
COMMENTS

The last term is 999999999786876856487355368576387875784644 = 999999999893438428238^2.  Giovanni Resta, May 19 2013
There are no squares with 43, 44, or 45 digits. Indeed, numbers of 45 digits have sum of digits 1^1+2^2+...+9^9 = 285, which mod 9 is equal to 6. It is easy to verify that no power can be equal to 6 mod 9, hence there are no squares, cubes, etc. of 45 digits. Similarly, the numbers of 44 and 43 digits can only be obtained by omitting the single 1 or the two 2's, so mod 9 they are equal to 5 and 2, respectively. Again, 2 and 5 are not squares or cubes mod 9, but they can be powers with exponents k = 5, 7, 11, 13, 17, 19, 23, 25,... (numbers not divisible by 2 or 3). Since 10^(44/k) is at most 6.3*10^8 (for k=5) excluding higher powers by generating them is not a tremendous computational effort, which can be further reduced noticing that certain candidates can be excluded based on their last digits. For example, 9993^5 mod 10000 is 3193, which contains a 1. So no number ending in 9993 can be the base for a 5th power of 44 digits (which should lack the 1). Since 4th powers are squares too, they can have at most 42 digits, and since 10^(42/4) is about 3.16*10^10, it is not difficult to ascertain that no 4th powers belong to A108571.  Giovanni Resta, Jul 26 2015
