|
%I #6 Apr 09 2014 10:15:50
%S 1,1,2,2,4,2,5,2,7,2,8,2,9,2,10,3,11,2,12,2,14,4,12,2,18,3,12,5,16,2,
%T 18,2,18,5,14,6,26,2,15,4,23,2,22,2,22,9,18,2,29,3,22,5,22,2,29,7,27,
%U 5,21,2,37,2,21,11,26,7,31,2,25,6,33,2,41,2,25,13,27,7,32,2,36,9,26,2,45
%N a(0) = 1, a(n) = number of a(k), for 0 <= k <= n-1, that divide n.
%e a(6) = 5 because a(0), a(1), a(2), a(3) and a(5) divide 6.
%K easy,nonn
%O 0,3
%A _Leroy Quet_, Aug 04 2000
|
