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Problem 876

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

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

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

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

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