MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 (ørmer'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).

share|cite|improve this answer

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 $N (2 \log 2 \log 3)/(\log N)^2 (1+O(1/\log N))$

share|cite|improve this answer
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.

for a in [0..LogInt(n,2)] do
  for b in [0..LogInt(n,3)] do
    if(2^a*3^b<=n) then count:=count+1;;

Edit: using the above code I searched Sloane's website and uncovered this. Specifically, these are called 3-smooth numbers. There's formulae there too.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.