Problem 876

Triplet Tricks

Starting with three numbers $a, b, c$, at each step do one of the three operations:

change $a$ to $2(b + c) - a$;
change $b$ to $2(c + a) - b$;
change $c$ to $2(a + b) - c$.

Define $f(a, b, c)$ to be the minimum number of steps required for one number to become zero. If this is not possible then $f(a, b, c)=0$.

For example, $f(6,10,35)=3$:
$$(6,10,35) \to (6,10,-3) \to (8,10,-3) \to (8,0,-3).$$
However, $f(6,10,36)=0$ as no series of operations leads to a zero number.

Also define $F(a, b)=\sum_{c=1}^\infty f(a,b,c)$. You are given $F(6,10)=17$ and $F(36,100)=179$.

Find $\displaystyle\sum_{k=1}^{18}F(6^k,10^k)$.

三元戏法

从三个数$a, b, c$开始，每一步进行以下三种操作之一：

将$a$替换为$2(b + c) - a$；
将$b$替换为$2(c + a) - b$；
将$c$替换为$2(a + b) - c$。

定义$f(a, b, c)$为使其中一个数变为零所需的最少步数。如果无法做到这一点，则$f(a, b, c)=0$。

例如，$f(6,10,35)=3$：
$$(6,10,35) \to (6,10,-3) \to (8,10,-3) \to (8,0,-3).$$
而$f(6,10,36)=0$，因为不存在使得其中一个数变为零的操作序列。

再定义$F(a, b)=\sum_{c=1}^\infty f(a,b,c)$。已知$F(6,10)=17$，$F(36,100)=179$。

求$\displaystyle\sum_{k=1}^{18}F(6^k,10^k)$。