Sorry, I'm not very good in writing but here is a more expanded version of my earlier post:

 

1) Number of digits in (k+1)^2 can be equal or one more than k^2 for any integer k>0. This is proved by showing that (k+1)^2/k^2=1+2/k+1/k^2 < 10.

 

2) If a=0.149162536496481...  is rational, then there is a repeating string of digits after some number of digits. Lets take a large number k0 so that the digits of k0^2 are fully inside the repeating region. Suppose P is the smallest multiple of the length of the repeating string that is larger or equal k0. Since the number of digits in the sequence k0^2, (k0+1)^2... can increase only by one (due to 1), then there is a number k>=k0 that the number of digits in k^2 is P.

 

4) If (k+1)^2 has the same length as k^2, then a repeating pattern is not possible. If (k+1)^2 is one digit longer than k^2, then (k+1)^2 consists of the same digits as k^2 plus the last (unit) digit, which means  (k+1)^2=10*k^2+m, where m<10. But from 1) (k+1)^2/k^2<10 so this is not possible. Therefore, a repeating pattern in a is not possible.