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%I #4 Oct 15 2013 22:30:56
%S 1,1,2,2,3,3,3,4,4,4,4,5,5,5,6,6,6,6,7,7,7,7,7,8,8,8,8,9,9,9,9,9,10,
%T 10,10,10,10,11,11,11,11,11,12,12,12,12,12,12,13,13,13,13,13,14,14,14,
%U 14,14,14,15,15,15,15,15,16,16,16,16,16,16,17,17,17,17,17,17,18,18,18,18
%N Largest m such that 2^n / primorial(m) >= 1.
%C Primorial order of powers of 2.
%F a(n)=Max{s| q(s)=Sign[Floor[2^n/A002110(n)]]=1}
%e for n=1 and 2, when 2 and 4 is divided by 2 gives quotient = 1 or 2, but when divided by 6 q<1, so the largest suitable primorial is the first; thus a(1)=a(2)=1. n=11, 2^11=2048. The largest primorial P, such that 2048/P > 1 is P=210, the 4th = A002110(4). So a(11)=4.
%Y Cf. binary order (A029837) of primorials, A045716
%Y Cf. A002110, A045716, A029837.
%K nonn
%O 0,3
%A _Labos Elemer_, Nov 22 2000
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