User blog comment:Ikosarakt1/Apocalyptic function/@comment-25418284-20130322220331/@comment-5529393-20130323022900

Hmm, it may well be that for some n the last n digits of powers of 2 will always contain a 666 for a sufficiently large power. The last n digits of a power of 2 appears to have a cycle of length 10^{n-1}, so, applying probabilistic arguments again, we get

\((1 - (.999)^{n-2})^{10^{n-1}} \approx (1-(.999)^n)^{10^n} = (1 - 1/(1000^n))^{10^n}\)

\(= (1 - 1/x^3)^x /approx e^{1/x^2} \rightarrow 1\) as x goes to infinity.

So it gets increasingly likely that the last n digits of powers of 2 will eventually always contain a 666 as n goes to infinity. But, computationally it seems very difficult, since just for n digits we will have to check probably 10^{n-1} different numbers.