# Average distance between numbers of the form $2^{a}3^{b}$

I want to order all numbers of the form $2^a3^b$. I need to find the average distance between a random consecutive pair.

For example, in case of a random consecutive pair $2^{n'}$ and $2^{n'+1}$, the distance is $2^{n'}$. Similarly, whats the distance between $2^{a_1}3^{b_1}$ and $2^{a_2}3^{b_2}$ such that there is no number of the form $2^{a}3^{b}$ between them.

Seems like an interesting problem.

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You can approximate how many of these numbers are less than N as the area of the right triangle with base log N/log 2 and height log N/log 3. The average distance should be on the order of N divided by this number. – S. Carnahan Jan 25 '10 at 10:17
Is it sonehow related to the Collatz problem? – Nurdin Takenov Jan 25 '10 at 11:58
Might Størmer's theorem (en.wikipedia.org/wiki/Størmer's_theorem) help here? – sdcvvc Jan 25 '10 at 14:06

More generally, if you fix a set of primes $p_1, \ldots, p_r$, and let $n_1=1 < n_2 < \cdots$ be the set of positive integers composed of primes in this set, then one has (effectively) $$n_{i+1}-n_i > \frac{n_i}{\left( \log n_i \right)^{c_1}}$$ and $$n_{i+1}-n_i < \frac{n_i}{\left( \log n_i \right)^{c_2}},$$ for constants $c_1 \geq r-1$ and $c_2 \leq r-1$. These follow from results on lower bounds for linear forms in logarithms and were proved by Tijdeman in the early 1970s (Compositio).

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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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For the record: I have no plans to post my comment in answer form. – S. Carnahan Jan 25 '10 at 14:28

Judging from some computer calculations $\lim_{n \rightarrow \infty} n/|\{(a,b):2^a \cdot 3^b\}| \rightarrow \infty$. This is the GAP code I used below; you can modify the initial value of n to quite a large value and it will still work.

n:=100000;;
count:=0;;