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

Author

Benoit Cloitre, Oct 25 2004

Keywords

• nonn

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

Crossrefs

Cf. A003586, A027641, A027642.

Programs

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 *)

PARI

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

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.