以$0$结尾的迭代幂次基的同余速度
Congruence speed of tetration bases ending with $0$
AI总结：
本文针对以$0$结尾（即$10$的倍数）的迭代幂次基，推导出其在任意给定超指数高度下的同余速度公式，填补了该特定情形下同余速度计算的空白。
摘要：
对于每个非负整数$a$和正整数$b$，迭代幂次$^{b}a$的同余速度是指$^{b}a$与$^{b+1}a$相同的最右侧数字个数同$^{b-1}a$与$^{b}a$相同的最右侧数字个数之差。在十进制中，若给定基$a$不是$10$的倍数，随着$b:=b(a)$变得足够大，同余速度的值不再依赖于$b$；否则，对于$b$的任意单位增量，$^{b}a$新增的最右侧零的个数会急剧增加，正因如此，我们此前未描述$a$为$10$的倍数时的同余速度。这篇简短的笔记填补了这一空白，给出了上述$a$值在任意给定超指数高度下的同余速度公式。
英文摘要：
For every non-negative integer $a$ and positive integer $b$, the congruence speed of the tetration $^{b}a$ is the difference between the number of the rightmost digits of $^{b}a$ that are the same as those of $^{b+1}a$ and the number of the rightmost digits of $^{b-1}a$ that are the same as those of $^{b}a$. In the decimal numeral system, if the given base $a$ is not a multiple of $10$, as $b:=b(a)$ becomes sufficiently large, we know that the value of the congruence speed does not depend on $b$ anymore, otherwise the number of the new rightmost zeros of $^{b}a$ drastically increases for any unit increment of $b$ and, for this reason, we have not previously described the congruence speed of $a$ when it is a multiple of $10$. This short note fills the gap by giving the formula for the congruence speed of the mentioned values of $a$ at any given height of the hyperexponent.
机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。
