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.

A052851 Expansion of e.g.f. 1/2 - (1/2)*(1+4*log(1-x))^(1/2).

Original entry on oeis.org

0, 1, 3, 20, 220, 3424, 69008, 1706256, 49956240, 1689497376, 64799254752, 2778906776832, 131756614920192, 6843405231815424, 386414606189283072, 23567401521343170048, 1543994621969805135360, 108137637714495023354880, 8062825821198926369725440
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

Previous name was: A simple grammar.

Crossrefs

Programs

  • Maple
    spec := [S,{B=Cycle(Z),S=Prod(B,C),C=Sequence(S)},labeled]: seq(combstruct[count](spec,size=n), n=0..20);
  • Mathematica
    CoefficientList[Series[1/2-1/2*(1+4*Log[1-x])^(1/2), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Sep 30 2013 *)
  • Maxima
    a(n):=sum(stirling1(n,k)*k!*binomial(2*k-2,k-1)/k*(-1)^(n+k), k,1,n); /* Vladimir Kruchinin, May 12 2012 */

Formula

E.g.f.: 1/2 - (1/2)*(1-4*log(-1/(-1+x)))^(1/2).
a(n) = Sum_{k=1..n} Stirling1(n,k)*k!*C(2*k-2,k-1)/k*(-1)^(n+k). - Vladimir Kruchinin, May 12 2012
a(n) ~ n^(n-1)/(sqrt(2)*exp(3*n/4)*(exp(1/4)-1)^(n-1/2)). - Vaclav Kotesovec, Sep 30 2013
From Seiichi Manyama, Sep 09 2024: (Start)
E.g.f. satisfies A(x) = (-log(1 - x)) / (1 - A(x)).
E.g.f.: Series_Reversion( 1 - exp(-x * (1 - x)) ). (End)

Extensions

New name using e.g.f., Vaclav Kotesovec, Sep 30 2013

A371315 E.g.f. satisfies A(x) = -log(1 - x)/(1 - A(x))^3.

Original entry on oeis.org

0, 1, 7, 110, 2796, 98754, 4469334, 246741984, 16079405784, 1208082769560, 102810760773096, 9774841791650880, 1026870593449179264, 118121793328191431232, 14766518531481521488704, 1993367920121834019649920, 288988424345833831094150016
Offset: 0

Views

Author

Seiichi Manyama, Mar 18 2024

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=1, n, (4*k-2)!/(3*k-1)!*abs(stirling(n, k, 1)));

Formula

a(n) = Sum_{k=1..n} (4*k-2)!/(3*k-1)! * |Stirling1(n,k)|.
E.g.f.: Series_Reversion( 1 - exp(-x * (1 - x)^3) ). - Seiichi Manyama, Sep 08 2024
Showing 1-2 of 2 results.