Growth of the set $\lbrace 2^n+3^m \rbrace$ in the integers Let $A=\lbrace 2^n+3^m \ | \ n,m\in{\mathbb N}\rbrace$, and denote by $a_k$ the $k$-th element of $A$ in order from smallest to greatest ; thus $a_1=2,a_2=3,a_3=4,a_4=5,a_5=7\ldots$. Is it known whether $\frac{a_{k+1}}{a_k}$ has a limit when $k\to\infty$ ?
Weaker question : is it true that $\frac{a_{k+1}}{a_k} \leq \frac{3}{2}$ for every $k$ ? I have checked this for $k\leq 1500$.
My thoughts : if we write $a_k=2^n+3^m$, then $a_k<2^{n+1}+3^m$ so $a_{k+1} \leq 2^{n+1}+3^m \leq 2a_k$, so if the limit exists it is $\leq 2$. Numerical evidence suggests that the $\liminf$ is $1$.
 A: Note that 
$$
\lim_{n \to \infty} \frac{2^n + 3^{m+1}}{2^n + 3^m} = \lim_{n \to \infty} \frac{1 + \frac{3^{m+1}}{2^n}}{1 + \frac{3^m}{2^n}} = 1
$$
Thus, we may conclude that each element $a_k$ is part of a subsequence $\{a_{k_1},a_{k_2}, \dots\}$ in which 
$$
\frac{a_{k_{j + 1}}}{a_{k_j}} \to 1
$$
From this, we may conclude that $a_{k+1}/a_k \to 1$ as desired.  In particular, take $k(j)$ to mean $k_j$, and recall that the $\{a_k\}$ form an increasing sequence.  Note, then, that for any $a_i$ in the stretch $\{a_{k(j)},a_{k(j)+1}, \dots, a_{k(j+1) - 1}\}$, we have
$$
\frac{a_{i+1}}{a_i} \leq \frac{a_{k(j+1)}}{a_{k(j)}}
$$
it follows that $\limsup_{i \to \infty} \frac{a_i}{a_{i+1}} \leq \lim_{j \to \infty} \frac{a_{k(j)}}{a_{k(j+1)}}$ for any sequence $a_{k(j)}$.
A: See, how many numbers $a_k$ are there below some huge $N$? At least this includes all of those for which $2^n\le{N\over2}$ and $3^m\le{N\over2}$ (and probably more), which is already $O(\log^2N)$, which pretty much forbids the existence of a limit > 1.
