A282211 Coefficients in q-expansion of (6*E_2^2*E_4 - 8*E_2*E_6 + 3*E_4^2 - E_2^4)/6912, where E_2, E_4, E_6 are the Eisenstein series shown in A006352, A004009, A013973, respectively.
0, 1, 24, 108, 448, 750, 2592, 2744, 7680, 9477, 18000, 15972, 48384, 30758, 65856, 81000, 126976, 88434, 227448, 137180, 336000, 296352, 383328, 292008, 829440, 484375, 738192, 787320, 1229312, 731670, 1944000, 953312, 2064384, 1724976
Offset: 0
Examples
a(6) = 1^4*6^3 + 2^4*3^3 + 3^4*2^3 + 6^4*1^3 = 2592.
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..1000
- Joerg Arndt, On computing the generalized Lambert series, arXiv:1202.6525v3 [math.CA], (2012).
Crossrefs
Programs
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Mathematica
a[0]=0;a[n_]:=(n^3)*DivisorSigma[1,n];Table[a[n],{n,0,33}] (* Indranil Ghosh, Feb 21 2017 *) -
PARI
a(n) = if (n==0, 0, n^3*sigma(n)); \\ Michel Marcus, Feb 21 2017
Formula
G.f.: phi_{4, 3}(x) where phi_{r, s}(x) = Sum_{n, m>0} m^r * n^s * x^{m*n}.
a(n) = n^3*A000203(n) for n > 0. - Seiichi Manyama, Feb 19 2017
G.f.: A(q) = Sum_{n >= 1} n^3*q^n*(q^(3*n) + 11*q^(2*n) + 11*q^n + 1)/(1 - q^n)^5. A faster converging series may be found by applying the operator x*d/dx once to equation 5 in Arndt, setting x = 1, and then applying the operator q*d/dq three times to the resulting equation. - Peter Bala, Jan 21 2021
Sum_{k=1..n} a(k) ~ c * n^5, where c = Pi^2/30 = 0.328986... . - Amiram Eldar, Dec 08 2022
From Amiram Eldar, Oct 31 2023: (Start)
Multiplicative with a(p^e) = p^(3*e) * (p^(e+1)-1)/(p-1).
Dirichlet g.f.: zeta(s-3)*zeta(s-4). (End)
G.f.: A(q) = Sum_{n >= 1} n^4*q^n*(q^(2*n) + 4*q^n + 1)/(1 - q^n)^4. - Mamuka Jibladze, Aug 27 2024
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