A098229 a(n) = 6*c(m,1) where m = A003586(n) is the n-th 3-smooth number, c(m,k) = {(m^(2*k)-1)*B(2*k)}, {x} denotes the fractional part of x and B(k) is the k-th Bernoulli number.

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, 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, 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

Offset

1,2

Comments

If m is a 3-smooth number (i.e., of form 2^i*3^j for i,j >= 0), the value of c(m,k) is independent of k.

Links

Amiram Eldar, Table of n, a(n) for n = 1..10000

Formula

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

Mathematica

s[n_] := If[Times @@ ({2, 3}^IntegerExponent[n, {2, 3}]) == n, 6 * FractionalPart[(n^2-1)/6], Nothing]; Array[s, 125000] (* Amiram Eldar, May 03 2025 *)

Prog

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

Crossrefs

Cf. A003586, A027641, A027642.

Sequence in context: A240225 A283893 A112427 * A346388 A128151 A071166

Adjacent sequences: A098226 A098227 A098228 * A098230 A098231 A098232

Keyword

nonn

Author

Benoit Cloitre, Oct 25 2004

Status

approved