Eric Angelini <eric.angelini@skynet.be> wrote:

Start with n, then let n=2n and iterate. When the string n reappears, stop the iterations and restart the procedure with the smallest integer not yet in S.

Question: is there a new n that will restart S at some point and _not_ lead to a new stop?

No. Since no non-zero power of 2 equals a power of 10, i.e. log(10)/log(2) is irrational, any sequence in which each term is the double of the previous will start with every decimal number infinitely many times. Powers of 2 (or powers of 2 times a constant) are dense on the interval created by prepending every integer with a decimal point (.1, .2, .3, ... .8, .9, .10, .11, ...). For instance there's a power of 2 that begins with 2020: 2^4180. The American independence year, 1776, starts 2^2087, and the Belgian independence year, 1830, starts 2^1323. See A018856. You won't find a power of 2 that begins with all the decimal digits of pi, but you can always find one that begins with the first n decimal digits of pi: 5, 98, 872, 10871, 55046, etc. (not in OEIS).