I'd like to get the digits of the Nth number by using the given numbers which are 3 and 9. From these numbers, I've got the numbers like 3, 9, 33, 39, 93, 99, 333 ... in order. So the digits of the 4th number is 2 and that of the 7th number is 3. What I want to do is to calculate the digits of the Nth number from this logic. Any suggestions? Thanks.
(You may assume that N is 1000000)
+What I've found was that there are 2 1-digit numbers (=2^1), 4 2-digit numbers (=2^2), and 8 3-digit numbers(=2^3). So I tried to apply the concept of geometric sequence but wasn't able to make it.
You already know that quantity of numbers with length k is 2^k, so we can exploit this fact. Essentially this is implementation of int(log2(n+1)) without any math libraries:
def nlen(n):
lng = 0
lcnt = 1
overall = 0
while overall < n:
lcnt *= 2
overall += lcnt
lng += 1
return lng