444444....4444888888....8889

[Where there are 'k' Fours, '(k-1)' Eights and 'Exactly One' 9],

are always perfect squares.

(For example the sequence of numbers: 49, 4489, 444889, ....etc. and so on are always perfect squares).

I was thinking about 4((10^x)+2)*(x 1s) + 1 for how to do that sequence.

('x 1s' is the sum of (10^x)+(10^x-1)+...+(10^1))

Would that help at all or is it too complicated?

Posted by Gamer on 2003-04-06 04:44:16