asymptotic for a sequence coming from multiplicative sets 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.
 A: 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: 


*

*$B_n = \{ a^rb^s \in A_n | r \le i_n, s \le j_n\}$ 

*$C_n = \{ a^rb^s \in A_n | r < i_n, s > j_n\}$ 

*$D_n = \{ a^rb^s \in A_n | r > i_n, s < j_n\}$ 
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
$$  e^{\gamma_1 \sqrt[k]{n} } \le u_n \le e^{\gamma_2 \sqrt[k]{n} } $$
A: A note on the general case, as per frame95's answer. Geometrically, $n$ is  the number of lattice points with fundamental parallelepiped $\prod_{j=1}^k [0,\log a_j]$ within the orthogonal simplex with orthogonal edges $\log u_n$ (minus $1$, if you insist not considering $a_1^0a_2^0\dots a_k^0$ as $u_1$ in the count ). In any case, $n\sim \frac{(\log u_n)^k}{k!\gamma}  $ as $n\to\infty$, with $\gamma:=   \prod_{j=1}^k\log a_j   $ ,   whence $ \log u_n  \sim n^{1/k} (\gamma k!)^{1/k}$. 
Also note that $a_j$ need not be integer numbers, yet "multiplicatively independent".
