User blog comment:Cloudy176/Proving the bound for S(7)/@comment-29743408-20160831082555

I can give a more precise value of the time T taken by the machine on C(2) and the number N of 1s left on the tape.

Let K = 31068918 + ln(98/67)/ln(4).

Then: 4^(4^(4^(4^K))) < N < T < 4^(4^(4^(4^(K+4^(-K-2)))).

Let a be such that T = 10^(10^(10^(10^a))), and let b = K ln(4)/ln(10) + ln(ln(4)/ln(10))/ln(10).

Then a and b agree up to their 18705352nd decimal, that is |a-b| < 10^(-18705352).

The proof is not difficult. The following lemma can be useful:

For all x > 0, 4^(10^(10^x)) > 10^(10^(10^(x-10^(-x)))).