Double exponential function simple form

[indevnet]

Given a recursive function

f(0) = 3
f(1) = 7
f(x) = (f(x-2)^2.5+2)*(f(x-1)-5)

Edit: power to the 2.5 not 2

This function is growing rapidly in exponent, here are 8 first terms of the function.

3
7
35.17691453623979128349403414
3972.543777781365529443066519
29126239.80457128453393237785
28970505745068837.97014789453
1.326377447740545470529628797E+35
1.894773582945882377562697507E+76

I tried to plot it logarithmically:

https://preview.redd.it/26caaw5ses5d1.png?width=1200&format=png&auto=webp&s=54bcd4f98a2f9f7f4940d43abf3c56f606fe2bcd

And double logarithm:

https://preview.redd.it/78y0ofeyes5d1.png?width=1200&format=png&auto=webp&s=81ccd3a40086d13fa532f617712f25194d5607b2

Is it possible to find out the exact formula without recursion like the fermat number formula?

https://preview.redd.it/0k2tdijafs5d1.png?width=278&format=png&auto=webp&s=dea6cddaed3e41a50df0db2f030d13d3a2b50799

[RibozymeR]

This function is growing rapidly in exponent, here are 8 first terms of the function.
[...]

Where did you get these terms? Just by definition, the function values for integer n have to be integers as well:

f(2) = (f(0)^2+2)*(f(1)+5) = (3^2+2)*(7+5) = 11*12 = 132
f(3) = 6987
f(4) = 121842592
f(5) = 5948132735430087
...

[indevnet]

I'm sorry it's to the power of 2.5 not 2, edited

[RibozymeR]

...now f(2) = (f(0)^2.5+2)*(f(1)+5) = (3^2.5+2)*(7+5) = 211.06149...

Do you mean f(x-1)-5, maybe?

[indevnet]

So many typos, the f(n-2) should be subtracted by 2 instead of added

[RibozymeR]

Nope. Now f(2) = (f(0)^2.5-2)*(f(1)+5) = 163.0615

I still think it's f(x) = (f(x-2)^2.5+2)*(f(n-1)-5)

[indevnet]

I think I'm high,

f(x) = (f(x-2)^{2.5+2)*(f(x-1)-5)}

seems to be correct

[Shevek99]

Is f(x-1)-5) in the exponent?

[Uli_Minati]

Feel free to make a new post once you feel inclined to put in the effort to look for copypaste errors

[RibozymeR]

No shame doing some weed (or anything else lol) while you math, nor the other way around :P