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%I #30 Jan 22 2024 13:02:49
%S 1,10,30,84,130,300,350,680,819,1300,1342,2520,2210,3500,3900,5456,
%T 4930,8190,6878,10920,10500,13420,12190,20400,16275,22100,22140,29400,
%U 24418,39000,29822,43680,40260,49300,45500,68796,50690,68780,66300,88400,68962,105000,79550,112728,106470
%N a(n) = n * sigma_2(n).
%C Moebius transform of A027847.
%H Amiram Eldar, <a href="/A328259/b328259.txt">Table of n, a(n) for n = 1..10000</a>
%H Joerg Arndt, <a href="http://arxiv.org/abs/1202.6525">On computing the generalized Lambert series</a>, arXiv:1202.6525v3 [math.CA], (2012).
%F G.f.: Sum_{k>=1} k^3 * x^k / (1 - x^k)^2.
%F G.f.: Sum_{k>=1} k * x^k * (1 + 4 * x^k + x^(2*k)) / (1 - x^k)^4.
%F Dirichlet g.f.: zeta(s - 1) * zeta(s - 3).
%F Sum_{k=1..n} a(k) ~ zeta(3) * n^4 / 4. - _Vaclav Kotesovec_, Oct 09 2019
%F Multiplicative with a(p^e) = (p^(3*e+2) - p^e)/(p^2 - 1). - _Amiram Eldar_, Dec 02 2020
%F 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
%F 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
%F 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
%t Table[n DivisorSigma[2, n], {n, 1, 45}]
%t nmax = 45; CoefficientList[Series[Sum[k^3 x^k/(1 - x^k)^2, {k, 1, nmax}], {x, 0, nmax}], x] // Rest
%o (PARI) a(n) = n*sigma(n, 2); \\ _Michel Marcus_, Dec 02 2020
%Y Cf. A001157, A027847, A038040, A064987, A275585, A281372, A294362.
%K nonn,mult,easy
%O 1,2
%A _Ilya Gutkovskiy_, Oct 09 2019
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