User blog comment:B1mb0w/Growth Rate of the S Function/@comment-25601061-20160708203814/@comment-10262436-20160714103649

Can we check your results for:

S(n, S(T(1),1,1), 1) = f_{phi(w,0)2}(n)

S(n, S(T(1),n+2,1), 1) = f_{phi(n, phi(w,0)+1)}(n)

I get

S(T(1),1,1) == phi(n-2,0).2   but lets call this phi(w,0).2 for now

S(T(1),3,1) == phi(1,phi(w,0)+1)

S(T(1),3,T(0)) == phi(2,phi(w,0)+1)

S(T(1),3,S(T(0),1,1)) == phi^2(2,phi(w,0)+1)

S(T(1),3,S(T(0),2,1)) == phi(3,phi(w,0)+1)

The function S(T(0),p,q) grows faster than phi^q(p,phi(w,0)+1) until

S(T(1),4,1) == phi(w,1)

Do you agree ?