A power tower of a number $x$ is typified by 


$$ x^{x^{x^{x^{x^{x^{x^{x^{x^x}}}}}}}}.$$ 


Here, however, we take the liberty of referring to the set $T$ of "$\{2,3\}$-power towers"; i.e., numbers


$$x\_1^{x\_2^{x\_3^{ \cdots\cdots^{x\_k}}}},$$


where each $x\_h$ is $2$ or $3,$ and $k \geq 2.$ Let $T\_2$ be the subset of $T$ consisting of towers rising from $x\_1=2.$ Let $R$ be the sequence of ranks of towers in $T\_2$ when all the towers in $T$ are jointly ranked. For example, $7 \in R$ means that the $7$th smallest element in $T$ is a power of $2$, not of $3$. (The term *jointly ranked* is borrowed from statistics: if the numbers in two or more sets are combined and arranged in nondecreasing order, they are said to be jointly ranked.) 


The first $15$ terms of $R$ are $$1, 2, 4, 7, 8, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29.$$ What are the next terms? 


Note that $T$ can be obtained recursively from $t\_2 = \{2^2,2^3,3^2,3^3\}$ by defining


$$t\_n =2^{t\_{n-1}} \cup 3^{t\_{n-1}}$$


for $n \geq 3;$ then $T$ is the union of the sets $t\_n$ for $n \geq 2.$


For a top-first version of the problem, change $x\_1=2$ to $x\_k=2,$ where $k$ is the height of the tower. Then the first $17$ terms are $$1,3,4,6,10,11,12,15,16,19,20,23,24,25,26,27,28,\ldots.$$ Here, too, the question is: what are the next terms?


Added later: Thanks, Yaakov, you are right, so my question is, what are the positions of the numbers in $T\_2$ in the sequence in the sequence $(1,2,3,\ldots)$. I have the first $30$ positions (or ranks) and would like to see a method for finding more terms.


It may help to see a list of the first $20$ towers ranked:


$$4 = 2^2$$
$$8 = 2^3$$
$$9 = 3^2$$
$$16 = 2^{2^{2}}$$
$$27 = 3^3$$
$$81 = 3^{2^{2}}$$
$$256 = 2^{2^{3}}$$
$$512 = 2^{3^{2}}$$
$$6561 = 3^{2^{3}}$$
$$19683 = 3^{3^{2}}$$ 
Continuing with tuple notation instead of tower notation:
$(2,2,2,2), (3,2,2,2), (2,3,3), (3,3,3), (2,3,2,2), (3,3,2,2), (2,2,2,3), (3,2,2,3), (2,2,3,2), (3,2,3,2), (2,3,2,3).$


My method, so far, has been by computer sort, which reaches overflow pretty quickly. Surely there must be a more insightful method. A related question: what is the position (or rank) of $(2,2,2,2,2,2)?$ 



Power tower made of $2$s and $3$s: too high, too soon?
Let $s\_1=2^2$, $s\_2=2^3$, $s\_3=3^2$ and so on. For $i\ge 5$, it holds that $s\_{i+1}\ge 2s\_i$. This can be proved by induction. Then $s\_{2i+3}=2^{s\_i}$ and $s\_{2i+4}=3^{s\_i}$. In particular, all the remaining elements of $R$ are precisely the odd numbers larger than the ones shown, and the solution of the top-first version of the problem consists of sequences of consecutive numbers doubling in length. The rank of $(2,2,2,2,2,2)$ can be found out simply by enumerating the sequences up to it.


