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何進法でも成り立つ連分数

157

任意の基数において、
$\frac{1}{1} + \frac{z^{1}}{111} + \frac{z^{1 + 1}}{11111} + \frac{z^{1 + 1 + 1}}{1111111} + \dots = \frac{1}{1 - \frac{1^{1 + 1} 1 z}{111 - \frac{11^{1 + 1} 10 z}{11111 - \frac{111^{1 + 1} 100 z}{1111111 - ⋱}}}}$

等比数列の和の公式により、
$\overset{n}{\overset{⏞}{1 \dots 1}} = \frac{10^{n} - 1}{10 - 1}$
よって、
$\begin{aligned} (左辺) & = \sum_{n = 0}^{\infty} \frac{z^{n}}{\frac{10^{(1 + 1) n + 1} - 1}{10 - 1}} = \sum_{n = 0}^{\infty} \frac{z^{n}}{\frac{10}{10 - 1} 100^{n} - \frac{1}{10 - 1}} \end{aligned}$

https://mathlog.info/articles/3898 の定理1より、
$\sum_{n = 0}^{\infty} \frac{z^{n}}{a r^{n} - b q^{n}} = {\begin{matrix} K \end{matrix}}_{n = 0}^{\infty} \frac{v_{n}}{a r^{n} - b q^{n}}$
${\begin{aligned} v_{0} & = 1 \\ v_{(1 + 1) n - 1} & = - (r q)^{n - 1} {(a r^{n - 1} - b q^{n - 1})}^{1 + 1} z \\ v_{(1 + 1) n} & = - a b (r q)^{n - 1} {(r^{n} - q^{n})}^{1 + 1} z \end{aligned}$

これに $r = 100, q = 1, a = \frac{10}{10 - 1}, b = \frac{1}{10 - 1}$ を代入すると、
${\begin{aligned} v_{0} & = 1 \\ v_{(1 + 1) n - 1} & = - 100^{n - 1} {(\frac{10 \cdot 100^{n - 1}}{10 - 1} - \frac{1}{10 - 1})}^{1 + 1} z & = - 10^{(1 + 1) n - (1 + 1)} {(\frac{10^{(1 + 1) n - 1} - 1}{10 - 1})}^{1 + 1} z \\ v_{(1 + 1) n} & = - \frac{10}{(10 - 1)^{1 + 1}} 100^{n - 1} {(100^{n} - 1)}^{1 + 1} z & = - 10^{(1 + 1) n - 1} {(\frac{10^{(1 + 1) n} - 1}{10 - 1})}^{1 + 1} z \end{aligned}$
$\begin{aligned} v_{n} & = - 10^{n - 1} {(\frac{10^{n} - 1}{10 - 1})}^{1 + 1} z \\ = - 10^{n - 1} {(\overset{n}{\overset{⏞}{1 \dots 1}})}^{1 + 1} z (n \geq 1) \end{aligned}$
よって、
$\begin{aligned} (左辺) & = \sum_{n = 0}^{\infty} \frac{z^{n}}{\frac{10}{10 - 1} 100^{n} - \frac{1}{10 - 1}} = {\begin{matrix} K \end{matrix}}_{n = 0}^{\infty} \frac{v_{n}}{\frac{10}{10 - 1} 100^{n} - \frac{1}{10 - 1}} = {\begin{matrix} K \end{matrix}}_{n = 0}^{\infty} \frac{v_{n}}{\frac{10^{(1 + 1) n + 1} - 1}{10 - 1}} \\ = \frac{1}{1 + {\begin{matrix} K \end{matrix}}_{n = 1}^{\infty} \frac{- 10^{n - 1} {(\overset{n}{\overset{⏞}{1 \dots 1}})}^{1 + 1} z}{\overset{(1 + 1) n + 1}{\overset{⏞}{1 \dots 1}}}} \\ = (右辺) \end{aligned}$