For all k and the mth term:2k^(m-1), for any k and m>0; 2^(2-m)*k^(m-1), for any k and m<2;

Hence the solutions sought are:

a(2) = {2,4,6,8,10},

a(5)={2,32,162,512,1250}

Edited on March 14, 2012, 10:51 am

blackjack

flooble's webmaster puzzle