AI 中文总结
研究1/p数字均值问题，对于1/p关于基b的周期数字展开，在已知偶数长度周期数字均值及特定奇数长度均值的基础上，解决了将其结果推广到任意奇数长度l的问题。
AI 中文摘要
设p≥3为素数，b≥2为整数且p不整除b。则1/p关于基b有周期数字展开。周期长度l是b模p的（乘法）阶。若l为偶数，一个周期数字的均值为(b - 1)l/2。奇数长度l的情况更有趣。若l = (p - 1)/2^m为奇数，之前已给出一个周期数字的均值，该均值涉及广义伯努利数。但不清楚如何将此结果推广到任意奇数长度l。本文解决了这种情况。
英文摘要
Let $p\ge 3$ be a prime and $b\ge 2$ an integer such that $p$ does not divide $b$. Then $1/p$ has a periodic digit expansion with respect to the basis $b$. The length $l$ of the period is the (multiplicative) order of $b$ mod $p$. If $l$ is even, then the mean value of the digits of a period is just $(b-1)l/2$. The case of an odd length $l$ is more interesting. If $l=(p-1)/2^m$ is odd, the mean value of the digits of a period was given previously. This mean value involves generalized Bernoulli numbers. However, it is not clear how this result can be generalized to an arbitrary odd length $l$. In the present note we settle this case.
