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.

Showing 1-4 of 4 results.

A289063 Coefficients in expansion of E_6^2/Product_{k>=1} (1-q^k)^24.

Original entry on oeis.org

1, -984, 196884, 21493760, 864299970, 20245856256, 333202640600, 4252023300096, 44656994071935, 401490886656000, 3176440229784420, 22567393309593600, 146211911499519294, 874313719685775360, 4872010111798142520, 25497827389410525184, 126142916465781843075
Offset: 0

Views

Author

Seiichi Manyama, Jun 23 2017

Keywords

Comments

Convolution square of A007242. - Michael Somos, Mar 31 2019

Examples

			G.f. = (1-q)^984 * (1-q^2)^286752 * (1-q^3)^102360024 * ...
G.f. = 1 - 984*q + 196884*q^2 + 21493760*q^3 + 864299970*q^4 + 20245856256*q^5 + ... .
		

Crossrefs

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[(1 - 504*Sum[DivisorSigma[5,k]*x^k, {k, 1, nmax}])^2 / Product[(1 - x^k)^24, {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Jul 09 2017 *)
    a[ n_] := SeriesCoefficient[ q Series[ 1728 (KleinInvariantJ[Log[q] / (2 Pi I)] - 1), {q, 0, n}], {q, 0, n}]; (* Michael Somos, Mar 31 2019 *)
  • PARI
    {a(n) = my(A, U1, U2); if( n<0, 0, A = x * O(x^n); U1 = eta(x + A)^24; U2 = eta(x^2 + A)^24; polcoeff( (U1 - 512*x * U2)^2 * (U1 + 64*x * U2) / (U1^2 * U2), n))}; /* Michael Somos, Mar 31 2019 */

Formula

G.f.: Product_{k>=1} (1-q^k)^A289061(k).
a(n) = A000521(n-1) for n = 0 and n > 1.
a(n) ~ exp(4*Pi*sqrt(n)) / (sqrt(2) * n^(3/4)). - Vaclav Kotesovec, Jul 09 2017
G.f.: q * (j(q) - 1728) where j(q) is a modular function. - Michael Somos, Mar 31 2019

A288995 a(n) = 12 * (A288968(n) - 2).

Original entry on oeis.org

264, 4152, 77064, 1551576, 33343752, 745374264, 17140046088, 402328199064, 9593786367240, 231629451811896, 5648880427214088, 138910500564007128, 3439808201626085640, 85686183717823968312, 2145402754204531455240, 53956201350487199168280
Offset: 1

Views

Author

Seiichi Manyama, Jun 23 2017

Keywords

Crossrefs

Related to E_{k+2}/E_k: this sequence (k=2), A192731 (k=4), A289061 (k=6).
Cf. A008683, A288877 (E_4*E_2), A288968.
Cf. A289062.

Formula

a(n) = (1/n) * Sum_{d|n} A008683(n/d) * A288877(d).

A289344 Coefficients in expansion of E_2^(1/2)/Product_{k>=1} (1-q^k).

Original entry on oeis.org

1, -11, -118, -1473, -23635, -434861, -8659573, -181387821, -3936961298, -87743843970, -1996149058302, -46163368994680, -1082012001849499, -25646334881233711, -613664275728573585, -14803437882920457712, -359626550280367615329
Offset: 0

Views

Author

Seiichi Manyama, Jul 03 2017

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[Sqrt[1 - 24*Sum[DivisorSigma[1, k]*x^k, {k, 1, nmax}]] / Product[1 - x^k, {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Jul 03 2017 *)

Formula

G.f.: Product_{k>=1} (1-q^k)^(A288995(k)/24).
a(n) ~ c / (n^(3/2) * r^n), where r = A211342 = 0.03727681029645165815098078565... is the root of the equation Sum_{k>=1} A000203(k) * r^k = 1/24 and c = -0.309300289625571303778321676728514880378401177270067457514896529... - Vaclav Kotesovec, Jul 03 2017

A289350 Coefficients in expansion of E_2/Product_{k>=1} (1-q^k)^2.

Original entry on oeis.org

1, -22, -115, -350, -940, -2124, -4615, -9130, -17575, -32100, -57239, -98512, -166595, -274350, -445055, -708124, -1112002, -1719410, -2629450, -3970230, -5937238, -8785502, -12889630, -18741250, -27045445, -38724088, -55074057, -77791320, -109215025
Offset: 0

Views

Author

Seiichi Manyama, Jul 03 2017

Keywords

Crossrefs

E_2^(m/2)/Product_{k>=1} (1-q^k)^m: A289344 (m=1), this sequence (m=2), A289062 (m=24).
Cf. A006352 (E_2), A066186, A288995.

Programs

  • Mathematica
    nmax = 50; CoefficientList[Series[(1 - 24*Sum[DivisorSigma[1, k]*x^k, {k, 1, nmax}]) / Product[(1 - x^k)^2, {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Jul 08 2017 *)

Formula

G.f.: Product_{k>=1} (1-q^k)^(A288995(k)/12).
a(n) ~ -3^(1/4) * exp(2*Pi*sqrt(n/3)) / n^(1/4). - Vaclav Kotesovec, Jul 08 2017
Showing 1-4 of 4 results.