I am looking for a way to construct an ordered set like
$\{2^3, 2^4, 3^3, 2^5, 2^6, 3^4, 5^3, 2^7...\} = \{8, 16, 27, 32, 64, 81, 125, 128...\}$
Preferably, but not necessarily, with all bases $>1$, all exponents $>2$, and no duplicates (like $4^3 = 2^6$).
A helpful construction would be something like the following so I can proof stuff for the entire set, by induction:
First element = $2^3$. Given element $n$, the element $n+1$ is... or has the properties ...
Is there a known way to do it? Is there a known reason why it would be impossible?
(... hope I have not mistranslated any words from german ...)
A common strategy for this sorry of problem is to start with the numbers $a^b$ where $a+b=5$, ordered, say, in increasing order of $a$, then move on to totals of 6, etc. So your sequence would start $ 2^3,3^2,2^4,3^3,4^2,2^5,...$.
The rule for getting to the next number is then "increase the base and decrease the exponent by 1 unless the exponent is 2 in which case...".
This, of course, counts duplicates, and the rule relies on you knowing the base and the exponent not just the resulting number. I would think a rule that meets these criteria would be pretty difficult to come up with.