Estimation and disproof of tetrational growth[]
The estimation is incorrect.
Proof (using optimized version of a program):
h(x,y,0) = x^2^x
h(x,0,l) = x for all values of l by internal "for" loop
h(x,y,l) = v(x,y,0) where v(x,y,y) = x and v(x,y,i) = v(x,y,i+1) * h(v(x,y,i+1),i,l-1) * h(v(x,y,i+1),i,l) (note: multiplying and not iterating)
harper = h(a,a,11) with a = 9*2^99999999.
Starting evaluation: let's set h(x,i,l) to x^(a(i)) for l >= 1. Then:
v(x,y,y) = x
v(x,y,y-1) = x*h(x,y-1,l-1)*h(x,y-1,l) = x^(2*a(y-1)+1) (because it doesn't depend from l).
For l = 1, v(x,1,0) = x*h(x,0,0)*h(x,0,1) = x*x*x = x^3
v(x,y,y-2) = x^(2*a(y-1)+1)*h(x^(2*a(y-1)+1),y-2,l-1)*h(x^(2*a(y-1)+1),y-2,l) = x^(2*a(y-1)+1)*x^(2*a(y-2)*(2*a(y-1)+1)) = x^((2*a(y-2)+1)*(2*a(y-1)+1))
etc.
v(x,y,0) = x^a_y where a_y is the product of all 2*a_i+1 starting with i = 1 and finishing with y-1. From previous text it's clear that a_1 = 3. Now we have a table:
a_1 = 3
b_2 = 2*3+1 = 7 and a_2 = 3*7 = 21
b_3 = 2*3*7+1 = 43 and a_3 = 21*43 = 903
b_4 = 2*3*7+1 = 1807 and a_4 = 903*1807 = 1631721
and so on. b_n is exactly Sylvester's sequence and a_n is one half less than the half of b_(n+1).
(It is indeed confirmed by direct simulation of the code that for l != 0 f(x,1,l) is x^3, f(x,2,l) is x^21, f(x,3,l) is x^903 and f(x,4,l) is x^1631721. Further f(x,y,l) for fixed x and l are causing overflow.)
It's not difficult to show from here that:
a_1 = 3 and a_n = 2*a_(n-1)^2 + a_(n-1).
By omitting linear term we can get that a_n >= 0.5*6^2^(n-1) (because 3, 2*3^2, 2*2^2*3^4 = 2^3*3^4, 2*2^6*3^8 = 2^7*3^8 and so on, one 2 is lacking from here).
So f(x,y,l) for l != 0 is x^a_y with a_y described above and:
harper = h(9*2^99999999,9*2^99999999,11) = (9*2^99999999)^(0.5*6^2^(9*2^99999999-1)) which is about 2^2^2^2^(10^8+2). This is much less than 2^^(2^2^(10^8+2)). Tetramur (talk) 16:54, 11 September 2022 (UTC)