Also, there's a completely different way to attack the problem of finding perfect squares using the dyadic form of q: (Prime Exponents). I discovered this while reading NuVoc about q:
fps1 =: 13 :'y#~0=+/"1]2|_ q:y' fps1 2+i.100 4 9 16 25 36 49 64 81 100 fps1 8200+i.1000 8281 8464 8649 8836 9025 %: 8281 8464 8649 8836 9025 91 92 93 94 95 fps1 ] #~ 0 = [: +/"1 [: ] 2 | _ q: ] The secret is: if all exponents of the prime factors of an integer are even, the number is a perfect square. Skip Skip Cave Cave Consulting LLC ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
