cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A282099 Coefficients in q-expansion of (E_2^2*E_4 - 2*E_2*E_6 + E_4^2)/1728, where E_2, E_4, E_6 are the Eisenstein series shown in A006352, A004009, A013973, respectively.

Original entry on oeis.org

0, 1, 36, 252, 1168, 3150, 9072, 16856, 37440, 61317, 113400, 161172, 294336, 371462, 606816, 793800, 1198336, 1420146, 2207412, 2476460, 3679200, 4247712, 5802192, 6436872, 9434880, 9844375, 13372632, 14900760, 19687808, 20511990, 28576800, 28630112, 38347776
Offset: 0

Views

Author

Seiichi Manyama, Feb 06 2017

Keywords

Comments

Multiplicative because A001158 is. - Andrew Howroyd, Jul 25 2018

Examples

			a(6) = 1^5*6^2 + 2^5*3^2 + 3^5*2^2 + 6^5*1^2 = 9072.
		

Crossrefs

Cf. A282097 (phi_{3, 2}), this sequence (phi_{5, 2}).
Cf. A006352 (E_2), A004009 (E_4), A013973 (E_6), A282208 (E_2^2*E_4), A282096 (E_2*E_6), A008410 (E_8 = E_4^2).
Cf. A001158 (sigma_3(n)), A281372 (n*sigma_3(n)), this sequence (n^2*sigma_3(n)), A282213 (n^3*sigma_3(n)).

Programs

  • Mathematica
    a[0]=0;a[n_]:=(n^2)*DivisorSigma[3,n];Table[a[n],{n,0,32}] (* Indranil Ghosh, Feb 21 2017 *)
    nmax = 40; CoefficientList[Series[Sum[k^5*x^k*(1 + x^k)/(1 - x^k)^3, {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Aug 02 2025 *)
  • PARI
    a(n) = if (n==0, 0, n^2*sigma(n, 3)); \\ Michel Marcus, Feb 21 2017

Formula

G.f.: phi_{5, 2}(x) where phi_{r, s}(x) = Sum_{n, m>0} m^r * n^s * x^{m*n}.
a(n) = (A282208(n) - 2*A282096(n) + A008410(n))/1728. - Seiichi Manyama, Feb 19 2017
a(n) = n^2*A001158(n) for n > 0. - Seiichi Manyama, Feb 19 2017
Sum_{k=1..n} a(k) ~ Pi^4 * n^6 / 540. - Vaclav Kotesovec, May 09 2022
From Amiram Eldar, Oct 30 2023: (Start)
Multiplicative with a(p^e) = p^(2*e) * (p^(3*e+3)-1)/(p^3-1).
Dirichlet g.f.: zeta(s-2)*zeta(s-5). (End)
G.f.: Sum_{k>=1} k^5*x^k*(1 + x^k)/(1 - x^k)^3. - Vaclav Kotesovec, Aug 02 2025