Problem 905

Now I Know

Three epistemologists, known as $A$, $B$, and $C$, are in a room, each wearing a hat with a number on it. They have been informed beforehand that all three numbers are positive and that one of the numbers is the sum of the other two.

Once in the room, they can see the numbers on each other’s hats but not on their own. Starting with $A$ and proceeding cyclically, each epistemologist must either honestly state “I don’t know my number” or announce “Now I know my number!” which terminates the game.

For instance, if their numbers are $A=2, B=1, C=1$ then $A$ declares “Now I know” at the first turn. If their numbers are $A=2, B=7, C=5$ then “I don’t know” is heard four times before $B$ finally declares “Now I know” at the fifth turn.

Let $F(A,B,C)$ be the number of turns it takes until an epistemologist declares “Now I know”, including the turn this declaration is made. So $F(2,1,1)=1$ and $F(2,7,5)=5$.

Find $\displaystyle \sum_{a=1}^7 \sum_{b=1}^{19} F(a^b, b^a, a^b + b^a)$.

现在我知道了

三位认识论学者$A$、$B$和$C$待在一间房间里，每人戴着一顶写有数的帽子。他们事先被告知，所有三个数都是正数，且其中一个数等于其他两个数之和。

在房间里时，他们可以看到别人帽子上的数，但看不到自己的。从$A$开始，每位认识论学者轮流循环发言，每次发言必须诚实地声明“我不知道我的数”或宣布“现在我知道我的数了！”在后一种情况下，游戏结束。

例如，如果他们帽子上的数分别是$A=2, B=1, C=1$，那么$A$在第一轮就可以宣布“现在我知道了”。如果他们的帽子上的数分别是$A=2, B=7, C=5$，那么最初的四轮发言都是声明“我不知道”，直到$B$最终在第五轮宣布“现在我知道了”。

令$F(A,B,C)$表示直到某位认识论学者首次宣布“现在我知道了”所需的轮数（包括宣布的那一轮），因此$F(2,1,1)=1$，$F(2,7,5)=5$。

求$\displaystyle \sum_{a=1}^7 \sum_{b=1}^{19} F(a^b, b^a, a^b + b^a)$。