
COMMENTS

a(9) > 10^11. It can be expected that a(n) exists for all n, at an order of magnitude a(n) ~ 10^m with m/log_10(m) ~ n.
A061910 considers only positive integers, and here we also consider this case. If 0 were allowed, then the first 4 terms would be 0.


EXAMPLE

For k = 1, 2 and 3, k^2 is a singledigit number and thus equal to its sum of digits, which therefore is a square. Therefore a(n) = 1 starts the first run of n consecutive integers with this property, for n = 1, 2 and 3.
However, the square of k = 4 has digit sum 7 which is not a square, and the same is the case for k = 5, 7 and 8. (Only k = 6 would have the required property.)
The consecutive integers { 9, 10, 11, 12, 13, 14, 15 } have squares 81, 100, 121, 169, 196, 225 which all have a digit sum (9, 1, 4, 16, 16 and 9) which is a square. Therefore a(n) = 9 starts the first run of n consecutive integers with this property, for n = 4 through 7.
(Actually, 10^(3m2)^21 starts a run of 7 such numbers, for any m >= 1.)
The first run of 8 such numbers is (46045846, ..., 46045853), whence a(8) = 46045846.
