%I #9 Jun 21 2014 04:08:02
%S 1,1,1,2,1,1,5,1,1,4,1,1,4,2,1,9,1,1,30,1,1,11,5,1,4,1,2,9,2,1
%N Number of groups of order L(n).
%C This is to A140987 as Lucas numbers A000032 are to Fibonacci numbers A000045.
%H <a href="http://www.gap-system.org/Packages/grpconst.html">GAP package GrpConst</a>
%F a(n) = A000001(A000032(n)).
%e a(6) = 5 because the Lucas number L(6) (starting at L(0) = 2) is 18 and there are 5 groups of order 18.
%Y Cf. A000001, A000032, A000045, A140987.
%K more,nonn
%O 0,4
%A _Jonathan Vos Post_, Aug 02 2008
%E a(16)-a(29) from _Eric M. Schmidt_, Jun 20 2014