Fourth Loneliest Number

Algebra Level 4

Consider the recurrence relation $f_n (x) = x^{f_{n-1}(x)}$ with $f_0(x) = 16$ . Find the number of integer solutions of $n$ in between 1 and 1000 inclusive such that $f_n(\sqrt[4]2) = 16$ .

The answer is 1000.

1 solution

Janardhanan Sivaramakrishnan
Jul 28, 2015

We need to solve for $n$ in the equation

$f_n(\sqrt[4]{2})=(\sqrt[4]{2})^{f_{n-1}(\sqrt[4]{2})}=16=(\sqrt[4]{2})^{16}$

This means that if $f_{n-1}(\sqrt[4]{2})=16$ , then $f_{n}(\sqrt[4]{2})=16$ .

Since, $f_0(x)=16$ for all x, $f_1(\sqrt[4]{2})=16$ and by induction

$n=1,2,3,\cdots,999,1000$ are all valid solutions.

Fourth Loneliest Number

The answer is 1000.

1 solution

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