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a(n)=6*c(n,1) where n runs through the 3-smooth numbers (see comment).
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%I #6 Mar 30 2012 18:39:23

%S 0,3,2,3,5,3,2,5,3,5,5,2,3,5,5,5,3,5,2,5,5,3,5,5,5,5,2,3,5,5,5,5,5,3,

%T 5,5,2,5,5,5,3,5,5,5,5,5,5,3,2,5,5,5,5,5,5,3,5,5,5,5,5,2,5,5,3,5,5,5,

%U 5,5,5,5,5,3,5,5,2,5,5,5,5,5,5,3,5,5,5,5,5,5,5,5,2,5,3,5,5,5,5,5,5,5,5,5,5

%N a(n)=6*c(n,1) where n runs through the 3-smooth numbers (see comment).

%C If n is a 3-smooth number, (i.e. of form 2^i*3^j for i,j>=0) the value c(n,k)={(n^(2k)-1)*B(2k)} is independent of k where {x} denotes the fractional part of x and B(k) is the k-th Bernoulli's number.

%F a(1)=0; for k>0, a(2^k)=3 a(3^k)=2; for i>0 and j>0 a(2^i*3^j)=5

%o (PARI) m=7;for(n=1,1000000,if(gcd(n,6^100)==n,print1(6*frac((n^(2*m)-1)*bernfrac(2*m)),",")))

%Y Cf. A003586.

%K nonn

%O 1,2

%A _Benoit Cloitre_, Oct 25 2004