co.combinatorics - Average distance between numbers of the form $2^{a}3^{b}$

The point of this answer is simply to repeat Scott's answer in a more visible place. If he wants to post it himself, I'll delete my post:

Let $a_i$ be the sequence of these integers, sorted into order. Let $a_r leq N < a_{r+1}$. We want to estimate

$$frac{1}{r-1} sum (a_{i+1} - a_i) = (a_r-a_1)/(r-1).$$

As Scott explains,

$$r=(1/2) cdot (log N/log 2 + O(1)) cdot (log N/log 3 + O(1)) = (log N)^2/(2 log 2 log 3) + O(log N) .$$

Also, there is clearly a power of $2$ between $N$ and $N/2$, so $a_r sim N$.

So the average distance is $sim N/(log N)^2$. If you work harder, you can probably tighten up the bounds to show that it is $N (2 log 2 log 3)/(log N)^2 (1+O(1/log N))$

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