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 1-3 of 3 results.

A354237 Expansion of e.g.f. 1 / (1 - log(1 + 2*x) / 2).

Original entry on oeis.org

1, 1, 0, 2, -8, 64, -592, 6768, -90624, 1395840, -24292608, 471453696, -10094066688, 236340378624, -6007053852672, 164713554069504, -4846361933021184, 152300800682754048, -5091189648734748672, 180386551596145508352, -6752521487083688165376
Offset: 0

Views

Author

Ilya Gutkovskiy, Jun 06 2022

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[1/(1 - Log[1 + 2 x]/2), {x, 0, nmax}], x] Range[0, nmax]!
    Table[Sum[StirlingS1[n, k] k! 2^(n - k), {k, 0, n}], {n, 0, 20}]
  • PARI
    my(x='x + O('x^20)); Vec(serlaplace(1/(1-log(1+2*x)/2))) \\ Michel Marcus, Jun 06 2022

Formula

a(n) = Sum_{k=0..n} Stirling1(n,k) * k! * 2^(n-k).
a(0) = 1; a(n) = Sum_{k=1..n} binomial(n,k) * (k-1)! * (-2)^(k-1) * a(n-k).
a(n) ~ n! * (-1)^(n+1) * 2^(n+1) / (n * log(n)^2) * (1 - (4 + 2*gamma)/log(n) + (12 + 12*gamma + 3*gamma^2 - Pi^2/2)/log(n)^2 + (2*Pi^2*gamma - 32 + 4*Pi^2 - 24*gamma^2 - 8*zeta(3) - 4*gamma^3 - 48*gamma)/log(n)^3 + (80 - 20*Pi^2*gamma + 40*zeta(3)*gamma - 5*Pi^2*gamma^2 + 160*gamma + 5*gamma^4 + 80*zeta(3) + 40*gamma^3 + Pi^4/12 - 20*Pi^2 + 120*gamma^2)/log(n)^4), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Jun 06 2022

A354751 Expansion of e.g.f. 1 / (1 - log(1 + 4*x) / 4).

Original entry on oeis.org

1, 1, -2, 14, -152, 2264, -42832, 982512, -26484096, 820207488, -28692711168, 1118821622016, -48112717347840, 2261868010650624, -115400220781209600, 6350152838136428544, -374874781697133871104, 23632196147497381625856, -1584445791263626895228928
Offset: 0

Views

Author

Ilya Gutkovskiy, Jun 06 2022

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 18; CoefficientList[Series[1/(1 - Log[1 + 4 x]/4), {x, 0, nmax}], x] Range[0, nmax]!
    Table[Sum[StirlingS1[n, k] k! 4^(n - k), {k, 0, n}], {n, 0, 18}]
  • PARI
    my(x='x + O('x^20)); Vec(serlaplace(1/(1-log(1+4*x)/4))) \\ Michel Marcus, Jun 06 2022

Formula

a(n) = Sum_{k=0..n} Stirling1(n,k) * k! * 4^(n-k).
a(0) = 1; a(n) = Sum_{k=1..n} binomial(n,k) * (k-1)! * (-4)^(k-1) * a(n-k).

A354752 a(n) = Sum_{k=0..n} Stirling1(n,k) * k! * n^(n-k).

Original entry on oeis.org

1, 1, 0, 6, -152, 6670, -451152, 43685208, -5741360256, 984176280288, -213379094227200, 57100689621382176, -18489130293293779968, 7125765731670143814672, -3223822934974620319272960, 1692009521117003600170128000, -1019755541584493644326799048704
Offset: 0

Views

Author

Ilya Gutkovskiy, Jun 06 2022

Keywords

Crossrefs

Programs

  • Mathematica
    Unprotect[Power]; 0^0 = 1; Table[Sum[StirlingS1[n, k] k! n^(n - k), {k, 0, n}], {n, 0, 16}]
    Join[{1}, Table[n! SeriesCoefficient[1/(1 - Log[1 + n x]/n), {x, 0, n}], {n, 1, 16}]]
  • PARI
    a(n) = sum(k=0, n, stirling(n, k, 1) * k! * n^(n-k)); \\ Michel Marcus, Jun 06 2022

Formula

a(n) = n! * [x^n] 1 / (1 - log(1 + n*x) / n) for n > 0.
a(n) ~ (-1)^(n+1) * n! * n^(n-2). - Vaclav Kotesovec, Jun 06 2022
Showing 1-3 of 3 results.