Find f(32) given f(x) = f(x-1) + 2^(f(x-1)+1) and f(0) = 5

Question

I would like to find f(32) given:

f(x) = f(x-1) + 2^(f(x-1)+1)

f(0) = 5

I figured that this function grows tetrationally (if that is a word) but I don't really know

Where I got this equation from:

In the collectable trading card game Magic: The Gathering, there are these cards:

Miirym, sentinel wyrm, Astral Dragon, Parallel Lives

If Astral Dragon where to come into play with Miirym and Parallel Lives already on the battlefield and all of the tokens Astral Dragon created where copies of Parallel Lives, you would end up with f(32) Parallel Lives in play.

Edit: I am not looking for a specific answer, just a rough estimate of how large that number is. I know it is larger than 2^^32 bit is it Larger than 2^^33? 2^^34?

Answer 1

I used Hypercalc at the suggestion of Gottfried Helms

The answer is 30 PT (3.553934904655 × 10^20)

Which means a power tower of 10s, 30 high with (3.553934904655 × 10^20) on top ie: 10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^(3.553934904655 × 10^20)

Answer 2

Since $f(0) = 5$, let's substitute $x = 1$ in your definition, in order to get

$$f(1) = f(0) + 2^{f(0) + 1}$$

And since $f(0) = 5$ we get

$$f(1) = 5 + 2^{6} = 69$$

From here we are able to derive every other value for $f(x)$. Say we want to go on with $f(2)$, then $x = 2$ hence

$$f(2) = f(2-1) + 2^{f(2-1)+1} \rightarrow f(2) = f(1) + 2^{f(1) + 1}$$

From before, $f(1) = 69$ hence

$$f(2) = 69 + 2^{70}$$

Just to be clear,

$$2^{70} = 1180591620717411303424$$

and $$f(2) = 1180591620717411303493$$

Going up to $f(32)$ will give you a hugely large result...

Just to say, $f(3)$ will be this large:

$$f(3) = f(2) + 2^{f(2) + 1}$$

That is

$$f(3) = 1180591620717411303493 + 2^{1180591620717411303494}$$

(overflow.)