{\pi m}{k})\right)&k\ is\ even \\ \end{cases} \\ \end{align} $

$ \begin{align} \prod_{n=m+1}^\infty\left(1-\frac{m^{2k}}{n^{2k}}\right)&=\prod_{n=m+1}^\infty\prod_{t=0}^{k-1}\left(1-\frac{e^{\frac{2\pi it}{k}}m^2}{n^2}\right) \\ &=\frac{\displaystyle \prod_{n=m+1}^\infty\left(1-\frac{m^2}{n^2}\right)\prod_{n=1}^\infty\prod_{t=1}^{k-1}\left(1-\frac{e^{\frac{2\pi it}{k}}m^2}{n^2}\right)}{\displaystyle \prod_{n=1}^m\left(\sum_{t=0}^{k-1}\frac{m^{2t}}{n^{2t}}\right)} \\ &=\frac{\displaystyle \prod_{n=1,n\neq m}^\infty\left(1-\frac{m^2}{n^2}\right)\prod_{n=1}^\infty\prod_{t=1}^{k-1}\left(1-\frac{e^{\frac{2\pi it}{k}}m^2}{n^2}\right)}{\displaystyle k\prod_{n=1}^{m-1}\left(1-\frac{m^{2k}}{n^{2k}}\right)} \\ &=\frac{\displaystyle (-1)^{m-1}\prod_{n=1}^\infty\prod_{t=1}^{k-1}\left(1-\frac{e^{\frac{2\pi it}{k}}m^2}{n^2}\right)}{\displaystyle 2k\prod_{n=1}^{m-1}\left(1-\frac{m^{2k}}{n^{2k}}\right)} \\ &=\frac{\displaystyle (-1)^{m-1}\prod_{t=1}^{k-1}\frac{\sin\pi e^{\frac{\pi ik}{n}}m}{\pi e^{\frac{\pi it}{k}}m}}{\displaystyle 2k\prod_{n=1}^{m-1}\left(1-\frac{m^{2k}}{n^{2k}}\right)} \\ &=\frac{\displaystyle (-1)^{m-1}\prod_{t=1}^{k-1}\sin\pi e^{\frac{\pi it}{k}}m}{\displaystyle 2km^{k-1}(\pi i)^{k-1}\prod_{n=1}^{m-1}\left(1-\frac{m^{2k}}{n^{2k}}\right)} \\ &=\begin{cases} \frac{1}{2k(m\pi)^{k-1}\prod_{n=1}^{m-1}\left(\frac{m^{2k}}{n^{2k}}-1\right)}\prod_{t=1}^{\frac{k-1}{2}}&\\ \left(\sin^2(\pi m\cos\frac{\pi t}{k})+\sinh^2(\pi m\sin\frac{\pi t}{k})\right)&k\ is\ odd \\ \frac{\sinh(\pi m)}{2k(m\pi)^{k-1}\prod_{n=1}^{m-1}\left(\frac{m^{2k}}{n^{2k}}-1\right)}\prod_{t=1}^{\frac{k}{2}-1}&\\ \left(\sin^2(\pi m\cos\frac{\pi t}{k})+\sinh^2(\pi m\sin\frac{\pi t}{k})\right)&k\ is\ even \\ \end{cases} \\ \end{align} $

自然数nをk個の整数の平方の和で表す方法を$ r_k(n) $とすると、 $ \sum_{n=0}^\infty r_k(n)x^n=\vartheta_3^k(x) $

また、$ \phi^3(q)=\sum_{n=0}^\infty (-1)^n(2n+1)q^{\frac{n(n+1)}{2}} $

$ \vartheta_3(0,e^{\pi i \tau})=\frac{\eta^2(\frac{1}{2}(\tau+1))}{\eta(\tau+1)} $

$ \vartheta_3(q)=\frac{\phi(q^2)^5}{\phi(q)^2\phi(q^4)^2} $

$ \vartheta_4(q)=\frac{\phi(q)^2}{\phi(q^2)} $

$ \vartheta_2(q)=q^{\frac{1}{4}}\frac{\phi(q^4)^2}{\phi(q^2)^2} $

$ \vartheta_3(e^{-\pi\sqrt{2}})=\frac{\Gamma(\frac{1}{8})}{2\Gamma(\frac{1}{4})}\sqrt{\displaystyle \frac{\Gamma(\frac{1}{4})}{2^{\frac{1}{4}}\pi}} $

$ \vartheta_3(e^{-\pi\sqrt{6}})=\left[\frac{\Gamma(\frac{1}{24})\Gamma(\frac{5}{24})\Gamma(\frac{7}{24})\Gamma(\frac{11}{24})}{16\sqrt{6}(18+12\sqrt{2}-10\sqrt{3}-7\sqrt{6})\pi^3}\right]^{\frac{1}{4}} $

$ \vartheta_4'(e^{-\pi})=\frac{e^\pi[2\Gamma^4(\frac{3}{4})-\pi^2]}{2^{\frac{13}{4}}\pi^{\frac{3}{4}}}\Gamma^5(\frac{3}{4}) $

$ \vartheta_3^2(x)=1+4\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{1-x^{2n+1}} $

$ \vartheta_3^4(x)=1+8\sum_{n=1}^\infty \frac{nx^{2n-1}}{1+(-1)^nx^{2n+1}} $

$ \frac{\vartheta_2(-e^{-\pi\sqrt{3}})}{\vartheta_3(-e^{-\pi\sqrt{3}})}=\sqrt[4]{4\sqrt{3}-7} $

$ \vartheta_3(z,q)=\vartheta_3(2z,q^4)+\vartheta_2(2z,q^4) $

$ \vartheta_4(z,q)=\vartheta_3(2z,q^4)-\vartheta_2(2z,q^4) $

$ \vartheta_4^4=\vartheta_3^4-\vartheta_2^4 $

$ \prod_{k=1}^\infty\left(\frac{1-n^{-2k}}{1+n^{-2k}}\right)^2=\frac{\vartheta_1'(n^{-1})}{\vartheta_2(n^{-1})} $

$ \prod_{k=1}^\infty\left[1+(-1)^{k-1}\frac{b}{k+a}\right]=\frac{\sqrt{\pi}\Gamma(a+1)}{2^a\Gamma(\frac{1}{2}(2+b-a))\Gamma(\frac{1}{2}(1+b+a))} $

$ \vartheta_3'(q)=\sqrt{\displaystyle \frac{\vartheta_2(q)}{\vartheta_3(q)}} $

$ \sum_{n=1}^\infty \frac{\lambda(n)q^n}{1-q^n} = \sum_{n=1}^\infty q^{n^2} = \frac{1}{2}\left(\vartheta_3(q)-1\right), $

$ \varphi(e^{-\pi x}) = \vartheta(0; {\mathrm{i}}x) = \theta_3(0;e^{-\pi x}) = \sum_{n=-\infty}^\infty e^{-x \pi n^2} $

$ \varphi\left(e^{-\pi} \right) = \frac{\sqrt[4]{\pi}}{\Gamma(\frac{3}{4})} $

$ \varphi\left(e^{-2\pi} \right) = \frac{\sqrt[4]{6\pi+4\sqrt2\pi}}{2\Gamma(\frac{3}{4})} $

$ \varphi\left(e^{-3\pi}\right) = \frac{\sqrt[4]{27\pi+18\sqrt3\pi}}{3\Gamma(\frac{3}{4})} $

$ \varphi\left(e^{-4\pi}\right) =\frac{\sqrt[4]{8\pi}+2\sqrt[4]{\pi}}{4\Gamma(\frac{3}{4})} $

$ \varphi\left(e^{-5\pi} \right) =\frac{\sqrt[4]{225\pi+ 100\sqrt5 \pi}}{5\Gamma(\frac{3}{4})} $

$ \varphi\left(e^{-6\pi}\right) = \frac{\sqrt[3]{3\sqrt{2}+3\sqrt[4]{3}+2\sqrt{3}-\sqrt[4]{27}+\sqrt[4]{1728}-4}\cdot \sqrt[8]{243{\pi}^2}}{6\sqrt[6]{1+\sqrt6-\sqrt2-\sqrt3}{\Gamma(\frac{3}{4})}} $

$ \phi(q)^2=\frac{1}{2}\sum_{m,n=-\infty}^\infty((-1)^n-(-1)^m)q^{\frac{3m^2+3n^2+4n+1}{4}} $

$ \phi(q)^4=\frac{1}{2}\sum_{m,n=-\infty}^\infty((-1)^{m+n})(2n+1)q^{\frac{3m^2+n^2+n-m}{2}} $

$ \phi(q)^6=\frac{1}{4}\sum_{m,n=-\infty}^\infty((-1)^n)(n^2-m^2)q^{\frac{m^2+n^2-1}{4}} $

$ \phi(q)^8=-\frac{1}{4}\sum_{m,n=-\infty}^\infty m^2(3n+2)(1+(-1)^{m+n})q^{\frac{m^2+3n^2+4n}{4}} $

$ \phi(q)^{10}=\frac{1}{2}\sum_{m,n=-\infty}^\infty(3m+2)(3n+2)^3((-1)^n-(-1)^m)q^{\frac{3m^2+3n^2+4m+4n+1}{4}} $


$ \sum_{n=1}^\infty \frac{n^{-11}}{e^{2n\pi}-1}=-\frac{1}{2}\zeta (11)+\frac{1453\pi}{851350500} $

$ \sum_{n=1}^\infty \frac{n^{-7}}{e^{2n\pi}-1}=-\frac{1}{2}\zeta (7)+\frac{19\pi}{113400} $

$ \sum_{n=1}^\infty \frac{n^{-3}}{e^{2n\pi}-1}=-\frac{1}{2}\zeta (3)+\frac{7\pi}{360} $

$ \sum_{n=1}^\infty \frac{n^{-1}}{e^{2n\pi}-1}=-\frac{\pi}{12}-\frac{1}{2} \log \frac{\omega}{\pi \sqrt{2}} $

$ \sum_{n=1}^\infty \frac{n}{e^{2n\pi}-1}=\frac{1}{24}-\frac{1}{8\pi} $

$ \sum_{n=1}^\infty \frac{n^{3}}{e^{2n\pi}-1}=-\frac{1}{240}+\frac{1}{80}(\frac{\omega}{\pi})^4 $

$ \sum_{n=1}^\infty \frac{n^{5}}{e^{2n\pi}-1}=\frac{1}{504} $

$ \sum_{n=1}^\infty \frac{n^{7}}{e^{2n\pi}-1}=-\frac{1}{480}+\frac{3}{160}(\frac{\omega}{\pi})^8 $

$ \sum_{n=1}^\infty \frac{n^{9}}{e^{2n\pi}-1}=\frac{1}{264} $

$ \sum_{n=1}^\infty \frac{n^{11}}{e^{2n\pi}-1}=-\frac{691}{65520}+\frac{89}{1040}(\frac{\omega}{\pi})^{12} $

$ \sum_{n=1}^\infty \frac{n^{13}}{e^{2n\pi}-1}=\frac{1}{24} $

$ \sum_{n=1}^\infty \frac{n^{15}}{e^{2n\pi}-1}=-\frac{3617}{16320}+\frac{43659}{5440}(\frac{\omega}{\pi})^{16} $

$ \sum_{n=1}^\infty \frac{n^{17}}{e^{2n\pi}-1}=\frac{43867}{28728} $

$ \sum_{n=1}^\infty \frac{n^{4k+1}}{e^{2n\pi}-1}=\frac{B_{4k+2}}{8k+4} (k \geq 1) $

$ \sum_{n=1}^\infty \frac{n^{4k-1}}{e^{2n\pi}-1}=\frac{1}{8k}(B_{4k}-(\frac{\omega}{\pi})^{4k} H_{4k}) (k \geq 1) $

$ \sum_{n=1}^\infty \frac{n^{-(4k-1)}}{e^{2n\pi}-1} $

$ =-\frac{1}{2}\zeta (4k-1)+\frac{(2\pi)^{4k-1}}{4(4k+4)!} (\binom{4k}{2k}(-1)^k B_{2k}^2 -2\sum_{m=0}^{k-1}\binom{4k}{2m}(-1)^m B_{2m} B_{4k-2m}) (k \geq 1) $

$ \pi = 72 \sum_{n=1}^\infty \frac{1}{n(e^{n\pi}-1)} -96 \sum_{n=1}^\infty \frac{1}{n(e^{2n\pi}-1)} + 24\sum_{n=1}^\infty \frac{1}{n(e^{4n\pi}-1)} $

$ \vartheta_1'=\pi\vartheta_2\vartheta_3\vartheta_4 $

$ \begin{align} \vartheta_1(v;\tau) &=-\sum_{n=-\infty}^{\infty}{e^{{\pi}i{\tau}\left(n+1/2\right)^2}e^{2{\pi}i(n+1/2)(v+1/2)}}\\ &=-ie^{{\pi}i{\tau}/4}e^{{\pi}iv}\sum_{n=-\infty}^{\infty}{e^{{\pi}i{\tau}n^2}e^{{\pi}i{\tau}n+2{\pi}ivn+{\pi}in}}\\ &=-ie^{{\pi}i{\tau}/4}e^{{\pi}iv}\prod_{m=1}^{\infty}{\left(1-e^{2m{\pi}i{\tau}}\right)\left(1-e^{2m{\pi}i{\tau}}e^{2{\pi}iv}\right)\left(1-e^{(2m-2){\pi}i{\tau}}e^{-2{\pi}iv}\right)}\\ &=-ie^{{\pi}i{\tau}/4}e^{{\pi}iv}(1-e^{-2{\pi}iv})\prod_{m=1}^{\infty}{\left(1-e^{2m{\pi}i{\tau}}\right)\left(1-e^{2m{\pi}i{\tau}}e^{2{\pi}iv}\right)\left(1-e^{2m{\pi}i{\tau}}e^{-2{\pi}iv}\right)}\\ &=2e^{{\pi}i{\tau}/4}\sin{\pi}v\prod_{m=1}^{\infty}{\left(1-e^{2m{\pi}i{\tau}}\right)\left(1-e^{2m{\pi}i{\tau}}e^{2{\pi}iv}\right)\left(1-e^{2m{\pi}i{\tau}}e^{-2{\pi}iv}\right)}\\ &=2e^{{\pi}i{\tau}/4}\sin{\pi}v\prod_{m=1}^{\infty}{\left(1-e^{2m{\pi}i{\tau}}\right)\left(1-2e^{2m{\pi}i{\tau}}\cos{2{\pi}v}+e^{4m{\pi}i{\tau}}\right)}\\ \end{align} $

$ \begin{align} \vartheta_2(v;\tau) &=2e^{{\pi}i{\tau}/4}\cos{\pi}v\prod_{m=1}^{\infty}{\left(1-e^{2m{\pi}i{\tau}}\right)\left(1+e^{2m{\pi}i{\tau}}e^{2{\pi}iv}\right)\left(1+e^{2m{\pi}i{\tau}}e^{-2{\pi}iv}\right)}\\ &=2e^{{\pi}i{\tau}/4}\cos{\pi}v\prod_{m=1}^{\infty}{\left(1-e^{2m{\pi}i{\tau}}\right)\left(1+2e^{2m{\pi}i{\tau}}\cos{2{\pi}v}+e^{4m{\pi}i{\tau}}\right)}\\ \end{align} $

$ \begin{align} \vartheta_3(v;\tau) &=\prod_{m=1}^{\infty}{\left(1-e^{2m{\pi}i{\tau}}\right)\left(1+e^{(2m-1){\pi}i{\tau}}e^{2{\pi}iv}\right)\left(1+e^{(2m-1){\pi}i{\tau}}e^{-2{\pi}iv}\right)}\\ &=\prod_{m=1}^{\infty}{\left(1-e^{2m{\pi}i{\tau}}\right)\left(1+2e^{(2m-1){\pi}i{\tau}}\cos{2{\pi}v}+e^{2(2m-1){\pi}i{\tau}}\right)}\\ \end{align} $

$ \begin{align} \vartheta_4(v;\tau) &=\prod_{m=1}^{\infty}{\left(1-e^{2m{\pi}i{\tau}}\right)\left(1-e^{(2m-1){\pi}i{\tau}}e^{2{\pi}iv}\right)\left(1-e^{(2m-1){\pi}i{\tau}}e^{-2{\pi}iv}\right)}\\ &=\prod_{m=1}^{\infty}{\left(1-e^{2m{\pi}i{\tau}}\right)\left(1-2e^{(2m-1){\pi}i{\tau}}\cos{2{\pi}v}+e^{2(2m-1){\pi}i{\tau}}\right)}\\ \end{align} $

$ i\vartheta_1\left(-\frac{i\pi}{4},e^{-2\pi}\right)=2^{\frac{5}{8}}\frac{e^{\frac{\pi}{32}}}{\pi^{3}{4}}\Gamma\left(\frac{5}{4}\right)\sqrt[16]{-68+323\sqrt{2}-56\sqrt{68-14\sqrt{2}}} $

$ i\vartheta_1\left(-\frac{i\pi}{4},e^{-4\pi}\right)=\frac{e^{\frac{\pi}{16}}}{\pi^{3}{4}}\Gamma\left(\frac{5}{4}\right)\sqrt[4]{2+6\sqrt{2}-4\sqrt{4+2\sqrt{2}}} $

$ \frac{\vartheta(e^{-\frac{\pi}{4}})}{\vartheta(e^{-\frac{\pi}{2}})}=\sqrt[8]{8+6\sqrt{2}} $

$ \frac{\vartheta(e^{-\frac{\pi}{3}})}{\vartheta(e^{-\frac{2\pi}{3}})}=\frac{(594100+420099\sqrt{2})^{\frac{1}{12}}(1+\sqrt{3}+\sqrt[4]{108})^{\frac{4}{3}}(-2-\sqrt{3}+\sqrt{9+6\sqrt{3}})}{2^{\frac{13}{24}}(-6+5\sqrt{2}+6\sqrt[4]{3}+9\sqrt[4]{18}-4\sqrt[4]{27}+\sqrt[4]{108}+7\sqrt{6})^{\frac{2}{3}}} $

$ \phi(e^{-\frac{\pi}{2}})=\frac{\sqrt[3]{\sqrt{2}-1}\sqrt[24]{4+3\sqrt{2}}e^{-\frac{\pi}{48}}\Gamma(\frac{1}{4})}{2^{\frac{5}{6}}\pi^{\frac{3}{4}}} $

$ E_4(\tau)=1+240\sum_{n=1}^\infty \frac {n^3q^n}{1-q^n} $
$ E_6(\tau)=1-504\sum_{n=1}^\infty \frac {n^5q^n}{1-q^n} $
$ E_8(\tau)=1+480\sum_{n=1}^\infty \frac {n^5q^n}{1-q^n} $
$ a=\theta_{2}(0; e^{\pi i\tau})=\vartheta_{10}(0; \tau) $
$ b=\theta_{3}(0; e^{\pi i\tau})=\vartheta_{00}(0; \tau) $
$ c=\theta_{4}(0; e^{\pi i\tau})=\vartheta_{01}(0; \tau) $
$ E_4(\tau)=\tfrac{1}{2}(a^8+b^8+c^8) $
$ E_6(\tau)=\tfrac{1}{2}\sqrt{\frac{(a^8+b^8+c^8)^3-54(abc)^8}{2}} $
$ E_8(\tau)=\tfrac{1}{2}(a^{16}+b^{16}+c^{16}) $


$ (1+240\sum_{n=1}^\infty \sigma_3(n) q^n)^2 = 1+480\sum_{n=1}^\infty \sigma_7(n) q^n, $

$ \begin{align} E_{8} &= E_4^2 \\ E_{10} &= E_4\cdot E_6 \\ 691 \cdot E_{12} &= 441\cdot E_4^3+ 250\cdot E_6^2 \\ E_{14} &= E_4^2\cdot E_6 \\ 3617\cdot E_{16} &= 1617\cdot E_4^4+ 2000\cdot E_4 \cdot E_6^2 \\ 43867 \cdot E_{18} &= 38367\cdot E_4^3\cdot E_6+5500\cdot E_6^3 \\ 174611 \cdot E_{20} &= 53361\cdot E_4^5+ 121250\cdot E_4^2\cdot E_6^2 \\ 77683 \cdot E_{22} &= 57183\cdot E_4^4\cdot E_6+20500\cdot E_4\cdot E_6^3 \\ 236364091 \cdot E_{24} &= 49679091\cdot E_4^6+ 176400000\cdot E_4^3\cdot E_6^2 + 10285000\cdot E_6^4 \end{align}. $

$ L(q)=1-24\sum_{n=1}^\infty \frac {nq^n}{1-q^n}=E_2(\tau) $
$ M(q)=1+240\sum_{n=1}^\infty \frac {n^3q^n}{1-q^n}=E_4(\tau) $
$ N(q)=1-504\sum_{n=1}^\infty \frac {n^5q^n}{1-q^n}=E_6(\tau), $
$ q\frac{dL}{dq} = \frac {L^2-M}{12} $
$ q\frac{dM}{dq} = \frac {LM-N}{3} $
$ q\frac{dN}{dq} = \frac {LN-M^2}{2} $
$ \sum_{k=0}^n\sigma(k)\sigma(n-k)=\frac5{12}\sigma_3(n)-\frac12n\sigma(n) $
$ \sum_{k=0}^n\sigma_3(k)\sigma_3(n-k)=\frac1{120}\sigma_7(n) $
$ \sum_{k=0}^n\sigma(2k+1)\sigma_3(n-k)=\frac1{240}\sigma_5(2n+1) $
$ \sum_{k=0}^n\sigma(3k+1)\sigma(3n-3k+1)=\frac19\sigma_3(3n+2). $

$ \sigma_{m,n}=\sum_{m=1}^\infty\sum_{n=1}^\infty m^rn^sx^{mn} $

$ \prod_{n=1}^\infty \left\{1+\left(\frac{\alpha+\beta}{\alpha+n}\right)^3\right\}\left\{1+\left(\frac{\alpha+\beta}{\beta+n}\right)^3\right\}=\frac{\Gamma(1+\alpha)\Gamma(1+\beta)}{\Gamma(1+\alpha+2\beta)\Gamma(1+\beta+2\alpha)}\left\{\frac{\cosh\pi\sqrt{3}(\alpha+\beta)-\cos\pi(\alpha-\beta)}{2\pi^2(\alpha^2+\alpha\beta+\beta^2)}\right\} $

$ \prod_{n=1}^\infty \left\{1+\left(\frac{\alpha}{n}\right)^3\right\}\left\{1+3\left(\frac{\alpha}{\alpha+2n}\right)^3\right\}=\frac{\Gamma(\frac{1}{2}\alpha)}{\Gamma(\frac{1}{2}(1+\alpha))}\frac{\cosh\pi\sqrt{3}\alpha-\cos\pi\alpha}{2^{\alpha+2}\pi^{\frac{3}{2}}\alpha} $

$ \sin_q(\pi z)=q^{(\frac{z-1}{2})^2}\frac{(q^{2z};q^2)_\infty(q^{2-2z};q^2)_\infty}{(q;q^2)_\infty^2},\cos_q(z)=q^{z^2}\frac{(q^{2z+1};q^2)_\infty(q^{1-2z};q^2)_\infty}{(q;q^2)_\infty^2},\Gamma_q(z)=\frac{\phi(q)}{(q^z;q)_\infty}(1-q)^{1-z} $

$ Q(n)=\frac{\phi(q^n)}{q^{\frac{n}{24}}\phi(q^2n)},(Q(1)Q(3))^3+\frac{8}{(Q(1)Q(3))^3}=(\frac{Q(3)}{Q(1)})^6-(\frac{Q(1)}{Q(3)})^6 $

$ \frac{1}{2}[\vartheta_2(q^9)-\vartheta_2(q)]=-q^{\frac{1}{4}}\frac{\phi(q^{12})\phi(q^{18})^2)}{\phi(q^6)\phi(q^{36})} $

$ \frac{1}{2}[\vartheta_3(q^9)-\vartheta_3(q)]=-q\frac{\phi(q^6)^2\phi(q^9)\phi(q^{36})}{\phi(q^3)\phi(q^{12})\phi(q^{18})} $

$ \frac{1}{2}[\vartheta_4(q^9)-\vartheta_4(q)]=q\frac{\phi(q^3)\phi(q^{18})^2}{\phi(q^6)\phi(q^9)} $

$ \frac{1}{2}[3\vartheta_2(q^9)-\vartheta_2(q)]=-q^{\frac{1}{4}}\frac{\phi(q^2)^2\phi(q^{12})}{\phi(q^4)\phi(q^6)} $

$ \frac{1}{2}[3\vartheta_3(q^9)-\vartheta_3(q)]=\frac{\phi(q)\phi(q^4)\phi(q^6)^2}{\phi(q^2)\phi(q^3)\phi(q^{12})} $

$ \frac{1}{2}[3\vartheta_4(q^9)-\vartheta_4(q)]=\frac{\phi(q^2)^2\phi(q^3)}{\phi(q)\phi(q^6)} $

$ E_{2k}(\tau)=\frac{G_{2k}(\tau)}{2\zeta (2k)}= 1+\frac {2}{\zeta(1-2k)}\sum_{n=1}^{\infty} \frac{n^{2k-1} q^n}{1-q^n} = 1 - \frac{4k}{B_{2k}} \sum_{d,n \geq 1} n^{2k-1} q^{n d} $

dk =(2k+3)k!G2k+4 とすると、全ての n ≥ 0 に対し、dk は関係式

$ \sum_{k=0}^n {n \choose k} d_k d_{n-k} = \frac{2n+9}{3n+6}d_{n+2} $

$ \sigma_{r+1,s+1}(x)=x\frac{d}{dx}\sigma_{r,s}(x) $

$ 288\sigma_{1,2}(x)=Q-P^2 $

$ 720\sigma_{1,4}(x)=PQ-R $

$ 1008\sigma_{1,6}(x)=Q^2-PR $

$ 720\sigma_{1,8}(x)=Q(PQ-R) $

$ 1584\sigma_{1,10}(x)=3Q^3+2R^2-5PQR $

$ 65520\sigma_{1,12}(x)=P(441Q^3+250R^2)-691Q^2R $

$ 144\sigma_{1,14}(x)=Q(3Q^3+4R^2-7PQR) $

$ \sigma_{2,3}(x)=3PQ-2R-P^3 $

$ \sigma_{2,5}(x)=P^2Q-2PR+Q^2 $

$ \sigma_{2,7}(x)=2PQ^2-P^2R-QR $

$ \sigma_{2,9}(x)=9P^2Q^2-18PQR+5Q^3+4R^2 $

$ \sigma_{2,11}(x)=6PQ^3-5P^2QR+4PR^2-5Q^2R $

$ \sigma_{3,4}(x)=6P^2Q-8PR+3Q^2-P^4 $

$ \sigma_{3,6}(x)=6P^2Q^2-2P^3R-6PQR+Q^3+R^2 $

$ \sigma_{4,5}(x)=15PQ^2-20P^2R+10P^2Q-4QR-P^5 $

$ \sigma_{4,7}(x)=7(P^4Q-4P^3R+6P^2Q^2-4PQR)+3Q^3+4R^2 $

$ {_1\psi_1}\left[\begin{matrix}a\\b\end{matrix};q,z\right]=\sum_{n=-\infty}^{\infty}\frac{(a;q)_n}{(b;q)_n}z^{n}=\frac{(az;q)_\infty(q;q)_\infty\left(\frac{q}{az};q\right)_\infty\left(\frac{b}{a};q\right)_\infty}{(z;q)_\infty(b;q)_\infty\left(\frac{b}{az};q\right)_\infty\left(\frac{q}{a};q\right)_\infty}\qquad(|q|<1,|b/a|<|z|<1) $

x = λ(1−λ) と置くと

$ j(\tau) = \frac{256(1-x)^3}{x^2} $

を得る。

$ \lambda(\tau) = \frac{\theta_2^4(0,\tau)}{\theta_3^4(0,\tau)} = k^2(\tau) $

$ q=e^{\pi i \tau} $ と定義し直すと、 \vartheta(0; \tau) = \vartheta_{00}(0; \tau) = 1 + 2 \sum_{n=1}^\infty \left(e^{\pi i\tau}\right)^{n^2} = \sum_{n=-\infty}^\infty q^{n^2}</math>

$ a=\theta_{2}(0; q)=\vartheta_{10}(0; \tau) $
$ b=\theta_{3}(0; q)=\vartheta_{00}(0; \tau) $
$ c=\theta_{4}(0; q)=\vartheta_{01}(0; \tau) $

ここに $ \theta_{m} $$ \vartheta_{n} $ は記法を変えたものとした。すると、ヴァイエルシュトラス定数 g2, g3 とη(τ) に対して、

$ g_2(\tau) = \tfrac{2}{3}\pi^4 \left(a^8 + b^8 + c^8\right) $
$ g_3(\tau) = \tfrac{4}{27}\pi^6 \sqrt{\frac{(a^8+b^8+c^8)^3-54(abc)^8}{2}} $
$ \Delta = g_2^3-27g_3^2 = (2\pi)^{12}\eta(\tau)^{24} = (2\pi)^{12} \left(\tfrac{1}{2}a b c\right)^8 $

となる。このようにすると、j (τ) を早く計算できる形に書き換えることができる。

$ j(\tau) = 1728\frac{g_2^3}{g_2^3-27g_3^2} = 32 {(a^8 + b^8 + c^8)^3 \over (a b c)^8}. $

j-不変量は、基本領域の「角」

$ \tfrac{1}{2}\left(1 + i \sqrt{3}\right) $

では 0 となる。

以下に、いくつかの特殊値を示す(以下の最初の 4つはよく知られている。j は J/1728 である)。

$ \begin{align} j(i) &= j \left( \tfrac{1 + i}{2} \right) = 1 \\ j\left(\sqrt{2}i\right) &= \big(\tfrac{5}{3}\big)^3 \\ j(2i) &= \big(\tfrac{11}{2}\big)^3 \\ j\left(2\sqrt{2}i\right) &= \tfrac{125}{216} \left(19 + 13\sqrt{2} \right)^3\\ j(4i) &= \tfrac{1}{64} \left(724 + 513\sqrt{2} \right)^3\\ j\left( \tfrac{1 + 2i}{2} \right) &= \tfrac{1}{64} \left(724 - 513\sqrt{2} \right)^3\\ j\left( \tfrac{1 + 2\sqrt{2}i}{3} \right) &= \tfrac{125}{216} \left(19 - 13\sqrt{2} \right)^3\\ j(3i) &= \tfrac{1}{27} \left(2 + \sqrt{3}\right)^2 \left(21 + 20\sqrt{3}\right )^3 \\ j\left(2\sqrt{3}i\right) &= \tfrac{125}{16} \left(30 + 17\sqrt{3}\right)^3\\ j\left( \tfrac{1 + 7\sqrt{3}i}{2} \right) &= -\tfrac{64000}{7} \left(651 + 142\sqrt{21} \right)^3\\ j\left(\tfrac{1 + 3\sqrt{11}i}{10} \right) &= \tfrac{64}{27} \left(23 - 4\sqrt{33}\right)^2 \left(-77 + 15\sqrt{33} \right)^3\\ j\left(\sqrt{21}i\right) &= \tfrac{1}{32} \left(5 + 3\sqrt{3}\right)^2 \left(3 + \sqrt{7} \right)^2 \left( 65 + 34\sqrt{3} + 26\sqrt{7} + 15\sqrt{21}\right)^3\\ j\left( \tfrac{\sqrt{30}i}{1} \right) &= \tfrac{1}{16} \left(10 + 7\sqrt{2} + 4\sqrt{5} + 3\sqrt{10} \right)^4 \left( 55 + 30\sqrt{2} + 12\sqrt{5} + 10\sqrt{10} \right)^3\\ j\left( \tfrac{\sqrt{30}i}{2} \right) &= \tfrac{1}{16} \left(10 + 7\sqrt{2} - 4\sqrt{5} - 3\sqrt{10} \right)^4 \left( 55 + 30\sqrt{2} - 12\sqrt{5} - 10\sqrt{10} \right)^3\\ j\left( \tfrac{\sqrt{30}i}{5} \right) &= \tfrac{1}{16} \left(10 - 7\sqrt{2} + 4\sqrt{5} - 3\sqrt{10} \right)^4 \left( 55 - 30\sqrt{2} + 12\sqrt{5} - 10\sqrt{10} \right)^3\\ j\left( \tfrac{\sqrt{30}i}{10} \right) &= \tfrac{1}{16} \left(10 - 7\sqrt{2} - 4\sqrt{5} + 3\sqrt{10} \right)^4 \left( 55 - 30\sqrt{2} - 12\sqrt{5} + 10\sqrt{10} \right)^3\\ j\left(\tfrac{1+\sqrt{31}i}{2}\right)&=\left(1-\left(1+\tfrac{\sqrt{19}}{2}\left(\left(\tfrac{13-\sqrt{93}}{13+\sqrt{93}}\right)^{1/2}\left(\tfrac{\sqrt{31}+\sqrt{27}}{\sqrt{31}-\sqrt{27}}\right)^{1/3}+\left(\tfrac{13+\sqrt{93}}{13-\sqrt{93}}\right)^{1/2}\left(\tfrac{\sqrt{31}-\sqrt{27}}{\sqrt{31}+\sqrt{27}}\right)^{1/3}\right)\right)^2\right)^3\\ j(\sqrt{70}i) &= \left(1 + \tfrac{9}{4}\left(303 + 220\sqrt{2} + 139\sqrt{5} + 96\sqrt{10}\right)^2 \right)^3\\ j(\sqrt{94}i) &= \left(1 + \tfrac{9}{64}\left(2454 + 1736\sqrt{2} + \left(546 + 384\sqrt{2}\right)\sqrt{9 + 8\sqrt{2}} + \left(527 + 373\sqrt{2} + \left(117 + 83\sqrt{2}\right)\sqrt{9 + 8\sqrt{2}}\right)\sqrt{3 + 4\sqrt{2} + 3\sqrt{9 + 8\sqrt{2}}}\right)^2\right)^3\\ j(7i) &= \left( 1 + \tfrac{9}{4}\sqrt{21+8\sqrt{7}} \left(30 + 11\sqrt{7} + \left (6+\sqrt{7} \right )\sqrt{21+8\sqrt{7}}\right)^2 \right)^3\\ j(8i) &= \left( 1 + \tfrac{9}{4} \sqrt[4]{2} \left (1 + \sqrt{2} \right) \left(123 + 104\sqrt[4]{2} + 88\sqrt{2} + 73\sqrt[4]{8}\right)^2 \right)^3\\ j(10i) &= \left(1 + \tfrac{9}{8}\left(2402 + 1607\sqrt[4]{5} + 1074\sqrt[4]{25} + 719\sqrt[4]{125}\right)^2 \right)^3\\ j \left( \frac{5 \, i}{2} \right) &= \left(1 + \tfrac{9}{8}\left(2402 - 1607\sqrt[4]{5} + 1074\sqrt[4]{25} - 719\sqrt[4]{125}\right)^2 \right)^3\\ j(2\sqrt{58}i) &= \left(1+\tfrac{9}{256}\left(1+\sqrt{2}\right)^5\left(5+\sqrt{29}\right)^5\left(793+907\sqrt{2}+237\sqrt{29}+103\sqrt{58}\right)^2\right)^3\\ j\left( \tfrac{1 + \sqrt{1435}i}{2} \right) &= \left( 1 - 9 \left ( 9892538 + 4424079\sqrt{5} + 1544955\sqrt{41} + 690925\sqrt{205} \right )^2 \right)^3\\ j\left( \tfrac{1 + \sqrt{1555}i}{2} \right) &= \left( 1 - 9 \left ( 22297077 + 9971556\sqrt{5} + \left ( 3571365 + 1597163\sqrt{5} \right ) \sqrt{\tfrac{31 + 21\sqrt{5}}{2}} \right)^2 \right)^3\\ \end{align} $

2014年にはいくつかの特殊値が計算された

$ \begin{align} j \left( \frac{5 \, i + 1}{2} \right) &= \left( \frac{2927 - 1323 \, \sqrt{5}}{2} \right)^3,\\ j \left( 5 \, i \right) &= \left( \frac{2927 + 1323 \, \sqrt{5}}{2} \right)^3,\\ j \left( \frac{5 \, i + 2}{4} \right) &= \Bigg( \frac{\left( 1 + \sqrt{5} \, \right)^{37}}{2^{39}} \Bigg( 1190448488 - 858585699 \, \sqrt{2} + 540309076 \, \sqrt{5} - 374537880 \, \sqrt{10} \, - \, \sqrt[4]{5} \left( 693172512 - 595746414 \, \sqrt{2} + 407357424 \, \sqrt{5} - 240819696 \, \sqrt{10} \, \right) \Bigg) \Bigg)^3,\\ j \left( \frac{10 \, i + 1}{2} \right) &= \Bigg( \frac{\left( 1 + \sqrt{5} \, \right)^{37}}{2^{39}} \Bigg( 1190448488 - 858585699 \, \sqrt{2} + 540309076 \, \sqrt{5} - 374537880 \, \sqrt{10} \, + \, \sqrt[4]{5} \left( 693172512 - 595746414 \, \sqrt{2} + 407357424 \, \sqrt{5} - 240819696 \, \sqrt{10} \, \right) \Bigg) \Bigg)^3,\\ j \left( \frac{5 \, i}{4} \right) &= \Bigg( \frac{\left( 1 + \sqrt{5} \, \right)^{37}}{2^{39}} \Bigg( 1190448488 + 858585699 \, \sqrt{2} + 540309076 \, \sqrt{5} + 374537880 \, \sqrt{10} \, - \, \sqrt[4]{5} \left( 693172512 + 595746414 \, \sqrt{2} + 407357424 \, \sqrt{5} + 240819696 \, \sqrt{10} \, \right) \Bigg) \Bigg)^3,\\ j(20 \, i) &= \Bigg( \frac{\left( 1 + \sqrt{5} \, \right)^{37}}{2^{39}} \Bigg( 1190448488 + 858585699 \, \sqrt{2} + 540309076 \, \sqrt{5} + 374537880 \, \sqrt{10} \, + \, \sqrt[4]{5} \left( 693172512 + 595746414 \, \sqrt{2} + 407357424 \, \sqrt{5} + 240819696 \, \sqrt{10} \, \right) \Bigg) \Bigg)^3. \end{align} $

これ以前に示したすべての値は実数である。複素共役のペアは、$ j(10 i) $$ j(5 i/2) $ に対し、参考文献のように値に沿って、上記のように対称的になっていると推察される。

$ \begin{align} j \left( \frac{5 \, i \pm 1}{4} \right) &= \left(1 - \tfrac{9}{8}\left((2402 - 1074\sqrt{5}) \, i \pm (1607 - 719\sqrt{5}) \sqrt[4]{5} \right)^2 \right)^3. \end{align} $

4つの特殊値は、2つの複素共役のペアにより与えられる $ \begin{align} j \left( \frac{4 \left( 5 \, i \pm 1 \right)}{13} \right) = \Bigg( \frac{\left( 1 - \sqrt{5} \, \right)^{37}}{2^{39}} \Bigg( 1190448488 - 858585699 \, \sqrt{2} - 540309076 \, \sqrt{5} + 374537880 \, \sqrt{10} \, \pm \, \textit{i} \, \sqrt[4]{5} \left( 693172512 - 595746414 \, \sqrt{2} - 407357424 \, \sqrt{5} + 240819696 \, \sqrt{10} \, \right) \Bigg) \Bigg)^3,\\ j \left( \frac{5 \left( 4 \, i \pm 1 \right)}{17} \right) = \Bigg( \frac{\left( 1 - \sqrt{5} \, \right)^{37}}{2^{39}} \Bigg( 1190448488 + 858585699 \, \sqrt{2} - 540309076 \, \sqrt{5} - 374537880 \, \sqrt{10} \, \pm \, \textit{i} \, \sqrt[4]{5} \left( 693172512 + 595746414 \, \sqrt{2} - 407357424 \, \sqrt{5} - 240819696 \, \sqrt{10} \, \right) \Bigg) \Bigg)^3 \end{align} $


$ f(s)=\sum_{n=1}^\infty\frac{\tau(n)}{n^s}=\prod_p\frac{1}{1-\tau(p)p^{-s}+p{11}-2s} $

$ \tau(n)=\frac{65}{756}\sigma_{11}(n)+\frac{691}{756}\sigma_5(n)-\frac{691}{3}\sum_{k=1}^{n-1}\sigma_5(k)\sigma_5(n-k) $

$ \sum_{n=1}^\infty\tau(n)x^n=x\prod{n=1}^\infty(1-x^n)^24 $

$ \tau_(p^{n+1})=\tau(p)\tau(p^n)-p^{11}\tau(p^{n-1}) $

$ \begin{align} \tau(n)&\equiv\sigma_{11}(n)(mod\ 2^8) \ for n odd\\ \tau(n)&\equiv n^2\sigma_{7}(n)(mod\ 3^3) \\ \tau(n)&\equiv n\sigma_{9}(n)(mod\ 5^2) \\ \tau(n)&\equiv n\sigma_{3}(n)(mod\ 7) \\ \tau(n)&\equiv\sigma_{11}(n)(mod\ 691) \\ \tau(n)&\equiv\begin{cases} \sigma_{11}(n)(mod\ 2^{11}) &if n\equiv1\ (mod\ 8) \\ 1217\sigma_{11}(n)(mod\ 2^{13}) &if n\equiv3\ (mod\ 8) \\ 1537\sigma_{11}(n)(mod\ 2^{12}) &if n\equiv5\ (mod\ 8) \\ 705\sigma_{11}(n)(mod 2^{14}) &if n\equiv7\ (mod\ 8) \\ \end{cases} \\ \tau(n)&\equiv n^{-610}\sigma_{1231}(n)\begin{cases} (mod\ 3^6) &if n\equiv 1\ (mod\ 3) \\ (mod\ 3^7) &if n\equiv 2\ (mod\ 3) \\ \end{cases} \\ \tau(n)&\equiv n^{-30}\sigma_{71}(n)\begin{cases} (mod\ 7) &if n\equiv 0,1,2,4\ (mod\ 7) \\ (mod\ 7^2) &if n\equiv 3,5,6\ (mod\ 7) \\ \end{cases} \\ \end{align} $

Examples

Order 2

Some order 2 mock theta functions were studied by (McIntosh 2007).

$ A(q) = \sum_{n\ge 0} \frac{q^{(n+1)^2}(-q;q^2)_n}{(q;q^2)^2_{n+1}} = \sum_{n\ge 0} \frac{q^{n+1}(-q^2;q^2)_n}{(q;q^2)_{n+1}} $ オンライン整数列大辞典の数列 A006304
$ B(q) = \sum_{n\ge 0} \frac{q^{n(n+1)}(-q^2;q^2)_n}{(q;q^2)^2_{n+1}} = \sum_{n\ge 0} \frac{q^{n}(-q;q^2)_n}{(q;q^2)_{n+1}} $ オンライン整数列大辞典の数列 A153140
$ \mu(q) = \sum_{n\ge 0} \frac{(-1)^nq^{n^2}(q;q^2)_n}{(-q^2;q^2)^2_{n}} $ オンライン整数列大辞典の数列 A006306

The function μ was found by Ramanujan in his lost notebook.

These are related to the functions listed in the section on order 8 functions by

$ U_0(q) - 2U_1(q) = \mu(q) $
$ V_0(q) - V_0(-q) = 4qB(q^2) $
$ V_1(q) + V_1(-q) = 2A(q^2) $

Order 3

$ f(q) = \sum_{n\ge 0} {q^{n^2}\over (-q; q)_n^2} = {2\over \prod_{n>0}(1-q^n)}\sum_{n\in Z}{(-1)^nq^{n(3n+1)/2}\over 1+q^n} $
$ \phi(q) = \sum_{n\ge 0} {q^{n^2}\over (-q^2;q^2)_n} = {1\over \prod_{n>0}(1-q^n)}\sum_{n\in Z}{(-1)^n(1+q^n)q^{n(3n+1)/2}\over 1+q^{2n}} $
$ \psi(q) = \sum_{n > 0} {q^{n^2}\over (q;q^2)_n} = {1\over \prod_{n>0}(1-q^{4n})}\sum_{n\in Z}{(-1)^nq^{6n(n+1)}\over 1-q^{4n+1}} $
$ \chi(q) = \sum_{n\ge 0} {q^{n^2}\over \prod_{1\le i\le n}(1-q^i+q^{2i})} = {1\over 2 \prod_{n>0}(1-q^n)}\sum_{n\in Z}{(-1)^n(1+q^n)q^{n(3n+1)/2}\over 1-q^n+q^{2n}} $
$ \omega(q) = \sum_{n\ge 0} {q^{2n(n+1)}\over (q;q^2)^2_{n+1}} = {1\over \prod_{n>0}(1-q^{2n})}\sum_{n\ge 0}{(-1)^nq^{3n(n+1)} {1+q^{2n+1}\over 1-q^{2n+1}}} $
$ \nu(q) = \sum_{n\ge 0} {q^{n(n+1)}\over (-q;q^2)_{n+1}} = {1\over \prod_{n>0}(1-q^n)}\sum_{n\ge 0}{(-1)^nq^{3n(n+1)/2}{1-q^{2n+1}\over 1+q^{2n+1}}} $
$ \rho(q) = \sum_{n\ge 0} {q^{2n(n+1)}\over \prod_{0\le i\le n}(1+q^{2i+1}+q^{4i+2})} = {1\over \prod_{n>0}(1-q^{2n})}\sum_{n\ge 0}{(-1)^nq^{3n(n+1)} {1-q^{4n+2}\over 1+q^{2n+1}+q^{4n+2}}} $

The first four of these form a group with the same shadow (up to a constant), and so do the last three. More precisely, the functions satisfy the following relations (found by Ramanujan and proved by Watson):

$ \begin{align} 2\phi(-q) - f(q) &= f(q) + 4\psi(-q) = \theta_4(0,q)\prod_{r > 0}\left(1 + q^r\right)^{-1} \\ 4\chi(q) - f(q) &= 3\theta_4^2(0, q^3)\prod_{r > 0}\left(1 - q^r\right)^{-1} \\ 2\rho(q) + \omega(q) &= 3\left(\frac{1}{2}q^{-\frac{3}{8}}\theta_2(0, q^\frac{3}{2})\right)^2\prod_{r > 0}\left(1 - q^{2r}\right)^{-1} \\ \nu(\pm q) \pm q\omega\left(q^2\right) &= \frac{1}{2}q^{-\frac{1}{4}}\theta_2(0, q)\prod_{r > 0}\left(1 + q^{2r}\right) \\ f\left(q^8\right) \pm 2q\omega(\pm q) \pm 2q^3\omega\left(-q^4\right) &= \theta_3(0, \pm q)\theta_3^2\left(0, q^2\right)\prod_{r > 0}\left(1 - q^{4r}\right)^{-2} \\ f(q^8) + q\omega(q) - q\omega(-q) &= \theta_3(0, q^4) \theta_3^2(0, q^2)\prod_{r > 0}\left(1 - q^{4r}\right)^{-2} \end{align} $

Order 5

$ f_0(q) = \sum_{n\ge 0} {q^{n^2}\over (-q;q)_{n}} $
$ f_1(q) = \sum_{n\ge 0} {q^{n^2+n}\over (-q;q)_{n}} $
$ \phi_0(q) = \sum_{n\ge 0} {q^{n^2}(-q;q^2)_{n}} $
$ \phi_1(q) = \sum_{n\ge 0} {q^{(n+1)^2}(-q;q^2)_{n}} $
$ \psi_0(q) = \sum_{n\ge 0} {q^{(n+1)(n+2)/2}(-q;q)_{n}} $
$ \psi_1(q) = \sum_{n\ge 0} {q^{n(n+1)/2}(-q;q)_{n}} $
$ \chi_0(q) = \sum_{n\ge 0} {q^{n}\over (q^{n+1};q)_{n}} = 2F_0(q)-\phi_0(-q) $
$ \chi_1(q) = \sum_{n\ge 0} {q^{n}\over (q^{n+1};q)_{n+1}} = 2F_1(q)+q^{-1}\phi_1(-q) $
$ F_0(q) = \sum_{n\ge 0} {q^{2n^2}\over (q;q^2)_{n}} $
$ F_1(q) = \sum_{n\ge 0} {q^{2n^2+2n}\over (q;q^2)_{n+1}} $
$ \Psi_0(q) = -1 + \sum_{n \ge 0} { q^{5n^2}\over(1-q)(1-q^4)(1-q^6)(1-q^9)...(1-q^{5n+1})} $
$ \Psi_1(q) = -1 + \sum_{n \ge 0} { q^{5n^2}\over(1-q^2)(1-q^3)(1-q^7)(1-q^8)...(1-q^{5n+2}) } $

Order 6

$ \phi(q) = \sum_{n\ge 0} {(-1)^nq^{n^2}(q;q^2)_n\over (-q;q)_{2n}} $
$ \psi(q) = \sum_{n\ge 0} {(-1)^nq^{(n+1)^2}(q;q^2)_n\over (-q;q)_{2n+1}} $
$ \rho(q) = \sum_{n\ge 0} {q^{n(n+1)/2}(-q;q)_n\over (q;q^2)_{n+1}} $
$ \sigma(q) = \sum_{n\ge 0} {q^{(n+1)(n+2)/2}(-q;q)_n\over (q;q^2)_{n+1}} $
$ \lambda(q) = \sum_{n\ge 0} {(-1)^nq^{n}(q;q^2)_n\over (-q;q)_{n}} $
$ 2\mu(q) = \sum_{n\ge 0} {(-1)^nq^{n+1}(1+q^n)(q;q^2)_n\over (-q;q)_{n+1}} $
$ \gamma(q) = \sum_{n\ge 0} {q^{n^2}(q;q)_n\over (q^3;q^3)_{n}} $
$ \phi_{-}(q) = \sum_{n\ge 1} {q^{n}(-q;q)_{2n-1}\over (q;q^2)_{n}} $
$ \psi_{-}(q) = \sum_{n\ge 1} {q^{n}(-q;q)_{2n-2}\over (q;q^2)_{n}} $

Order 7

  • $ \displaystyle F_0(q) = \sum_{n\ge 0}{q^{n^2}\over (q^{n+1};q)_n} $
  • $ \displaystyle F_1(q) = \sum_{n\ge 0}{q^{n^2}\over (q^{n};q)_n} $
  • $ \displaystyle F_2(q) = \sum_{n\ge 0}{q^{n(n+1)}\over (q^{n+1};q)_{n+1}} $

These three mock theta functions have different shadows, so unlike the case of Ramanujan's order 3 and order 5 functions, there are no linear relations between them and ordinary modular forms. The corresponding weak Maass forms are

$ \displaystyle M_1(\tau) = q^{-1/168}F_1(q) + R_{7,1}(\tau) $
$ \displaystyle M_2(\tau) = -q^{-25/168}F_2(q) + R_{7,2}(\tau) $
$ \displaystyle M_3(\tau) = q^{47/168}F_3(q) + R_{7,3}(\tau) $

where

$ R_{p,j}(\tau) = \sum_{n\equiv j\bmod p}{12\choose n}\sgn(n)\beta(n^2y/6p)q^{-n^2/24p} $

and

$ \beta(x) = \int_x^\infty u^{-1/2}e^{-\pi u} du $

is more or less the complementary error function. Under the metaplectic group, these three functions transform according to a certain 3-dimensional representation of the metaplectic group as follows

$ M_j(-1/\tau) = \sqrt{\tau/7i}\sum_{k=1}^32\sin(6\pi jk/7)M_k(\tau) $
$ M_1(\tau+1) = e^{-2\pi i/168} M_1(\tau) $, $ M_2(\tau+1) = e^{-2\times 25\pi i/168} M_2(\tau) $, $ M_3(\tau+1) = e^{-2\times 121\pi i/168} M_3(\tau). $

In other words, they are the components of a level 1 vector-valued harmonic weak Maass form of weight 1/2.

Order 8

$ S_0(q) = \sum_{n\ge 0} {q^{n^2} (-q;q^2)_n \over (-q^2;q^2)_n} $
$ S_1(q) = \sum_{n\ge 0} {q^{n(n+2)} (-q;q^2)_n \over (-q^2;q^2)_n} $
$ T_0(q) = \sum_{n\ge 0} {q^{(n+1)(n+2)} (-q^2;q^2)_n \over (-q;q^2)_{n+1}} $
$ T_1(q) = \sum_{n\ge 0} {q^{n(n+1)} (-q^2;q^2)_n \over (-q;q^2)_{n+1}} $
$ U_0(q) = \sum_{n\ge 0} {q^{n^2} (-q;q^2)_n \over (-q^4;q^4)_n} $
$ U_1(q) = \sum_{n\ge 0} {q^{(n+1)^2} (-q;q^2)_n \over (-q^2;q^4)_{n+1}} $
$ V_0(q) = -1+2\sum_{n\ge 0} {q^{n^2} (-q;q^2)_n \over (q;q^2)_n} = -1+2\sum_{n\ge 0} {q^{2n^2} (-q^2;q^4)_n \over (q;q^2)_{2n+1}} $
$ V_1(q) = \sum_{n\ge 0} {q^{(n+1)^2} (-q;q^2)_n \over (q;q^2)_{n+1}} = \sum_{n\ge 0} {q^{2n^2+2n+1} (-q^4;q^4)_n \over (q;q^2)_{2n+2}} $

Order 10

  • $ \phi(q)=\sum_{n\ge 0}{q^{n(n+1)/2}\over (q;q^2)_{n+1}} $
  • $ \psi(q)=\sum_{n\ge 0}{q^{(n+1)(n+2)/2}\over (q;q^2)_{n+1}} $
  • $ \Chi(q)=\sum_{n\ge 0}{(-1)^nq^{n^2}\over (-q;q)_{2n}} $
  • $ \chi(q)=\sum_{n\ge 0}{(-1)^nq^{(n+1)^2}\over (-q;q)_{2n+1}} $

$ A-B+C-D+E-F\cdots=\cfrac{A}{1+\cfrac{B}{A-B+\cfrac{AC}{B-C+\cfrac{BD}{C-D+\cfrac{CE}{D-E+\cdots}}}}} $

$ \frac{1}{A}-\frac{1}{B}+\frac{1}{C}-\frac{1}{D}+\frac{1}{E}-\frac{1}{F}\cdots=\cfrac{1}{A+\cfrac{A^2}{B-A+\cfrac{B^2}{C-B+\cfrac{C^2}{D-C+\cfrac{D^2}{E-D+\cdots}}}}} $

$ \begin{align} &\frac{1}{A}-\frac{1}{AB}+\frac{1}{ABC}-\frac{1}{ABCD}+\frac{1}{ABCDE}-\frac{1}{ABCDEF}\cdots \\ =&\cfrac{1}{A+\cfrac{A}{B-1+\cfrac{B}{C-1+\cfrac{C}{D-1+\cfrac{D}{E-1+\cdots}}}}} \end{align} $

$ \begin{align} &A-Bz+Cz^2-Dz^3+Ez^4-Fz^5\cdots \\ =&\cfrac{A}{1+\cfrac{Bz}{A-Bz+\cfrac{ACz}{B-Cz+\cfrac{BDz}{C-Dz+\cfrac{CEz}{D-Ez+\cdots}}}}} \end{align} $

$ \begin{align} &\frac{A}{L}-\frac{By}{Mz}+\frac{Cy^2}{Nz^2}-\frac{Dy^3}{Oz^3}+\frac{Ey^4}{Pz^4}\cdots \\ =&\cfrac{A}{L+\cfrac{BL^2y}{AMz-BLy+\cfrac{ACM^2yz}{BNz-CMy+\cfrac{BDN^2yz}{COz-DYy\cdots}}}} \end{align} $

$ \begin{align} &\frac{A}{L}-\frac{ABy}{LMz}+\frac{ABCy^2}{LMNz^2}-\frac{ABCDy^3}{LMNOz^3}\cdots \\ =&\cfrac{Az}{Lz+\cfrac{BLyz}{Mz-By+\cfrac{CMyz}{Nz-Cy+\cfrac{DNyz}{Oz-Dy\cdots}}}} \end{align} $

$ \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}} $

$ \begin{align} &\sum_{n=-\infty}^\infty q^{P_{p,n}}=\sum_{n=-\infty}^\infty q^{\frac{(p-2)n^2-(p-4)n}{2}} \\ =&\prod_{m=1}^\infty (1-q^{(p-2)m})(1+q^{(p-2)(m-\frac{1}{2})+\frac{p-4}{2}})(1+q^{(p-2)(m-\frac{1}{2})-\frac{p-4}{2}}) \\ =&\prod_{m=1}^\infty (1-q^{(p-2)m})(1+q^{pm-2m-1})(1+q^{pm-2m-p+3}) \\ =&(q^{p-2};q^{p-2})_\infty(-q^{p-3};q^{p-2})_\infty(-q;q^{p-2})_\infty \end{align} $

$ \begin{align} &\sum_{n=-\infty}^\infty q^{P_{p,n}}=\sum_{n=-\infty}^\infty q^{\frac{(p-2)n^2-(p-4)n}{2}} \\ =&\prod_{m=1}^\infty (1-q^{(p-2)m})(1+q^{(p-2)(m-\frac{1}{2})+\frac{p-4}{2}})(1+q^{(p-2)(m-\frac{1}{2})-\frac{p-4}{2}}) \\ =&\prod_{m=1}^\infty (1-q^{(p-2)m})(1+q^{pm-2m-1})(1+q^{pm-2m-p+3}) \\ =&(q^{p-2};q^{p-2})_\infty(-q^{p-3};q^{p-2})_\infty(-q;q^{p-2})_\infty \end{align} $

$ \sum_{n=-\infty}^\infty (-1)^nq^{P_{p,n}}=(q^{p-2};q^{p-2})_\infty(q^{p-3};q^{p-2})_\infty(q;q^{p-2})_\infty $

このことから、三角数定理、四角数定理、五角数定理、六角数定理、八角数定理をつくることができる。

$ \left(\sum_{n=-\infty}^\infty q^{n^2}\right)^2=\sum_{n=0}^\infty q^{4\sum_{2|d|n}(-1)^{\frac{d-1}{2}}} $

$ \left(\sum_{n=-\infty}^\infty q^{n^2}\right)^4=\sum_{n=0}^\infty q^{8\sum_{2|d|n}d} $

$ \begin{align} &\sum_{n,m=-\infty}^{\infty}{q^{n^2+m^2}z^{n}y^{m}} \\ =&\prod_{m=1}^{\infty}{\left(1-q^{2m}\right)^2\left(1+q^{2m-1}z\right)\left(1+q^{2m-1}z^{-1}\right)\left(1+q^{2m-1}y\right)\left(1+q^{2m-1}y^{-1}\right)} \end{align} $

ワトソンの五重積は$ z=q^{\frac{1}{6}} $のとき、ファルカシュの7重積は$ z=q^{\frac{1}{5}} $のときに有益な結果が出る可能性がある。