# 2 To The 2 To The 2 To The 2 To The Ad Infinitum

$\large 2,2^2,2^{2^2},2^{2^{2^2}},\ldots$ The sequence above can be recursively defined by $a_1=2, a_{k+1}=2^{a_{k}}$

Given a positive integer $n$, a new sequence is created by dividing each term $\{a_k\}$ by $n$ and writing down the remainder of the division. We denote this new sequence by $\{b_{k}\}$.

For a given $n$, it is known that the terms of $\{b_{k}\}$ eventually become constant. Let $f(n)$ denote the index at which the constant values begin, i.e. $f(n)$ is the smallest number $i$ for which $b_{i-1} \neq b_i = b_{i+1} = b_{i+2} = \cdots$.

Find $f(2016)+f(2015)$.

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