Let's call S the (infinite) string that is made by concatenating the consecutive positive integers (starting from 1) written down in base 10.
Thus, S = 1234567891011121314151617181920212223242...

It's easy to see that any number will show up an infinite number of times in S.

Let's call f(n) the starting position of the n^th occurrence of n in S.
For example, f(1)=1, f(5)=81, f(12)=271 and f(7780)=111111365.

Find f(3^k) for 1k13.