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a(n)=sigma[A097018(n)]/p[n]=A000203[A097018(n)]/A000040(n).
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%I #4 Oct 15 2013 22:32:33

%S 2,1,3,1,4,1,4,2,6,6,1,2,4,6,6,4,6,3,6,4,7,2,4,23,2,10,6,6,6,6,1,4,4,

%T 2,10,12,2,6,10,4,10,8,22,8,4,2,2,8,4,2,16,6,14,12,12,4,6,2,12,4,6,4,

%U 1,10,6,6,2,2,6,8,10,6,2,6,2,4,6,6,22,7,16,12,4,8,2,6,6,6,12,6,4,10,12,10,2

%N a(n)=sigma[A097018(n)]/p[n]=A000203[A097018(n)]/A000040(n).

%e n=11: a[11]=1: the quotient of the earliest number whose sigma is divisible by the 11th prime and p[11]=31.

%Y Cf. A000203, A000040, A097018.

%K nonn

%O 1,1

%A _Labos Elemer_, Aug 23 2004