メモ 1(随時更新予定)

1106

メモ
見つけた数式などいくつか羅列します。いつかきちんと記事に書きたい(随時更新)

$\begin{aligned} (\binom{n}{k}) := \frac{n!}{k! (n - k)!} \\ H_{n} := \frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{n} \\ ζ (s) := \sum_{n = 1}^{\infty} \frac{1}{n^{s}} \\ η (s) := \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1}}{n^{s}} \\ t (s) := \sum_{n = 0}^{\infty} \frac{1}{(2 n + 1)^{s}} \\ β (s) := \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{(2 n + 1)^{s}} \\ {L i}_{1} (z) = - \log (1 - z) \\ {L i}_{s + 1} (z) = \int_{0}^{z} \frac{{L i}_{s} (t)}{t} d t \\ {L i}_{s} (z) = \sum_{n = 1}^{\infty} \frac{z^{n}}{n^{s}} \\ k = (k_{1}, k_{2}, \dots, k_{r}) \\ β = (β_{1}, β_{2}, \dots, β_{r}) \\ ζ (k) = \sum_{0 < n_{1} < \dots < n_{r}} \frac{1}{n_{1}^{k_{1}} n_{2}^{k_{2}} \dots n_{r}^{k_{r}}} \\ t (k) = \sum_{0 < n_{1} < \dots < n_{r}} \frac{1}{(2 n_{1} - 1)^{k_{1}} (2 n_{2} - 1)^{k_{2}} \dots (2 n_{r} - 1)^{k_{r}}} \\ ζ (k; β) = \sum_{0 \leq n_{1} < \dots < n_{r}} \frac{1}{(n_{1} + β_{1})^{k_{1}} (n_{2} + β_{2})^{k_{2}} \dots (n_{r} + β_{r})^{k_{r}}} \\ ζ^{⋆} (k) = \sum_{0 < n_{1} \leq \dots \leq n_{r}} \frac{1}{n_{1}^{k_{1}} n_{2}^{k_{2}} \dots n_{r}^{k_{r}}} \\ t^{⋆} (k) = \sum_{0 < n_{1} \leq \dots \leq n_{r}} \frac{1}{(2 n_{1} - 1)^{k_{1}} (2 n_{2} - 1)^{k_{2}} \dots (2 n_{r} - 1)^{k_{r}}} \\ ζ (k_{1}, \dots, \bar{k_{i}}, \dots, k_{r}) = \sum_{0 < n_{1} < \dots < n_{r}} \frac{(- 1)^{n_{i}}}{n_{1}^{k_{1}} \dots n_{i}^{k_{i}} \dots n_{r}^{k_{r}}} \\ t (k_{1}, \dots, \bar{k_{i}}, \dots, k_{r}) = \sum_{0 < n_{1} < \dots < n_{r}} \frac{(- 1)^{n_{i}}}{(2 n_{1} - 1)^{k_{1}} \dots (2 n_{i} - 1)^{k_{i}} \dots (2 n_{r} - 1)^{k_{r}}} \end{aligned}$

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

$\begin{aligned} k = (k_{1}, k_{2}, \dots, k_{r}) \\ ζ (k) = (- 1)^{r} \sum_{\begin{array}{c} 0 < m_{1}, m_{2}, \dots, m_{r} \\ 0 \leq n_{1}, n_{2}, \dots, n_{r} \end{array}} \frac{(- 1)^{m_{1} + m_{2} + \dots + m_{r}} 2^{n_{1} + n_{2} + \dots + n_{r}}}{(m_{1} 2^{n_{1}})^{k_{1}} (m_{1} 2^{n_{1}} + m_{2} 2^{n_{2}})^{k_{2}} \dots (m_{1} 2^{n_{1}} + m_{2} 2^{n_{2}} + \dots + m_{r} 2^{n_{r}})^{k_{r}}} \end{aligned}$

$\begin{aligned} \frac{e^{π} - e^{- π}}{2 π} = \exp (\frac{ζ (2)}{1} - \frac{ζ (4)}{2} + \frac{ζ (6)}{3} - \frac{ζ (8)}{4} + \dots) \\ \frac{e^{\frac{π}{2}} + e^{- \frac{π}{2}}}{π} = \exp (\frac{η (2)}{1} - \frac{ζ (4)}{2} + \frac{η (6)}{3} - \frac{ζ (8)}{4} + \dots) \\ \frac{e^{\frac{π}{2}} + e^{- \frac{π}{2}}}{2} = \exp (\frac{t (2)}{1} - \frac{t (4)}{2} + \frac{t (6)}{3} - \frac{t (8)}{4} + \dots) \end{aligned}$

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

$\begin{aligned} \sin (\sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} ζ (4 n - 2)}{2 n - 1} z^{4 n - 2}) = \frac{\sin (\frac{π}{\sqrt{2}} z) \cosh (\frac{π}{\sqrt{2}} z) - \cos (\frac{π}{\sqrt{2}} z) \sinh (\frac{π}{\sqrt{2}} z)}{\sqrt{\cosh (\sqrt{2} π z) - \cos (\sqrt{2} π z)}} \\ \cos (\sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} ζ (4 n - 2)}{2 n - 1} z^{4 n - 2}) = \frac{\sin (\frac{π}{\sqrt{2}} z) \cosh (\frac{π}{\sqrt{2}} z) + \cos (\frac{π}{\sqrt{2}} z) \sinh (\frac{π}{\sqrt{2}} z)}{\sqrt{\cosh (\sqrt{2} π z) - \cos (\sqrt{2} π z)}} \\ \sin (\sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} t (4 n - 2)}{2 n - 1} z^{4 n - 2}) = \frac{\sqrt{2} \sin (\frac{\sqrt{2}}{4} π z) \sinh (\frac{\sqrt{2}}{4} π z)}{\sqrt{\cos (\frac{π}{\sqrt{2}} z) + \cosh (\frac{π}{\sqrt{2}} z)}} \\ \cos (\sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} t (4 n - 2)}{2 n - 1} z^{4 n - 2}) = \frac{\sqrt{2} \cos (\frac{\sqrt{2}}{4} π z) \cosh (\frac{\sqrt{2}}{4} π z)}{\sqrt{\cos (\frac{π}{\sqrt{2}} z) + \cosh (\frac{π}{\sqrt{2}} z)}} \\ \sin (\sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} η (4 n - 2)}{2 n - 1} z^{4 n - 2}) = 2 x \frac{\sinh (\frac{\sqrt{2}}{4} π z) \cosh (\frac{\sqrt{2}}{4} π z) - \sin (\frac{\sqrt{2}}{4} π z) \cos (\frac{\sqrt{2}}{4} π z)}{\sqrt{\cosh (\sqrt{2} π z) - \cos (\sqrt{2} π z)}} \\ \cos (\sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} η (4 n - 2)}{2 n - 1} z^{4 n - 2}) = 2 x \frac{\sinh (\frac{\sqrt{2}}{4} π z) \cosh (\frac{\sqrt{2}}{4} π z) + \sin (\frac{\sqrt{2}}{4} π z) \cos (\frac{\sqrt{2}}{4} π z)}{\sqrt{\cosh (\sqrt{2} π z) - \cos (\sqrt{2} π z)}} \end{aligned}$

$\begin{aligned} ζ ({2}^{r}) = \sum_{0 < n_{1} < \dots < n_{r}} \frac{1}{n_{1}^{2} n_{2}^{2} \dots n_{r}^{2}} = \frac{π^{2 r}}{(2 r + 1)!} \\ ζ ({4}^{r}) = \sum_{0 < n_{1} < \dots < n_{r}} \frac{1}{n_{1}^{4} n_{2}^{4} \dots n_{r}^{4}} = \frac{2^{2 r + 1} π^{4 r}}{(4 r + 2)!} \\ ζ ({6}^{r}) = \sum_{0 < n_{1} < \dots < n_{r}} \frac{1}{n_{1}^{6} n_{2}^{6} \dots n_{r}^{6}} = \frac{6 \cdot 2^{6 r} π^{6 r}}{(6 r + 3)!} \\ ζ ({8}^{r}) = \sum_{0 < n_{1} < \dots < n_{r}} \frac{1}{n_{1}^{8} n_{2}^{8} \dots n_{r}^{8}} = \frac{2^{4 r + 1} π^{8 r}}{(8 r + 4)!} {(2 + \sqrt{2})^{4 r + 2} + (2 - \sqrt{2})^{4 r + 2}} \\ ζ ({10}^{r}) = \sum_{0 < n_{1} < \dots < n_{r}} \frac{1}{n_{1}^{10} n_{2}^{10} \dots n_{r}^{10}} = \frac{10 \cdot 2^{10 r} π^{10 r}}{(10 r + 5)!} (1 + L_{10 r + 5}) (L_{n} は n 番目のリュカ数) \end{aligned}$

$\begin{aligned} ζ ({2 m}^{r}) & = \sum_{0 < n_{1} < \dots < n_{r}} \frac{1}{n_{1}^{2 m} n_{2}^{2 m} \dots n_{r}^{2 m}} \\ = \frac{(- i)^{m + 1}}{2^{m - 1}} \frac{(- 1)^{(m + 1) r} π^{2 m r}}{(2 m r + m)!} \overset{―}{\sum_{}^{}} (e^{\frac{i π}{m}} \pm e^{\frac{2 i π}{m}} \pm \dots \pm e^{\frac{m i π}{m}})^{2 m r + m} \end{aligned}$

ζ ({2 m}^{r})

の式にある

\overset{―}{\sum_{}^{}}

について

$f (k_{1} \pm k_{2} \pm \dots \pm k_{n})$
で表される $2^{n - 1}$ 個の数の総和を
$\sum_{}^{} f (k_{1} \pm k_{2} \pm \dots \pm k_{n}) と書くこととする。$

(例)
$\sum_{} f (A \pm B \pm C) = f (A + B + C) + f (A + B - C) + f (A - B + C) + f (A - B - C)$
また、各項について $\pm$ の部分で-を選んだ回数を $M$ で表して
$\overset{―}{\sum_{}^{}} f (k_{1} \pm k_{2} \pm \dots \pm k_{n}) := \sum_{}^{} f (k_{1} \pm k_{2} \pm \dots \pm k_{n}) (- 1)^{M}$

$\begin{aligned} t ({2}^{r}) = \sum_{0 < n_{1} < \dots < n_{r}} \frac{1}{(2 n_{1} - 1)^{2} (2 n_{2} - 1)^{2} \dots (2 n_{r} - 1)^{2}} = \frac{π^{2 r}}{2^{2 r} (2 r)!} \\ t ({4}^{r}) = \sum_{0 < n_{1} < \dots < n_{r}} \frac{1}{(2 n_{1} - 1)^{4} (2 n_{2} - 1)^{4} \dots (2 n_{r} - 1)^{4}} = \frac{π^{4 r}}{2^{2 r} (4 r)!} \\ t ({6}^{r}) = \sum_{0 < n_{1} < \dots < n_{r}} \frac{1}{(2 n_{1} - 1)^{6} (2 n_{2} - 1)^{6} \dots (2 n_{r} - 1)^{6}} = \frac{3 π^{6 r}}{4 (6 r)!} \end{aligned}$

気に入っている積分1
$\begin{array}{r} \int_{0}^{1} \frac{{L i}_{2} (\frac{1 + x}{2}) - \frac{π^{2}}{12} + \frac{1}{2} (\log 2)^{2}}{x} d x = \frac{13}{8} ζ (3) - \frac{π^{2}}{6} \log 2 \end{array}$

$\begin{aligned} Ω (n) : n の重複を含めた素因数の個数 \\ ω (n) : n の重複を含めない素因数の個数 \\ A (n) : n の重複を含めた素因数の和 \\ t (s) := \sum_{n = 1}^{\infty} \frac{1}{(2 n - 1)^{s}} \\ 正の整数 n とその素因数分解 n = p_{1}^{α_{1}} p_{2}^{α_{2}} \dots p_{ω (n)}^{α_{ω (n)}} に対し、 \\ Y (n) := \prod_{k = 1}^{ω (n)} α_{k} (ただし Y (1) = 1) \\ σ_{a} (n) : n の約数の a 乗和 (約数関数) \\ \sum_{n = 1}^{\infty} \frac{(- 1)^{Ω (n)}}{n^{s}} = \frac{ζ (2 s)}{ζ (s)} \\ \sum_{n = 1}^{\infty} \frac{2^{ω (n)}}{n^{s}} = \frac{ζ (s)^{2}}{ζ (2 s)} \\ \sum_{n = 1}^{\infty} \frac{(- 1)^{A (n)}}{n^{s}} = \frac{2^{s}}{2^{s} - 1} \frac{t (2 s)}{t (s)} \\ \sum_{n = 1}^{\infty} \frac{σ_{a} (n) (- 1)^{Ω (n)}}{n^{s}} = \frac{ζ (2 s) ζ (2 s - 2 a)}{ζ (s) ζ (s - a)} \\ \sum_{n = 1}^{\infty} \frac{Y (n^{2})}{n^{s}} = \frac{ζ (s)^{2} ζ (2 s)}{ζ (4 s)} \\ \sum_{n = 1}^{\infty} \frac{Y (n^{4})}{n^{s}} = \frac{ζ (s)^{4}}{ζ (2 s)^{2}} \\ (\sum_{n = 1}^{\infty} \frac{Y (n)}{n^{s}}) (\sum_{n = 1}^{\infty} \frac{Y (n^{3})}{n^{s}}) = \frac{ζ (s)^{4} ζ (2 s)}{ζ (6 s)} \\ \sum_{n = 1}^{\infty} \frac{σ_{a} (n)}{n^{s}} = ζ (s) ζ (s - a) \\ \sum_{n = 1}^{\infty} \frac{σ_{a} (n) σ_{b} (n)}{n^{s}} = \frac{ζ (s) ζ (s - a) ζ (s - b) ζ (s - a - b)}{ζ (2 s - a - b)} \\ \sum_{n = 0}^{\infty} \frac{(- 1)^{n} σ_{a} (2 n + 1)}{(2 n + 1)^{s}} = β (s) β (s - a) \\ \sum_{n = 0}^{\infty} \frac{(- 1)^{n} σ_{a} (2 n + 1) σ_{b} (2 n + 1)}{(2 n + 1)^{s}} = \frac{β (s) β (s - a) β (s - b) β (s - a - b)}{t (2 s - a - b)} \\ \sum_{0 < n_{1} < n_{2}} \frac{σ_{a} (n_{1}) σ_{a} (n_{2})}{n_{1}^{s} n_{2}^{s}} = \frac{1}{2} {(ζ (s) ζ (s - a))^{2} - \frac{ζ (2 s) ζ (2 s - a)^{2} ζ (2 s - 2 a)}{ζ (4 s - 2 a)}} \end{aligned}$

$\begin{aligned} \sum_{0 < n_{1} < n_{2} < n_{3}} \frac{1}{(2 n_{1} - 1) (2 n_{2} - 1) (2 n_{3} - 1) 5^{n_{3} - n_{2}}} = \frac{1}{20} ζ (3) \end{aligned}$

$\begin{aligned} t ({\bar{1}}^{r}) = \sum_{0 < n_{1} < \dots < n_{r}} \frac{(- 1)^{n_{1} + n_{2} + \dots + n_{r}}}{(2 n_{1} - 1) (2 n_{2} - 1) \dots (2 n_{r} - 1)} = \frac{(- 1)^{[\frac{1 + r}{2}]} π^{r}}{2^{2 r} r!} \\ t ({\bar{3}}^{r}) = \sum_{0 < n_{1} < \dots < n_{r}} \frac{(- 1)^{n_{1} + n_{2} + \dots + n_{r}}}{(2 n_{1} - 1)^{3} (2 n_{2} - 1)^{3} \dots (2 n_{r} - 1)^{3}} = \frac{3 π^{3 r}}{2^{3 r + 1} (3 r)!} (- 1)^{[\frac{1 + r}{2}]} \end{aligned}$

$\begin{aligned} ζ ({\bar{2}}^{r}) = \sum_{0 < n_{1} < \dots < n_{r}} \frac{(- 1)^{n_{1} + n_{2} + \dots + n_{r}}}{n_{1}^{2} n_{2}^{2} \dots n_{r}^{2}} = \frac{(- 1)^{[\frac{1 + r}{2}]} π^{2 r}}{2^{r} (2 r + 1)!} \\ ζ ({\bar{4}}^{r}) = \sum_{0 < n_{1} < \dots < n_{r}} \frac{(- 1)^{n_{1} + n_{2} + \dots + n_{r}}}{n_{1}^{4} n_{2}^{4} \dots n_{r}^{4}} = \frac{(- 1)^{[\frac{1 + r}{2}]} π^{4 r}}{2^{r} (4 r + 2)!} ((\sqrt{2} + 1)^{2 r + 1} - (\sqrt{2} - 1)^{2 r + 1}) \end{aligned}$

コネクターと級数
許容インデックス $k = (k_{1}, k_{2}, \dots, k_{r})$ に対して
$U (k) := \sum_{0 < n_{1} < n_{2} < \dots < n_{r}} \frac{n_{r}}{(n_{1} - \frac{1}{2})^{k_{1}} (n_{2} - \frac{1}{2})^{k_{2}} \dots (n_{r} - \frac{1}{2})^{k_{r}}} \frac{(\binom{2 n_{r}}{n_{r}})}{2^{2 n_{r}}}$
とすると $U (k)$ は双対性を持つ。すなわち、
$k$ の双対インデックス $k^{†}$ に対し
$U (k) = U (k^{†})$
が成り立つ。

多重ゼータ値の双対性2
コネクターと級数
$\begin{aligned} k = (k_{1}, k_{2}, \dots, k_{r}) \\ (a)_{n} : ポッホハマー記号, α = ({α}^{d e p (k^{†})}) \\ \sum_{0 < n_{1} < \dots < n_{r}} \frac{2^{2 n_{r}}}{n_{1}^{k_{1}} n_{2}^{k_{2}} \dots n_{r}^{k_{r}} (\binom{2 n_{r}}{n_{r}})} = 2^{w t (k)} t (k^{†}) \\ \sum_{0 < n_{1} < \dots < n_{r}} \frac{n_{r}!}{n_{1}^{k_{1}} n_{2}^{k_{2}} \dots n_{r}^{k_{r}} (α)_{n_{r}}} = ζ (k^{†}; α) \end{aligned}$

$\begin{aligned} k = (k_{1}, k_{2}, \dots, k_{r}) に対して \\ ς (k) := \sum_{0 \leq n_{1} < n_{2} < \dots < n_{r}} \frac{1}{(n_{1} + \frac{1}{2})^{k_{1}} (n_{2} + \frac{1}{2})^{k_{2}} \dots (n_{r} + \frac{1}{2})^{k_{r}}} \frac{(\binom{2 n_{r}}{n_{r}})}{2^{2 n_{r}}} \\ k^{'} := ((k_{↑})^{†})_{↓} とすると \\ ς (k) = ς (k^{'}) が成り立つ \end{aligned}$

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

$\sum_{0 < n_{1} < n_{2} < \dots < n_{r}} \frac{n_{r}}{(n_{1} - \frac{1}{2})^{2} (n_{2} - \frac{1}{2})^{2} \dots (n_{r} - \frac{1}{2})^{2}} \frac{(\binom{2 n}{n})}{2^{2 n}} = \frac{π^{2 r - 1}}{(2 r - 1)!}$

$\begin{aligned} M := M (n_{1}, n_{2}, \dots, n_{2 r}) = \sum_{k = 1}^{2 r} n_{k} (- 1)^{k} = n_{2 r} - n_{2 r - 1} + \dots + n_{2} - n_{1} \\ \sum_{0 < n_{1} < \dots < n_{2 r}} \frac{(- 1)^{M}}{(n_{1} - \frac{1}{2}) (n_{2} - \frac{1}{2}) \dots (n_{2 r} - \frac{1}{2}) \cdot 3^{M}} = \frac{(- 1)^{r} π^{2 r}}{3^{2 r} (2 r)!} \\ \sum_{0 < n_{1} < \dots < n_{2 r}} \frac{1}{(n_{1} - \frac{1}{2}) (n_{2} - \frac{1}{2}) \dots (n_{2 r} - \frac{1}{2}) 2^{M}} = \frac{2^{2 r}}{(2 r)!} (\log (1 + \sqrt{2}))^{2 r} \\ \sum_{0 < n_{1} < \dots < n_{2 r}} \frac{1}{(n_{1} - \frac{1}{2}) (n_{2} - \frac{1}{2}) \dots (n_{2 r} - \frac{1}{2}) 3^{M}} = \frac{2^{2 r}}{(2 r)!} (\log (\frac{\sqrt{2} + \sqrt{6}}{2}))^{2 r} \\ \sum_{0 < n_{1} < \dots < n_{2 r}} \frac{1}{(n_{1} - \frac{1}{2}) (n_{2} - \frac{1}{2}) \dots (n_{2 r} - \frac{1}{2}) 5^{M}} = \frac{2^{2 r}}{(2 r)!} (\log (ϕ))^{2 r} \end{aligned}$

$\begin{aligned} M^{'} := M^{'} (n_{1}, n_{2}, \dots, n_{2 r - 1}) = \sum_{k = 1}^{2 r - 1} n_{k} (- 1)^{k + 1} = n_{2 r - 1} - n_{2 r - 2} + \dots + n_{1} \\ \sum_{0 < n_{1} < \dots < n_{2 r - 1}} \frac{(- 1)^{M^{'}}}{(n_{1} - \frac{1}{2}) (n_{2} - \frac{1}{2}) \dots (n_{2 r - 1} - \frac{1}{2}) 3^{M^{'}}} = \frac{1}{\sqrt{3}} \frac{(- 1)^{r} π^{2 r - 1}}{3^{2 r - 1} (2 r - 1)!} \\ \sum_{0 < n_{1} < \dots < n_{2 r - 1}} \frac{1}{(n_{1} - \frac{1}{2}) (n_{2} - \frac{1}{2}) \dots (n_{2 r - 1} - \frac{1}{2}) 2^{M^{'}}} = \frac{2^{2 r - 1}}{(2 r - 1)!} \frac{(\log (1 + \sqrt{2}))^{2 r - 1}}{\sqrt{2}} \\ \sum_{0 < n_{1} < \dots < n_{2 r - 1}} \frac{1}{(n_{1} - \frac{1}{2}) (n_{2} - \frac{1}{2}) \dots (n_{2 r - 1} - \frac{1}{2}) 3^{M^{'}}} = \frac{2^{2 r - 1}}{(2 r - 1)!} \frac{(\log (\frac{\sqrt{2} + \sqrt{6}}{2}))^{2 r - 1}}{\sqrt{3}} \\ \sum_{0 < n_{1} < \dots < n_{2 r - 1}} \frac{1}{(n_{1} - \frac{1}{2}) (n_{2} - \frac{1}{2}) \dots (n_{2 r - 1} - \frac{1}{2}) 5^{M^{'}}} = \frac{2^{2 r - 1}}{(2 r - 1)!} \frac{(\log (ϕ))^{2 r - 1}}{\sqrt{5}} \end{aligned}$

$\begin{aligned} \sum_{0 \leq n < m} \frac{1}{m (n + m)} \frac{(\binom{2 m + 4 n}{m + 2 n}) (\binom{2 m}{m})}{2^{6 n} (\binom{2 m}{m})} = \frac{π^{3}}{6 Γ (\frac{3}{4})^{4}} \\ \sum_{0 \leq n < m < l} \frac{2^{2 l}}{m (n + m) (l + 2 n) (l + 3 n)} \frac{(\binom{2 m + 4 n}{m + 2 n})}{(\binom{2 l + 4 n}{l + 2 n}) (\binom{2 m}{m})} = \frac{π^{5}}{24 Γ (\frac{3}{4})^{4}} \end{aligned}$

$\begin{aligned} \sum_{n_{1} = 1}^{\infty} \sum_{n_{2} = 1}^{\infty} \sum_{n_{3} = 1}^{\infty} \frac{(n_{1} - 1)! (n_{2} - 1)! (n_{3} - 1)!}{(n_{1} + n_{2} + n_{3})!} = \sum_{n = 1}^{\infty} \sum_{m = 1}^{\infty} \frac{1}{n + m} \frac{(n - 1)! (m - 1)!}{(n + m)!} \\ \sum_{0 < n_{1}, n_{2}, \dots, n_{r}} \frac{(n_{1} - 1)! (n_{2} - 1)! \dots (n_{r} - 1)!}{(n_{1} + n_{2} + \dots + n_{r})!} = \sum_{0 < n_{1}, n_{2}, \dots, n_{r - 1}} \frac{1}{n_{1} + n_{2} + \dots + n_{r - 1}} \frac{(n_{1} - 1)! (n_{2} - 1)! \dots (n_{r - 1} - 1)!}{(n_{1} + n_{2} + \dots + n_{r - 1})!} \\ \sum_{0 < n_{1} . n_{2} . n_{3}} \frac{n_{1}! n_{2}! n_{3}!}{(n_{1} + n_{2} + n_{3})!} \frac{(\binom{2 n_{1}}{n_{1}}) (\binom{2 n_{2}}{n_{2}}) (\binom{2 n_{3}}{n_{3}})}{(\binom{2 n_{1} + 2 n_{2} + 2 n_{3}}{n_{1} + n_{2} + n_{3}})} \frac{n_{1} + n_{2} + n_{3} - \frac{1}{2}}{(n_{1} - \frac{1}{2}) (n_{2} - \frac{1}{2}) (n_{3} - \frac{1}{2})} = \sum_{0 < n, m} \frac{n! m!}{(n + m)!} \frac{(\binom{2 n}{n}) (\binom{2 m}{m})}{(\binom{2 n + 2 m}{n + m})} \frac{n + m - \frac{1}{2}}{(n - \frac{1}{2}) (m - \frac{1}{2}) (n + m - 1)} \end{aligned}$

AMZVメモ
$\begin{aligned} ζ (\bar{1}) & = - \ln 2 \\ ζ (\bar{1}, \bar{1}) & = \frac{1}{2} \ln^{2} 2 - \frac{1}{2} ζ (2) \\ ζ (\bar{1}, \bar{1}, \bar{1}) & = - \frac{1}{6} \ln^{3} 2 + \frac{π^{2}}{12} \ln 2 - \frac{1}{4} ζ (3) \\ ζ (\bar{1}, \bar{1}, \bar{1}, \bar{1}) & = \frac{1}{24} \ln^{4} 2 - \frac{π^{2}}{24} \ln^{2} 2 + \frac{5}{16} ζ (3) \ln 2 + \frac{π^{4}}{1440} \\ ζ (1, \bar{1}, 1, \bar{1}) & = \frac{1}{24} \ln^{4} 2 + \frac{π^{4}}{720} - \frac{5}{32} ζ (3) \ln 2 \end{aligned}$

$\begin{aligned} \sum_{\begin{array}{c} n < m \\ (n, m) \in (Z ∖ {0})^{2} \end{array}} \frac{1}{n^{3} m^{3}} = - ζ (6) \\ \sum_{\begin{array}{c} n < m \\ (n, m) \in (Z ∖ {0})^{2} \end{array}} \frac{1}{n^{2} m^{4}} = \frac{π^{6}}{378} \end{aligned}$

$\begin{aligned} \sum_{\begin{array}{c} n_{1} < n_{2} < \dots < n_{r} \\ (n_{1}, n_{2}, \dots, n_{r}) \in (Z ∖ {0})^{r} \end{array}} \frac{1}{n_{1}^{2} n_{2}^{2} \dots n_{r}^{2}} = \frac{2^{2 r + 1}}{(2 r + 2)!} π^{2 r} \\ \sum_{\begin{array}{c} n_{1} < n_{2} < \dots < n_{r} \\ (n_{1}, n_{2}, \dots, n_{r}) \in (Z ∖ {0})^{r} \end{array}} \frac{1}{n_{1}^{4} n_{2}^{4} \dots n_{r}^{4}} = \frac{2^{6 r + 4} + 2^{4 r + 3} (- 1)^{r}}{(4 r + 4)!} π^{4 r} \end{aligned}$

$\begin{aligned} \sum_{\begin{array}{c} n_{1} < n_{2} < \dots < n_{2 r - 1} \\ (n_{1}, n_{2}, \dots, n_{2 r - 1}) \in (Z ∖ {0})^{2 r - 1} \end{array}} \frac{(- 1)^{n_{1} + n_{2} + \dots + n_{2 r - 1}}}{n_{1}^{2} n_{2}^{2} \dots n_{2 r - 1}^{2}} = \frac{(- 1)^{r} 2^{2 r} π^{4 r - 2}}{(4 r)!} \\ \sum_{\begin{array}{c} n_{1} < n_{2} < \dots < n_{2 r} \\ (n_{1}, n_{2}, \dots, n_{2 r}) \in (Z ∖ {0})^{2 r} \end{array}} \frac{(- 1)^{n_{1} + n_{2} + \dots + n_{2 r}}}{n_{1}^{2} n_{2}^{2} \dots n_{2 r}^{2}} = \frac{2 π^{4 r}}{(4 r + 2)!} \end{aligned}$

$\begin{aligned} \sum_{\begin{array}{c} n_{1} < n_{2} < \dots < n_{r} \\ (n_{1}, n_{2}, \dots, n_{r}) \in Z^{r} \end{array}} \frac{1}{(n_{1} + \frac{1}{2})^{2} (n_{2} + \frac{1}{2})^{2} \dots (n_{r} + \frac{1}{2})^{2}} = \frac{2^{2 r - 1}}{(2 r)!} π^{2 r} \\ \sum_{\begin{array}{c} n_{1} < n_{2} < \dots < n_{r} \\ (n_{1}, n_{2}, \dots, n_{r}) \in Z^{r} \end{array}} \frac{1}{(n_{1} + \frac{1}{2})^{4} (n_{2} + \frac{1}{2})^{4} \dots (n_{r} + \frac{1}{2})^{4}} = \frac{2^{6 r - 2} + 2^{4 r - 1} (- 1)^{r}}{(4 r)!} π^{4 r} \end{aligned}$

$\begin{aligned} \frac{1}{\sqrt{2} \cos \frac{π}{4} (1 - x)} = \frac{1}{\sin \frac{π x}{4} + \cos \frac{π x}{4}} = 1 + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + \dots \\ a_{n} = \sum_{\begin{array}{c} w t (k) = n \end{array}} t (k) \\ k は \bar{1}, 2, \bar{3}, 4, \dots で構成されるインデックス \end{aligned}$

$\begin{aligned} \sum_{0 \leq n_{1}} \frac{1}{(2 n_{1} + 1)} \frac{{(\binom{2 n_{1}}{n_{1}})}^{2}}{2^{4 n_{1}}} & = \frac{4}{π} β (2) \\ \sum_{0 \leq n_{1} < n_{2}} \frac{1}{(2 n_{1} + 1)^{2} (2 n_{2} + 1)} \frac{{(\binom{2 n_{2}}{n_{2}})}^{2}}{2^{4 n_{2}}} & = \frac{π}{2} β (2) - \frac{4}{π} β (4) \\ \sum_{0 \leq n_{1} < n_{2} < n_{3}} \frac{1}{(2 n_{1} + 1)^{2} (2 n_{2} + 1)^{2} (2 n_{3} + 1)} \frac{{(\binom{2 n_{3}}{n_{3}})}^{2}}{2^{4 n_{3}}} & = \frac{π^{3}}{96} β (2) - \frac{π}{2} β (4) + \frac{4}{π} β (6) \end{aligned}$
$\begin{aligned} \sum_{0 \leq n_{1} < \dots < n_{r + 1}} \frac{1}{(2 n_{1} + 1)^{2} \dots (2 n_{r} + 1)^{2} (2 n_{r + 1} + 1)} \frac{{(\binom{2 n_{r + 1}}{n_{r + 1}})}^{2}}{2^{4 n_{r + 1}}} = \frac{4}{π} \sum_{k = 0}^{r} β (2 k + 2) \frac{(- 1)^{k} π^{2 r - 2 k}}{2^{2 r - 2 k} (2 r - 2 k)!} \end{aligned}$

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

$\frac{1}{4} \sum_{0 < n} \frac{2^{2 n}}{n^{4} (\binom{2 n}{n})} + 2 \sum_{0 \leq n < m} \frac{1}{(2 n + 1)^{3} (2 m) (2 m + 1)} = \frac{π^{2}}{4} \ln^{2} 2$

$\sum_{0 < n} \frac{2^{2 n}}{n^{3} (\binom{2 n}{n})} = 3 \sum_{0 < n} \frac{2^{2 n}}{n^{3}} \frac{(n!)^{3}}{(3 n)!} + 2 \sum_{0 < n < m} \frac{2^{2 n}}{n m^{2} (\binom{2 n}{n})} \frac{(2 m)! n!}{(2 m + n)!} = π^{2} \ln 2 - \frac{7}{2} ζ (3)$

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

$\begin{aligned} ζ (3) & = \sum_{n = 1}^{\infty} \sum_{m = 1}^{\infty} \frac{1}{n m^{2}} \frac{n! m!}{(n + m)!} \\ = 3 \sum_{n = 1}^{\infty} \sum_{m = 1}^{\infty} \frac{1}{n m^{2}} (\frac{n! m!}{(n + m)!})^{2} \\ ζ (3) & = 6 \sum_{\begin{array}{c} 0 < n_{1} < n_{2} \\ 0 < m \end{array}} \frac{1}{n_{1} n_{2} m} (\frac{n_{2}! m!}{(n_{2} + m)!})^{2} + 3 \sum_{\begin{array}{c} 0 < n_{1} < n_{2} \\ 0 < m \end{array}} \frac{1}{n_{1} n_{2}^{2}} (\frac{n_{2}! m!}{(n_{2} + m)!})^{2} \\ ζ (3) & = 2 \sum_{\begin{array}{c} 0 < n_{1} < n_{2} \\ 0 < m \end{array}} \frac{1}{n_{1} n_{2} m} (\frac{n_{2}! m!}{(n_{2} + m)!})^{2} + \sum_{\begin{array}{c} 0 < n_{1} < n_{2} \\ 0 < m \end{array}} \frac{1}{n_{1} m^{2}} (\frac{n_{2}! m!}{(n_{2} + m)!})^{2} \end{aligned}$

$ζ (3) + \sum_{\begin{matrix} 0 < n \\ 0 < m \end{matrix}} \frac{1}{n^{3}} (\frac{n! m!}{(n + m)!})^{2} = 2 \sum_{\begin{matrix} 0 < n \\ 0 < m \end{matrix}} \frac{1}{n^{2}} (\frac{n! m!}{(n + m)!})^{2} + \sum_{\begin{matrix} 0 < n \\ 0 < m \end{matrix}} \frac{1}{n m} (\frac{n! m!}{(n + m)!})^{2}$

$\begin{aligned} \frac{1}{1^{2 k}} - \frac{1}{3^{2 k}} - \frac{1}{5^{2 k}} + \frac{1}{7^{2 k}} + \frac{1}{9^{2 k}} - \frac{1}{11^{2 k}} - \frac{1}{13^{2 k}} + \dots = \frac{π}{2 \sqrt{2}} t^{⋆} ({\bar{1}}^{2 k - 1}) \\ \frac{1}{1^{2 k + 1}} + \frac{1}{3^{2 k + 1}} - \frac{1}{5^{2 k + 1}} - \frac{1}{7^{2 k + 1}} + \dots = \frac{π}{2 \sqrt{2}} t^{⋆} ({\bar{1}}^{2 k}) \end{aligned}$

$ζ^{⋆} ({3}^{k}) = \frac{3 π}{4} \sum_{n = 0}^{\infty} \frac{1}{(2 n + 1)^{3 k - 1}} \frac{(\binom{2 n}{n})}{2^{4 n}} \prod_{m = n + 1}^{\infty} \frac{1}{(1 + \frac{3 (2 n + 1)^{2}}{(2 m + 1)^{2}})} - \frac{3}{8} \sum_{n = 1}^{\infty} \frac{1}{n^{3 k} (\binom{2 n}{n})} \prod_{m = n + 1}^{\infty} \frac{1}{(1 + \frac{3 n^{2}}{m^{2}})}$

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

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

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

$2 π^{2} \log^{2} 2 = \sum_{0 < n_{1} < n_{2} < n_{3}} \frac{(\binom{2 n_{1}}{n_{1}})}{n_{1} 2^{2 n_{1}}} \frac{1}{n_{2}} \frac{2^{2 n_{3}}}{n_{3}^{2} (\binom{2 n_{3}}{n_{3}})} + 2 \sum_{0 < n_{1} < n_{2} < n_{3}} \frac{1}{n_{1}} \frac{(\binom{2 n_{2}}{n_{2}})}{n_{2} 2^{2 n_{2}}} \frac{2^{2 n_{3}}}{n_{3}^{2} (\binom{2 n_{3}}{n_{3}})}$
$14 ζ (1, 3) - 16 t (1, 3) = \sum_{0 < n_{1} < n_{2} < n_{3}} \frac{1}{n_{1}} \frac{2^{2 n_{2}}}{n_{2}^{2} (\binom{2 n_{2}}{n_{2}})} \frac{(\binom{2 n_{3}}{n_{3}})}{n_{3} 2^{2 n_{3}}} + 2 \sum_{0 < n_{1} < n_{2} \leq n_{3}} \frac{1}{n_{1}} \frac{2^{2 n_{2}}}{n_{2}^{2} (\binom{2 n_{2}}{n_{2}})} \frac{(- 1)^{n_{2} + n_{3}}}{n_{2} + n_{3}}$
$14 ζ (3) = \sum_{0 < n \leq m} \frac{1}{(n - \frac{1}{2})} \frac{2^{2 m}}{m^{2} (\binom{2 m}{m})}$
$16 t (1, 3) = \sum_{0 < n_{1} < n_{2} \leq n_{3}} \frac{(\binom{2 n_{1}}{n_{1}})}{n_{1} 2^{2 n_{1}}} \frac{1}{n_{2} - \frac{1}{2}} \frac{2^{2 n_{3}}}{n_{3}^{2} (\binom{2 n_{3}}{n_{3}})} - \sum_{0 < n_{1} < n_{2} < n_{3}} \frac{(\binom{2 n_{1}}{n_{1}})}{n_{1} 2^{2 n_{1}}} \frac{1}{n_{2}} \frac{2^{2 n_{3}}}{n_{3}^{2} (\binom{2 n_{3}}{n_{3}})}$

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

$Ξ (k_{1}, k_{2}, \dots, k_{r}) := \sum_{0 < n_{1} < \dots < n_{r}} \frac{1}{n_{1}^{k_{1}} \dots n_{r - 1}^{k_{r - 1}} {n_{r}}^{k_{r}}} \frac{(a)_{n_{r - 1}}}{n_{r - 1}!} \frac{n_{r}!}{(a)_{n_{r}}}$
$特に r = 1 のとき, Ξ (k) := \sum_{n = 1}^{\infty} \frac{n!}{n^{k} (a)_{n}}$
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)
$Ξ (k) = \sum_{k \in I_{r} (k, r)} Ξ (k)$

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

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

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

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

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

$\begin{aligned} \frac{2^{2 a}}{(\binom{2 a}{a})} \sum_{a \leq n_{1} \leq \dots \leq n_{r}} \frac{1}{n_{1} (n_{1} + \frac{1}{2}) \dots n_{r} (n_{r} + \frac{1}{2})} \frac{(\binom{2 n_{r}}{n_{r}})}{2^{2 n_{r}}} \\ = \sum_{0 < n_{1} \leq n_{2} < \dots < n_{2 r - 1} \leq n_{2 r}} \frac{1}{n_{1}} \frac{2^{2 n_{1}}}{(\binom{2 n_{r}}{n_{r}})} \frac{1}{n_{2} + \frac{1}{2}} \frac{(\binom{2 n_{2}}{n_{2}})}{2^{2 n_{2}}} \dots \frac{1}{n_{2 r - 1}} \frac{2^{2 n_{2 r - 1}}}{(\binom{2 n_{2 r - 1}}{n_{2 r - 1}})} \frac{1}{n_{2 r} + \frac{1}{2}} \frac{(\binom{2 n_{2 r}}{n_{2 r}})}{2^{2 n_{2 r}}} \frac{n_{2 r}! a!}{(n_{2 r} + a)!} \end{aligned}$

$β_{n} := \frac{(\binom{2 n}{n})}{2^{2 n}}$
$\begin{aligned} \sum_{0 \leq n < n_{1} < \dots < n_{r}} β_{n}^{3} \frac{1}{n_{1}^{3} \dots n_{r}^{3} β_{n_{r}}^{3}} = \sum_{0 \leq n < n_{1} < \dots < n_{2 r - 1}} β_{n}^{3} \frac{1}{n_{1}^{2} β_{n_{1}}} \frac{β_{n_{2}}}{n_{2}} \dots \frac{1}{n_{2 r - 3}^{2} β_{n_{2 r - 3}}} \frac{β_{n_{2 r - 2}}}{n_{2 r - 2}} \frac{1}{n_{2 r - 1}^{3} β_{n_{2 r - 1}}^{3}} \end{aligned}$

$k^{\lor}$ :Hoffman双対インデックス
$\begin{aligned} H_{N}^{⋆} (k; x) := \sum_{1 \leq n_{1} \leq \dots \leq n_{a} \leq N} \frac{1}{(n_{1} - x)^{k_{1}} \dots (n_{a} - x)^{k_{a}}} \frac{(- 1)^{n_{a} - 1} (1 - x)_{N}}{(1 - x)_{n_{a}} (N - n_{a})!} \\ G_{N}^{⋆} (k; x) := \sum_{1 \leq n_{1} \leq \dots \leq n_{a} \leq N} \frac{1}{(n_{1} - x)^{k_{1}} \dots (n_{a} - x)^{k_{a}}} \frac{(1 - x)_{n_{1} - 1}}{(n_{1} - 1)!} \end{aligned}$
このとき,次が成立する.
$H_{N}^{⋆} (k; x) = G_{N}^{⋆} (k^{\lor}; x)$

$k = (k_{1}, \dots, k_{r})$ :許容インデックス
$ϵ = (ϵ_{1}, \dots, ϵ_{r}) \in {0, 1}^{r}$
$ν_{a, b} (\begin{matrix} ϵ \\ k \end{matrix}) := \sum_{0 = n_{0} < n_{1} < \dots < n_{r}} \prod_{i = 1}^{r} \frac{1}{n_{i}^{k_{i}}} (\frac{n_{i}^{2}}{n_{i}^{2} - a^{2}} \frac{(1 - b)_{n_{i - 1}}}{n_{i - 1}!} \frac{n_{i - 1}!^{2}}{(1 - a)_{n_{i - 1}} (1 + a)_{n_{i - 1}}} \frac{n_{i}!}{(1 - b)_{n_{i}}} \frac{(1 - a)_{n_{i}} (1 + a)_{n_{i}}}{n_{i}!^{2}})^{ϵ_{i}}$
このとき,正整数 $a_{1}, \dots, a_{s}, b_{1}, \dots, b_{s}$ と $(ϵ_{1}, \dots, ϵ_{r}) \in {0, 1}^{r}$ について次が成立する。
$ν_{a, b} (\begin{matrix} {0}^{a_{1} - 1}, & ϵ_{1}, & \dots, & {0}^{a_{s} - 1}, & ϵ_{s} \\ {1}^{a_{1} - 1}, & b_{1} + 1, & \dots, & {1}^{a_{s} - 1}, & b_{s} + 1 \end{matrix}) = ν_{a, b} (\begin{matrix} {0}^{b_{s} - 1}, & ϵ_{s}, & \dots, & {0}^{b_{1} - 1}, & ϵ_{1} \\ {1}^{b_{s} - 1}, & a_{s} + 1, & \dots, & {1}^{b_{1} - 1}, & a_{1} + 1 \end{matrix})$

$k^{'} := ((k_{↑})^{†})_{↓}$
$ζ_{α, β, γ, δ} (k) := \frac{Γ (β) Γ (α + γ)}{Γ (α + β + γ)} \sum_{0 \leq n_{1} < \dots < n_{r}} \frac{(1 - δ)_{n_{1}}}{n_{1}!} \frac{1}{(n_{1} + γ)^{k_{1}} \dots (n_{r} + γ)^{k_{r}}} \frac{(α + γ)_{n_{r}}}{(α + β + γ)_{n_{r}}}$
このとき
$ζ_{α, β, γ, δ} (k) = ζ_{δ, γ, β, α} (k^{'})$

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

$\sum_{0 \leq m \leq n} \frac{{(\binom{2 m}{m})}^{2}}{2^{4 m}} \frac{1}{(2 n + 1)^{2}} = \frac{7}{2 π} ζ (3)$
$\sum_{0 \leq m \leq n} \frac{{(\binom{2 m}{m})}^{2}}{2^{4 m}} \frac{1}{(2 n + 1)^{3}} = \frac{4}{π} β (2)^{2}$

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

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

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

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

$\sum_{n = 1}^{\infty} \frac{2^{6 n}}{n^{3} {(\binom{2 n}{n})}^{3}} (\sum_{m = 0}^{n - 1} \frac{{(\binom{2 m}{m})}^{2}}{2^{4 m}})^{2} - π \sum_{n = 1}^{\infty} \frac{{(\binom{2 n}{n})}^{3}}{2^{6 n}} \sum_{m = 1}^{n} \frac{2^{2 m}}{m^{2} (\binom{2 m}{m})} = \frac{8 π^{2}}{Γ (\frac{3}{4})^{4}} β (2)$

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

$7 ζ (3) = \sum_{0 \leq n_{1} < n_{2}} \frac{1}{n_{1} + \frac{1}{2}} \frac{3 n_{2} - 1}{n_{2}^{3}} \frac{2^{4 n_{2}}}{{(\binom{2 n_{2}}{n_{2}})}^{3}} - \sum_{0 < n} \frac{1}{n^{3}} \frac{2^{4 n}}{{(\binom{2 n}{n})}^{3}}$

$\frac{π^{4}}{6} = \sum_{0 \leq n_{1} < n_{2}} \frac{1}{(n_{1} + \frac{1}{2})^{2}} \frac{3 n_{2} - 1}{n_{2}^{3}} \frac{2^{4 n_{2}}}{{(\binom{2 n_{2}}{n_{2}})}^{3}} - 4 \sum_{0 < n_{1} < n_{2}} \frac{1}{n_{1}^{2}} \frac{3 n_{2} - 1}{n_{2}^{3}} \frac{2^{4 n_{2}}}{{(\binom{2 n_{2}}{n_{2}})}^{3}}$

$\frac{π^{4}}{6} = \sum_{0 < m < n} \frac{1}{m} \frac{2^{4 n}}{n^{3} {(\binom{2 n}{n})}^{2}} + \sum_{0 < m < n} \frac{(\binom{2 m}{m})}{m^{2}} \frac{3 n - 1}{n^{3}} \frac{2^{4 n}}{{(\binom{2 n}{n})}^{3}} + 3 \sum_{0 < n_{1} < n_{2} < n_{3}} \frac{1}{n_{1}} \frac{(\binom{2 n_{2}}{n_{2}})}{n_{2}} \frac{3 n_{3} - 1}{n_{3}^{3}} \frac{2^{4 n_{3}}}{{(\binom{2 n_{3}}{n_{3}})}^{3}}$

$A (n) := \sin (\frac{π}{4} + \frac{n π}{2})$
$B (n) := \cos (\frac{π}{4} + \frac{n π}{2})$
$t^{⋆} (k_{1}, \dots, \bar{k_{i}}, \dots, k_{r}) := \sum_{0 \leq n_{1} \leq \dots \leq n_{r}} \frac{(- 1)^{n_{i}}}{(2 n_{1} + 1)^{k_{1}} \dots (2 n_{r} + 1)^{k_{r}}}$
$ζ^{⋆} (k_{1}, \dots, \bar{k_{i}}, \dots, k_{r}) := \sum_{0 < n_{1} \leq \dots \leq n_{r}} \frac{(- 1)^{n_{i} - 1}}{n_{1}^{k_{1}} \dots n_{r}^{k_{r}}}$

$\begin{array}{r} \end{array}$

$\frac{π}{4} t^{⋆} ({\bar{1}}^{2 k - 1}) = \sum_{n = 0}^{\infty} \frac{B (n)}{(2 n + 1)^{2 k}}$
$\frac{π}{4} t^{⋆} ({\bar{1}}^{2 k}) = \sum_{n = 0}^{\infty} \frac{A (n)}{(2 n + 1)^{2 k + 1}}$
$\begin{array}{r} \end{array}$

$\frac{π}{12} t^{⋆} ({\bar{3}}^{2 k - 1}) = \sum_{n = 0}^{\infty} \frac{B (n)}{(2 n + 1)^{6 k - 2}} \frac{1}{\cosh (\frac{\sqrt{3} π}{4} (2 n + 1)) + B (n)}$
$\frac{π}{12} t^{⋆} ({\bar{3}}^{2 k}) = \sum_{n = 0}^{\infty} \frac{A (n)}{(2 n + 1)^{6 k + 1}} \frac{1}{\cosh (\frac{\sqrt{3} π}{4} (2 n + 1)) + B (n)}$
$\begin{matrix} \end{matrix}$
$β_{n} := \frac{(\binom{2 n}{n})}{2^{2 n}}$
$ζ^{⋆} ({\bar{1}}^{2 k - 1}) = \sum_{n = 1}^{\infty} \frac{\sqrt{2} B (n - 1)}{n^{2 k - 1}} β_{[\frac{n}{2}]}$
$ζ^{⋆} ({\bar{1}}^{2 k}) = \sum_{n = 1}^{\infty} \frac{\sqrt{2} A (n - 1)}{n^{2 k}} β_{[\frac{n}{2}]}$

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

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

$\begin{aligned} \sum_{0 < n} (\frac{1}{n} - 3 n (\binom{2 n}{n}) \sum_{n < m} \frac{1}{m^{2} (\binom{2 m}{m})}) & = \frac{8}{3 \sqrt{3}} \sum_{0 < n} \frac{\sin (\frac{π n}{3})}{n^{2}} \\ \sum_{0 < n} (\frac{1}{n^{2}} - 3 (\binom{2 n}{n}) \sum_{n < m} \frac{1}{m^{2} (\binom{2 m}{m})}) & = \frac{π^{2}}{3} - \frac{12}{3 \sqrt{3}} \sum_{0 < n} \frac{\sin (\frac{π n}{3})}{n^{2}} \end{aligned}$

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

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

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

$\frac{π}{Γ (\frac{1}{4})^{4}} = \frac{1}{2^{10} \cdot 3} \sum_{0 \leq n} (- 1)^{n} \frac{28672 n^{5} - 14848 n^{4} + 1600 n^{3} + 32 n^{2} - 8 n - 1}{(n - \frac{1}{4})^{3}} \frac{1}{3^{3 n}} \frac{(\frac{1}{4})_{n}^{2}}{(\frac{7}{12})_{n} (\frac{11}{12})_{n}} \frac{{(\binom{4 n}{2 n})}^{3}}{2^{12 n}}$
$\frac{π}{Γ (\frac{3}{4})^{4}} = \frac{1}{2^{10} \cdot 3^{3}} \sum_{0 \leq n} (- 1)^{n} \frac{28672 n^{5} + 56832 n^{4} + 43584 n^{3} + 16000 n^{2} + 2760 n + 171}{(n + \frac{1}{2})^{3} (n + \frac{1}{12}) (n + \frac{5}{12})} \frac{1}{3^{3 n}} \frac{(\frac{3}{4})_{n}^{2}}{(\frac{1}{12})_{n} (\frac{5}{12})_{n}} \frac{{(\binom{4 n}{2 n})}^{3}}{2^{12 n}}$

$\begin{aligned} \int_{0}^{1} (x^{- x})^{{(x^{- x})}^{(x^{- x}) \dots}} d x = \frac{π^{2}}{6} \\ \int_{0}^{1} (x^{x})^{{(x^{x})}^{(x^{x}) \dots}} d x = \frac{π^{2}}{12} \\ \int_{0}^{1} (x^{- \frac{1}{2} x})^{{(x^{- \frac{1}{2} x})}^{(x^{- \frac{1}{2} x}) \dots}} d x = \frac{π^{2}}{6} - (\log 2)^{2} \\ \int_{0}^{1} (x^{- α x})^{{(x^{- α x})}^{(x^{- α x}) \dots}} d x = \frac{1}{α} {Li}_{2} (α) \end{aligned}$

$\int_{0}^{1} \int_{0}^{1} (x^{- y x})^{{(x^{- y x})}^{(x^{- y x}) \dots}} d x d y = ζ (3)$

$- \frac{1}{3 \sqrt{2}} \frac{π}{Γ (\frac{5}{8})^{2} Γ (\frac{7}{8})^{2}} = \sum_{0 \leq m < n} \frac{1}{m + \frac{1}{2}} \frac{(- 1)^{n} (2 n + \frac{1}{2}) {(\binom{2 n}{n})}^{3}}{2^{6 n}}$
$\frac{π}{6} = \sum_{0 \leq m < n} \frac{1}{(m + \frac{1}{2})^{2}} \frac{(- 1)^{n} (2 n + \frac{1}{2}) {(\binom{2 n}{n})}^{3}}{2^{6 n}} + \sum_{0 \leq m < n} \frac{1}{m + \frac{1}{2}} \frac{(- 1)^{n} {(\binom{2 n}{n})}^{3}}{2^{6 n}} + 3 \sum_{0 \leq n_{1} < n_{2} < n_{3}} \frac{1}{(n_{1} + \frac{1}{2}) (n_{2} + \frac{1}{2})} \frac{(- 1)^{n_{3}} (2 n_{3} + \frac{1}{2}) {(\binom{2 n_{3}}{n_{3}})}^{3}}{2^{6 n_{3}}}$

$\frac{1}{3} ζ (2, 3) = 2 \sum_{0 < n_{1} < n_{2}} \frac{(\binom{2 n_{1}}{n_{1}})}{n_{1}^{3} n_{2}^{2} (\binom{2 n_{2}}{n_{2}})} - 3 \sum_{0 < n_{1} < n_{2} < n_{3}} \frac{(\binom{2 n_{2}}{n_{2}})}{n_{1}^{2} n_{2} n_{3}^{2} (\binom{2 n_{3}}{n_{3}})}$

$8 π β (2) - \frac{64}{π} β (4) = 2 \sum_{0 \leq n_{1} < n_{2}} \frac{2^{4 n_{1}}}{(n_{1} + \frac{1}{2})^{4} (\binom{2 n_{1}}{n_{1}})} \frac{(6 n_{2} + 1) {(\binom{2 n_{2}}{n_{2}})}^{3}}{2^{8 n_{2}}} - 3 \sum_{0 \leq n_{1} < n_{2} < n_{3}} \frac{1}{(n_{1} + \frac{1}{2})^{2}} \frac{2^{4 n_{2}}}{(n_{2} + \frac{1}{2})^{2} (\binom{2 n_{2}}{n_{2}})} \frac{(6 n_{3} + 1) {(\binom{2 n_{3}}{n_{3}})}^{3}}{2^{8 n_{3}}}$