A328259 a(n) = n * sigma_2(n).
1, 10, 30, 84, 130, 300, 350, 680, 819, 1300, 1342, 2520, 2210, 3500, 3900, 5456, 4930, 8190, 6878, 10920, 10500, 13420, 12190, 20400, 16275, 22100, 22140, 29400, 24418, 39000, 29822, 43680, 40260, 49300, 45500, 68796, 50690, 68780, 66300, 88400, 68962, 105000, 79550, 112728, 106470
Offset: 1
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
- Joerg Arndt, On computing the generalized Lambert series, arXiv:1202.6525v3 [math.CA], (2012).
Programs
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Mathematica
Table[n DivisorSigma[2, n], {n, 1, 45}] nmax = 45; CoefficientList[Series[Sum[k^3 x^k/(1 - x^k)^2, {k, 1, nmax}], {x, 0, nmax}], x] // Rest -
PARI
a(n) = n*sigma(n, 2); \\ Michel Marcus, Dec 02 2020
Formula
G.f.: Sum_{k>=1} k^3 * x^k / (1 - x^k)^2.
G.f.: Sum_{k>=1} k * x^k * (1 + 4 * x^k + x^(2*k)) / (1 - x^k)^4.
Dirichlet g.f.: zeta(s - 1) * zeta(s - 3).
Sum_{k=1..n} a(k) ~ zeta(3) * n^4 / 4. - Vaclav Kotesovec, Oct 09 2019
Multiplicative with a(p^e) = (p^(3*e+2) - p^e)/(p^2 - 1). - Amiram Eldar, Dec 02 2020
G.f.: Sum_{n >= 1} q^(n^2)*( n^4 - (2*n^4 - 4*n^3 - 3*n^2 - n)*q^n - (8*n^3 - 4*n)*q^(2*n) + (2*n^4 + 4*n^3 - 3*n^2 + n)*q^(3*n) - n^4*q^(4*n) )/(1 - q^n)^4. Apply the operator x*d/dx twice, followed by the operator q*d/dq once, to equation 5 in Arndt and then set x = 1. - Peter Bala, Jan 21 2021
a(n) = Sum_{k = 1..n} sigma_3( gcd(k, n) ) = Sum_{d divides n} sigma_3(d) * phi(n/d). - Peter Bala, Jan 19 2024
a(n) = Sum_{1 <= i, j, k <= n} sigma_1( gcd(i, j, k, n) ) = Sum_{d divides n} sigma_1(d) * J_3(n/d), where the Jordan totient function J_3(n) = A059376(n). - Peter Bala, Jan 22 2024
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