二項係数の逆数が付いたMZV

360

自分用に適当にまとめてみる。
※記事の中でインデックスの深さを明記しない場合があるが、空気を読んで解釈してほしい。
※過度な一般化はしない。

本題

$\begin{array}{r} Z (k_{}; l_{}) = \sum_{0 = m_{0} < m_{1} < \dots < m_{a} 0 = n_{0} < n_{1} < \dots < n_{b}} \frac{2^{2 m_{a}}}{m^{k} (\binom{2 m_{a}}{m_{a}})} \frac{{(\frac{1}{2})}_{m_{a}} {(\frac{1}{2})}_{n_{b}}}{{(\frac{1}{2})}_{m_{a} + n_{b}}} \frac{1}{{(n - \frac{1}{2})}^{l}} \end{array}$

以下、 $C (m, n) = \frac{{(\frac{1}{2})}_{m} {(\frac{1}{2})}_{n}}{{(\frac{1}{2})}_{m + n}}$ とする。

$\begin{array}{r} Z (k_{↑}; l_{}) = Z (k_{}; l_{\to}) \end{array}$

部分分数分解により、
$\sum_{n_{b} < n_{b + 1}} \frac{1}{n_{b + 1} - \frac{1}{2}} \frac{{(\frac{1}{2})}_{n_{b + 1}}}{{(\frac{1}{2})}_{m_{a} + n_{b + 1}}} = \frac{1}{m_{a}} \frac{{(\frac{1}{2})}_{n_{b}}}{{(\frac{1}{2})}_{m_{a} + n_{b}}}$
両辺に ${(\frac{1}{2})}_{m_{a}}$ を掛けると
$\frac{1}{m_{a}} C (m_{a}, n_{b}) = \sum_{n_{b} < n_{b + 1}} \frac{1}{n_{b + 1} - \frac{1}{2}} C (m_{a}, n_{b + 1})$
より従う。

$Z (k_{\to}; l_{}) = Z (k_{}; l_{↑})$

$\frac{2^{2 n}}{(\binom{2 n}{n})} = \frac{n!}{{(\frac{1}{2})}_{n}}$ に注意すると、部分分数分解により、
$\sum_{m_{a} < m_{a + 1}} \frac{2^{2 m_{a + 1}}}{m_{a + 1} (\binom{2 m_{a + 1}}{m_{a + 1}})} \frac{{(\frac{1}{2})}_{m_{a + 1}}}{{(\frac{1}{2})}_{m_{a + 1} + n_{b}}} = \frac{2^{2 m_{a}}}{(n_{b} - \frac{1}{2}) (\binom{2 m_{a}}{m_{a}})} \frac{{(\frac{1}{2})}_{m_{a}}}{{(\frac{1}{2})}_{m_{a} + n_{b}}}$
両辺に ${(\frac{1}{2})}_{n_{b}}$ を掛けると
$\sum_{m_{a} < m_{a + 1}} \frac{2^{2 m_{a + 1}}}{m_{a + 1} (\binom{2 m_{a + 1}}{m_{a + 1}})} C (m_{a + 1}, n_{b}) = \frac{2^{2 m_{a}}}{(n_{b} - \frac{1}{2}) (\binom{2 m_{a}}{m_{a}})} C (m_{a}, n_{b})$
より従う。

輸送関係式

$\begin{aligned} Z (k_{↑}; l_{}) & = Z (k_{}; l_{\to}) \\ Z (k_{\to}; l_{}) & = Z (k_{}; l_{↑}) \end{aligned}$

$Z (k_{}; \emptyset_{})$ に対して輸送関係式を適用していくことで次が分かる。

$Z (k_{}; \emptyset_{}) = Z (\emptyset_{}; k_{}^{†})$

$Z (k_{}; \emptyset_{}) = \sum_{0 < n_{1} < \dots < n_{r}} \frac{2^{2 n_{r}}}{n^{k} (\binom{2 n_{r}}{n_{r}})}$ ， $Z (\emptyset_{}; k_{}) = \sum_{0 < n_{1} < \dots < n_{r}} \frac{1}{{(n - \frac{1}{2})}^{k}}$ より、最大の変数に二項係数の逆数が付いた級数をMtV (AMZV) で表せたことになる。

計算例

$\begin{aligned} \sum_{0 < m_{1} < m_{2}} \frac{2^{2 m_{2}}}{m_{1} m_{2}^{2} (\binom{2 m_{2}}{m_{2}})} & = \sum_{0 < m_{1} < m_{2}} \frac{2^{2 m_{2}}}{m_{1} m_{2} (\binom{2 m_{2}}{m_{2}})} \frac{1}{m_{2}} C (m_{2}, 0) \\ = \sum_{0 < m_{1} < m_{2}} \frac{2^{2 m_{2}}}{m_{1} m_{2} (\binom{2 m_{2}}{m_{2}})} \sum_{0 < n_{1}} \frac{1}{n_{1} - \frac{1}{2}} C (m_{2}, n_{1}) \\ = \sum_{0 < n_{1}} \frac{1}{n_{1} - \frac{1}{2}} \sum_{0 < m_{1}} \frac{1}{m_{1}} \sum_{m_{1} < m_{2}} \frac{2^{2 m_{2}}}{m_{2} (\binom{2 m_{2}}{m_{2}})} C (m_{2}, n_{1}) \\ = \sum_{0 < n_{1}} \frac{1}{n_{1} - \frac{1}{2}} \sum_{0 < m_{1}} \frac{1}{m_{1}} \frac{2^{2 m_{1}}}{(n_{1} - \frac{1}{2}) (\binom{2 m_{1}}{m_{1}})} C (m_{1}, n_{1}) \\ = \sum_{0 < n_{1}} \frac{1}{{(n_{1} - \frac{1}{2})}^{2}} \sum_{0 < m_{1}} \frac{2^{2 m_{1}}}{m_{1} (\binom{2 m_{1}}{m_{1}})} C (m_{1}, n_{1}) \\ = \sum_{0 < n_{1}} \frac{1}{{(n_{1} - \frac{1}{2})}^{3}} C (0, n_{1}) \\ = 7 ζ (3) \end{aligned}$

応用

次のような級数を考える。
$\begin{aligned} \sum_{0 < m_{1} < m_{2}} \frac{(\binom{2 m_{1}}{m_{1}})}{2^{2 m_{1}}} \frac{1}{m_{1} m_{2}^{2}} \frac{2^{2 m_{2}}}{(\binom{2 m_{2}}{m_{2}})} & = \sum_{0 < m_{1} < m_{2}} \frac{(\binom{2 m_{1}}{m_{1}})}{2^{2 m_{1}}} \frac{1}{m_{1} m_{2}} \frac{2^{2 m_{2}}}{(\binom{2 m_{2}}{m_{2}})} \sum_{0 < n_{1}} \frac{1}{n_{1} - \frac{1}{2}} C (m_{2}, n_{1}) \\ = \sum_{0 < n_{1}} \frac{1}{n_{1} - \frac{1}{2}} \sum_{0 < m_{1}} \frac{(\binom{2 m_{1}}{m_{1}})}{2^{2 m_{1}}} \frac{1}{m_{1}} \frac{2^{2 m_{1}}}{(n_{1} - \frac{1}{2}) (\binom{2 m_{1}}{m_{1}})} C (m_{1}, n_{1}) \\ = \sum_{0 < n_{1}} \frac{1}{{(n_{1} - \frac{1}{2})}^{2}} \sum_{0 < m_{1}} \frac{1}{m_{1}} C (m_{1}, n_{1}) \end{aligned}$
ここで次の補題を示す。

$\sum_{0 < m} \frac{1}{m} C (m, n) = - 2 \sum_{0 < m} \frac{(- 1)^{m}}{2 n - 1 + m}$

$a_{n} := \sum_{0 < m} \frac{1}{m} C (m, n)$ とする。 $1 < n$ のとき、
$\begin{aligned} a_{n - 1} - a_{n} & = \sum_{0 < m} \frac{{(\frac{1}{2})}_{m}}{m} \frac{{(\frac{1}{2})}_{n}}{{(\frac{1}{2})}_{m + n}} \frac{2 m}{2 n - 1} \\ = \frac{2}{2 n - 1} \sum_{0 < m} \frac{{(\frac{1}{2})}_{m} {(\frac{1}{2})}_{n}}{{(\frac{1}{2})}_{m + n}} \\ = \frac{1}{(2 n - 1) (n - 1)} \end{aligned}$
よって $a_{n} = - \frac{1}{(n - 1) (2 n - 1)} + a_{n - 1}$ であるから、これを帰納的に適用して
$a_{n} = a_{1} - \sum_{m = 1}^{n - 1} \frac{1}{m (2 m + 1)}$
定義から $a_{1} = \sum_{0 < m} \frac{1}{m (2 m + 1)}$ なので
$a_{n} = \sum_{n \leq m} \frac{1}{m (2 m + 1)}$
が分かり、結局
$\begin{aligned} \sum_{0 < m} \frac{1}{m} C (m, n) & = \sum_{n \leq m} \frac{1}{m (2 m - 1)} \\ = 2 \sum_{n \leq m} (\frac{1}{2 m} - \frac{1}{2 m + 1}) \\ = 2 \sum_{0 \leq m} \frac{(- 1)^{m}}{2 n + m} \\ = - 2 \sum_{0 < m} \frac{(- 1)^{m}}{2 n - 1 + m} \end{aligned}$

先程の式に補題を適用して、
$\begin{aligned} \sum_{0 < m_{1} < m_{2}} \frac{(\binom{2 m_{1}}{m_{1}})}{2^{2 m_{1}}} \frac{1}{m_{1} m_{2}^{2}} \frac{2^{2 m_{2}}}{(\binom{2 m_{2}}{m_{2}})} & = - 2 \sum_{0 < n_{1}} \frac{1}{{(n_{1} - \frac{1}{2})}^{2}} \sum_{0 < n_{2}} \frac{(- 1)^{n_{2}}}{2 n_{1} - 1 + n_{2}} \\ = - 8 \sum_{0 < n_{1}, n_{2}} \frac{(- 1)^{n_{2}}}{(2 n_{1} - 1)^{2} (2 n_{1} - 1 + n_{2})} \\ = - 4 \sum_{0 < n_{1}, n_{2}} \frac{(1 - (- 1)^{n_{1}}) (- 1)^{n_{2}}}{n_{1}^{2} (n_{1} + n_{2})} \\ = - 4 (ζ (\overset{―}{2}, \overset{―}{1}) - ζ (2, \overset{―}{1})) \\ = π^{2} \ln 2 - \frac{7}{2} ζ (3) \end{aligned}$
となる。これを一般化する。

$\begin{aligned} R (k) & := \sum_{0 < m_{1} < \dots < m_{a}} \frac{(\binom{2 m_{1}}{m_{1}})}{2^{2 m_{1}}} \frac{1}{m^{k}} \frac{2^{2 m_{a}}}{(\binom{2 m_{a}}{m_{a}})} \\ S (k) & := - 2 \sum_{0 < m_{1} < \dots < m_{a} 0 < m} \frac{1}{{(m - \frac{1}{2})}^{k}} \frac{(- 1)^{m}}{2 m_{a} - 1 + m} \end{aligned}$
とする (アルファベットは適当) 。定義から、 $R (k) = ζ (k)$ である。先程の例と同様に計算することで、
$\begin{aligned} R (k, k) & = \sum_{0 < m_{1} < \dots < m_{a}} \frac{1}{{(m - \frac{1}{2})}^{k^{†}}} \sum_{0 < m} \frac{1}{m^{k}} C (m, m_{a}) \\ = \sum_{0 < m_{1} < \dots < m_{a}} \frac{1}{{(m - \frac{1}{2})}^{k^{†}}} \sum_{m_{a} < m_{a + 1} < \dots < m_{a + k - 1}} \frac{1}{(m_{a + 1} - \frac{1}{2}) \dots (m_{a + k - 1} - \frac{1}{2})} \sum_{0 < m} \frac{1}{m} C (m, m_{a + k - 1}) \\ = S (k^{†}, {1}^{k - 1}) \end{aligned}$
$\begin{aligned} S (k) & = - 2 \sum_{0 < m_{1} < \dots < m_{a} 0 < m} \frac{1}{{(m - \frac{1}{2})}^{k}} \frac{(- 1)^{m}}{2 m_{a} - 1 + m} \\ = - 2^{1 + wt (k)} \sum_{0 < m_{1} < \dots < m_{a} 0 < m} \frac{1}{{(2 m - 1)}^{k}} \frac{(- 1)^{m}}{2 m_{a} - 1 + m} \\ = - 2^{1 + wt (k) - dep (k)} \sum_{0 < m_{1} < \dots < m_{a} 0 < m} \frac{(1 - (- 1)^{m})}{m^{k}} \frac{(- 1)^{m}}{m_{a} + m} \\ = 2^{1 + wt (k) - dep (k)} \sum_{0 < m_{1} < \dots < m_{a + 1}} \frac{(1 - (- 1)^{m_{1}}) \dots (1 - (- 1)^{m_{a}}) (- 1)^{m_{a + 1}}}{{m_{1}}^{k_{1}} \dots {m_{a}}^{k_{a}} m_{a + 1}} \end{aligned}$
よって、 $R (k)$ はAMZVで書けることが分かる (どうやらMtVの線形結合になりそうなのだが、示せなかった) 。

計算例 (おまけ)

$\begin{aligned} R (1, 2, 3, 4) & = S (1, 1, 2, 1, 2, 2) \\ = 2^{4} \sum_{0 < m_{1} < m_{2} < m_{3} < m_{4} < m_{5} < m_{6} < m_{7}} \frac{(1 - (- 1)^{m_{1}}) (1 - (- 1)^{m_{2}}) (1 - (- 1)^{m_{3}}) (1 - (- 1)^{m_{4}}) (1 - (- 1)^{m_{5}}) (1 - (- 1)^{m_{6}}) (- 1)^{m_{7}}}{m_{1} m_{2} m_{3}^{2} m_{4} m_{5}^{2} m_{6}^{2} m_{7}} \\ = 16 (ζ (1, 1, 2, 1, 2, 2, \overset{―}{1}) - ζ (\overset{―}{1}, 1, 2, 1, 2, 2, \overset{―}{1}) - ζ (1, \overset{―}{1}, 2, 1, 2, 2, \overset{―}{1}) - ζ (1, 1, \overset{―}{2}, 1, 2, 2, \overset{―}{1}) - ζ (1, 1, 2, \overset{―}{1}, 2, 2, \overset{―}{1}) - ζ (1, 1, 2, 1, \overset{―}{2}, 2, \overset{―}{1}) - ζ (1, 1, 2, 1, 2, \overset{―}{2}, \overset{―}{1}) + ζ (\overset{―}{1}, \overset{―}{1}, 2, 1, 2, 2, \overset{―}{1}) + ζ (\overset{―}{1}, 1, \overset{―}{2}, 1, 2, 2, \overset{―}{1}) + ζ (\overset{―}{1}, 1, 2, \overset{―}{1}, 2, 2, \overset{―}{1}) + ζ (\overset{―}{1}, 1, 2, 1, \overset{―}{2}, 2, \overset{―}{1}) + ζ (\overset{―}{1}, 1, 2, 1, 2, \overset{―}{2}, \overset{―}{1}) + ζ (1, \overset{―}{1}, \overset{―}{2}, 1, 2, 2, \overset{―}{1}) + ζ (1, \overset{―}{1}, 2, \overset{―}{1}, 2, 2, \overset{―}{1}) + ζ (1, \overset{―}{1}, 2, 1, \overset{―}{2}, 2, \overset{―}{1}) + ζ (1, \overset{―}{1}, 2, 1, 2, \overset{―}{2}, \overset{―}{1}) + ζ (1, 1, \overset{―}{2}, \overset{―}{1}, 2, 2, \overset{―}{1}) + ζ (1, 1, \overset{―}{2}, 1, \overset{―}{2}, 2, \overset{―}{1}) + ζ (1, 1, \overset{―}{2}, 1, 2, \overset{―}{2}, \overset{―}{1}) + ζ (1, 1, 2, \overset{―}{1}, \overset{―}{2}, 2, \overset{―}{1}) + ζ (1, 1, 2, \overset{―}{1}, 2, \overset{―}{2}, \overset{―}{1}) + ζ (1, 1, 2, 1, \overset{―}{2}, \overset{―}{2}, \overset{―}{1}) - ζ (\overset{―}{1}, \overset{―}{1}, \overset{―}{2}, 1, 2, 2, \overset{―}{1}) - ζ (\overset{―}{1}, \overset{―}{1}, 2, \overset{―}{1}, 2, 2, \overset{―}{1}) - ζ (\overset{―}{1}, \overset{―}{1}, 2, 1, \overset{―}{2}, 2, \overset{―}{1}) - ζ (\overset{―}{1}, \overset{―}{1}, 2, 1, 2, \overset{―}{2}, \overset{―}{1}) - ζ (\overset{―}{1}, 1, \overset{―}{2}, \overset{―}{1}, 2, 2, \overset{―}{1}) - ζ (\overset{―}{1}, 1, \overset{―}{2}, 1, \overset{―}{2}, 2, \overset{―}{1}) - ζ (\overset{―}{1}, 1, \overset{―}{2}, 1, 2, \overset{―}{2}, \overset{―}{1}) - ζ (\overset{―}{1}, 1, 2, \overset{―}{1}, \overset{―}{2}, 2, \overset{―}{1}) - ζ (\overset{―}{1}, 1, 2, \overset{―}{1}, 2, \overset{―}{2}, \overset{―}{1}) - ζ (\overset{―}{1}, 1, 2, 1, \overset{―}{2}, \overset{―}{2}, \overset{―}{1}) - ζ (1, \overset{―}{1}, \overset{―}{2}, \overset{―}{1}, 2, 2, \overset{―}{1}) - ζ (1, \overset{―}{1}, \overset{―}{2}, 1, \overset{―}{2}, 2, \overset{―}{1}) - ζ (1, \overset{―}{1}, \overset{―}{2}, 1, 2, \overset{―}{2}, \overset{―}{1}) - ζ (1, \overset{―}{1}, 2, \overset{―}{1}, \overset{―}{2}, 2, \overset{―}{1}) - ζ (1, \overset{―}{1}, 2, \overset{―}{1}, 2, \overset{―}{2}, \overset{―}{1}) - ζ (1, \overset{―}{1}, 2, 1, \overset{―}{2}, \overset{―}{2}, \overset{―}{1}) - ζ (1, 1, \overset{―}{2}, \overset{―}{1}, \overset{―}{2}, 2, \overset{―}{1}) - ζ (1, 1, \overset{―}{2}, \overset{―}{1}, 2, \overset{―}{2}, \overset{―}{1}) - ζ (1, 1, \overset{―}{2}, 1, \overset{―}{2}, \overset{―}{2}, \overset{―}{1}) - ζ (1, 1, 2, \overset{―}{1}, \overset{―}{2}, \overset{―}{2}, \overset{―}{1}) + \dots (もう無理)) \end{aligned}$