A341343 Dirichlet g.f.: Sum_{n>=1} a(n)/n^s = (zeta(s))^3 / (zeta(3*s))^2.
1, 3, 3, 6, 3, 9, 3, 8, 6, 9, 3, 18, 3, 9, 9, 9, 3, 18, 3, 18, 9, 9, 3, 24, 6, 9, 8, 18, 3, 27, 3, 9, 9, 9, 9, 36, 3, 9, 9, 24, 3, 27, 3, 18, 18, 9, 3, 27, 6, 18, 9, 18, 3, 24, 9, 24, 9, 9, 3, 54, 3, 9, 18, 9, 9, 27, 3, 18, 9, 27, 3, 48, 3, 9, 18, 18, 9, 27, 3, 27, 9, 9, 3, 54, 9
Offset: 1
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 1..10000
Programs
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PARI
{T(n,k) = if( n<0, 0, polcoeff( (1 + x + x^2)^n, k))}; \\ A027907 a(n)={my(f=factor(n));prod(k=1,#f[,1],sum(i=0,f[k,2],T(2,i)))}; for(j=1,75,print1(a(j),", ")) \\ Hugo Pfoertner, Feb 13 2021
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PARI
for(n=1, 100, print1(direuler(p=2, n, (1 - X^3)^2/(1 - X)^3)[n], ", ")) \\ Vaclav Kotesovec, Nov 20 2021
Formula
Multiplicative with a(p^e) = Sum_{i=0..e} trinomial(2,i) for prime p and e >= 0, where trinomial(n,k) = 0 if 2*n < k.
Let b(n), n > 0, be the Dirichlet inverse of a(n); b(n) is multiplicative with b(p^(3*e)) = 1 for e >= 0, and b(p^(3*e-2)) = -3*e and b(p^(3*e-1)) = 3*e for e > 0 and prime p.
Sum_{k=1..n} a(k) ~ n * (log(n)^2/2 + (3*gamma - 6*zeta'(3)/zeta(3) - 1)*log(n) + 1 - 3*gamma + 3*gamma^2 + 6*(1 - 3*gamma)*zeta'(3)/zeta(3) + 27*zeta'(3)^2 / zeta(3)^2 - 9*zeta''(3)/zeta(3) - 3*sg1) / zeta(3)^2, where gamma is the Euler-Mascheroni constant A001620, zeta(3) = A002117, zeta'(3) = -A244115, zeta''(3) = A340442 and sg1 is the first Stieltjes constant (see A082633). - Vaclav Kotesovec, Nov 20 2021
Comments