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 1-2 of 2 results.

A354259 Expansion of e.g.f. 1/sqrt(1 - 6 * log(1+x)).

Original entry on oeis.org

1, 3, 24, 330, 6354, 157482, 4772268, 170950392, 7066790676, 331108863372, 17340063707952, 1003726452207960, 63635982830437320, 4385439331442232840, 326404115258791793040, 26093904013675118381760, 2229931839713559043435920
Offset: 0

Views

Author

Seiichi Manyama, May 21 2022

Keywords

Crossrefs

Programs

  • Mathematica
    With[{nn=20},CoefficientList[Series[1/Sqrt[1-6Log[1+x]],{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Oct 06 2023 *)
  • PARI
    my(N=20, x='x+O('x^N)); Vec(serlaplace(1/sqrt(1-6*log(1+x))))
    
  • PARI
    my(N=20, x='x+O('x^N)); Vec(serlaplace(sum(k=0, N, binomial(2*k, k)*(3*log(1+x)/2)^k)))
    
  • PARI
    a(n) = sum(k=0, n, (3/2)^k*(2*k)!*stirling(n, k, 1)/k!);

Formula

E.g.f.: Sum_{k>=0} binomial(2*k,k) * (3 * log(1+x)/2)^k.
a(n) = Sum_{k=0..n} (3/2)^k * (2*k)! * Stirling1(n,k)/k!.
a(n) ~ n^n / (sqrt(3) * (exp(1/6)-1)^(n + 1/2) * exp(n - 1/12)). - Vaclav Kotesovec, Jun 04 2022

A354262 Expansion of e.g.f. 1/sqrt(1 + 8 * log(1-x)).

Original entry on oeis.org

1, 4, 52, 1112, 33192, 1272576, 59607552, 3298935552, 210638509824, 15241340093952, 1232504690492928, 110154484622208000, 10782300230031713280, 1147157496053856645120, 131810751499551281786880, 16266976762439018716323840, 2145960434809665656603320320
Offset: 0

Views

Author

Seiichi Manyama, May 21 2022

Keywords

Crossrefs

Programs

  • Mathematica
    With[{nn=20},CoefficientList[Series[1/Sqrt[1+8*Log[1-x]],{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Dec 14 2024 *)
  • PARI
    my(N=20, x='x+O('x^N)); Vec(serlaplace(1/sqrt(1+8*log(1-x))))
    
  • PARI
    my(N=20, x='x+O('x^N)); Vec(serlaplace(sum(k=0, N, binomial(2*k, k)*(-2*log(1-x))^k)))
    
  • PARI
    a(n) = sum(k=0, n, 2^k*(2*k)!*abs(stirling(n, k, 1))/k!);

Formula

E.g.f.: Sum_{k>=0} binomial(2*k,k) * (-2 * log(1-x))^k.
a(n) = Sum_{k=0..n} 2^k * (2*k)! * |Stirling1(n,k)|/k!.
a(n) ~ n^n / (2 * (exp(1/8)-1)^(n + 1/2) * exp(7*n/8)). - Vaclav Kotesovec, Jun 04 2022
Showing 1-2 of 2 results.