# Inspired by Patrick Corn and Myself

Number Theory Level pending

Let $$a_{n+1} = 2^{a_n}$$ and $$b_{n+1} = 3^{b_n}$$ both for $$n \ge 1.$$

If $$a_1 = 2$$ and $$b_1 = 3$$, then find the minimum $$l$$ and maximum $$k$$ such that

$$a_{n+k}<b_{n}<a_{n+l}$$ for all $$n \geq 2$$

Input your answer as $$k+l$$.

Inspiration

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