Imagine a game played with tokens. You start with nothing. Every turn, you receive $2^k$ tokens, where $k$ is the total number of tokens you already have. How many tokens do you have after the $n$th turn?
(As an example, the first turn you have zero tokens, so you receive one. The second turn you have one token, so you receive two. The third turn you have three tokens, so you receive eight. The fourth turn you have eleven tokens, so you receive 2048.)
I'm curious if there's a closed form for this. I suspect there isn't, so I'd also be satisfied with the asymptotic growth rate (the "big O"). It certainly seems to grow faster than exponential, but I can't figure out how much faster.
My best attempt at a recursive definition so far is $a_0 = 0$, $a_n = a_{n-1} + 2^{a_{n-1}}$, but this is rather inelegant and hard to do anything with.
(This question came up during a game of Magic: the Gathering, using two particular cards: Anointed Procession says that if you would "create" a token (put it into play), you "create" twice that many instead, and Mythos of Illuna creates a token that's a copy of another card. The virtual tabletop crashes on the fourth iteration, and I'm curious how ridiculous this could get if it kept going.)
The recursion is given by $a_0 = 0$ and $a_n = a_{n-1} + 2^{a_{n-1}}$
This doesn't seem to have a closed form and Wolfram doesn't think there is one either [link]
Since $a_{n-1} = o(2^{a_n-1})$, asymptotically, $a_n$ would grow by that function which looks like tetration.
$a(n) = \Omega(2^{a(n-1)})$ i.e. it grows at least as fast as tetration. Aymptotically, since the other $a_{n-1}$ is much smaller, it should hold that $a(n) = \Theta(2^{a(n-1)})$
In any case, the tetration function is not elementary so it should not have a closed form in elementary functions.