Let $a,b$ two multiplicatively independant positive integers with $1<a<b$ 
that is $a^mb^n\ne1$ for all $m,n\in\mathbb N\setminus\{0\}$
We sort out the set $E=\{a^mb^n\mid m,n\in \mathbb N\setminus\{0\}\}$ in a sequence $(u\_n)\_{n\in\mathbb N}$ by ordering $E$ by increasing order.
Can one give an asymptotic for $u\_n$?


Thanks in advance for any hint or solution.



asymptotic for a sequence coming from multiplicative sets
I found that $ \log u\_n \sim \sqrt{n} $, but I have not determined precisely the costants.


Let $ u\_n := a^{i\_n} b^{j\_n} $, and $\alpha = \log a / \log b, \ \beta = \alpha^{-1}$.   
 


Let's count how many integers of the form $a^rb^s$ are $< u\_n$, i.e. $|A\_n| := | E \cap [1,u\_n [ \, | $. 
  
 We partition $A\_n$ in three sets $B\_n, C\_n, D\_n$ such that:   



1. $B\_n = \{ a^rb^s \in A\_n | r \le i\_n, s \le j\_n\}$
2. $C\_n = \{ a^rb^s \in A\_n | r < i\_n, s > j\_n\}$
3. $D\_n = \{ a^rb^s \in A\_n | r > i\_n, s < j\_n\}$


Simple manipulations bring to $$ |B\_n| = i\_nj\_n + i\_n + j\_n, \ \ |C\_n| = \sum\_{k=1}^{i\_n} \lfloor j\_n + \alpha k \rfloor, \ \ |D\_n| = \sum\_{k=1}^{j\_n} \lfloor i\_n + \beta k \rfloor $$
So, recalling $|A\_n| = n$ and inequalities on floor-function, we have
$$ n \ge \alpha i\_n^2 + \beta j\_n^2 + 3i\_n j\_n + i\_n (1+ \alpha/2) + j\_n(1+\beta/2) $$ 
$$ n \le \alpha i\_n^2 + \beta j\_n^2 + 3i\_n j\_n + i\_n (2+ \alpha/2) + j\_n(2+\beta/2) $$ 
Passing to limits and using $ \displaystyle \lim\_{n \to \infty} i\_n = j\_n = + \infty$, we have
$$ (\*) \lim\_{n \to \infty} \frac{\alpha i\_n^2 + \beta j\_n^2 + 3i\_nj\_n}{n} = 1$$


Now, note that 
$$ \limsup\_{n \to \infty} \frac{\beta}{n} \left ( \frac{\log u\_n}{\log b} \right )^2 = \limsup\_{n \to \infty} \beta \frac{( j\_n + \alpha i\_n)^2}{n} = \limsup\_{n \to \infty} \frac{\beta j\_n^2 + \alpha i\_n^2 + 2i\_nj\_n}{n} \le 1$$
and that, analogously:
$$ \liminf\_{n \to \infty} \frac{5\beta}{4n} \left ( \frac{\log u\_n}{\log b} \right )^2 = \liminf\_{n \to \infty} \frac{1}{n}( \beta j\_n^2 + \alpha i\_n^2 + 2i\_nj\_n) + \frac{1}{4n} ( \beta j\_n^2 + \alpha i\_n^2 + 2i\_nj\_n) \ge \liminf\_{n \to \infty} \frac{ \beta j\_n^2 + \alpha i\_n^2 + 3i\_nj\_n}{n} = 1$$
where we have used that $x^2 + y^2 \ge 2xy \ \forall x,y \in \mathbb{R}$ .
This means, in conclusion, that $\forall \epsilon > 0 $ it definitely holds:
$$ e^{\sqrt{\gamma\_1(1-\epsilon)n} } \le u\_n \le e^{\sqrt{\gamma\_2(1+\epsilon)n} } $$
where $\gamma\_2 = \log a \log b, \gamma\_1 = (4/5) \gamma\_2$.


Hope it helps, 
Andrea


**Added:** I found a quite more rough method which applies to the general case.
Let $a\_1, \ldots, a\_k >1$ be positive integers. Define $E= \{ a\_1^{\beta\_1} \ldots a\_k^{\beta\_k}: \beta\_1, \ldots, \beta\_k \in \mathbb{N}\_0\}$, and $u\_n$ the $n-th$ term in the non-decreasing list of elements of $E$. Finally, we call $\rho(x) = | E \cap [1,x] |$ . By the fact that
$$ \{ a\_1^{\beta\_1} \ldots a\_k^{\beta\_k}: (\beta\_1, \ldots, \beta\_k) \in [1, (\log\_{a\_k} x )/k ]^k\} \subseteq E \cap [1,x] \subseteq \{ a\_1^{\beta\_1} \ldots a\_k^{\beta\_k}: (\beta\_1, \ldots, \beta\_k) \in [1, \log\_{a\_1}x]^k\} $$
we deduce that 
$$ \frac{(\log x )^k }{ (k\log a\_k)^k } \le \rho(x) \le \frac{(\log x )^k }{ (\log a\_1)^k } $$
Denote by $\gamma\_1 = { k\log a\_k }, \gamma\_2 = \log a\_1$. So, substituting $x = u\_n$ and using $\rho(u\_n) = n$ we have
$$ e^{\gamma\_1 \sqrt[k]{n} } \le u\_n \le e^{\gamma\_2 \sqrt[k]{n} } $$


