Let's define eleven-non-free numbers:
If we consider a number as a string, then if any substring inside is a (non-zero) power of 11, then this number is an eleven-non-free number.
For example, 1123 is an eleven-non-free number as 11 inside is 11^1. Also 12154 is one as 121 is 11^2. But 12345 is not, because we can't find any non-zero power of 11 inside.
So given a k, find the kth eleven-non-free number. For example the 13th such number is 211.
I don't know how to efficiently do it. the brute-force way is to increase i from 1 and check every number and count until the kth.
I guess we should consider strings with different length (1, 2, 3, 4, ...). then for each length, we try to fill in 11, 11^2, 11^3, etc and try to get all the combinations.
But it seems quite complicated as well.
Anyone?