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.

A033434 Third differences of Catalan numbers A000108.

Original entry on oeis.org

1, 4, 13, 43, 145, 497, 1727, 6071, 21554, 77180, 278426, 1010990, 3692213, 13553555, 49981875, 185082495, 687923790, 2565602160, 9598056630, 36008860650, 135446603370, 510706730274, 1929930236790, 7308166696118, 27727426756580, 105387411817352, 401231661076148, 1529970156473276, 5842655231153741, 22342874048993015
Offset: 0

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Keywords

Crossrefs

Cf. A000108.

Programs

  • Magma
    [(27*n^3+81*n^2+108*n+24)*Binomial(2*n, n)/((n+1)*(n+2)*(n+3)*(n+4)): n in [0..30]]; // Vincenzo Librandi, Feb 05 2014
    
  • Maple
    C:= n-> binomial(2*n,n)/(n+1);
    a:= n-> add((-1)^j*binomial(3,j)*C(n-j+3), j=0..3);
    seq(a(n), n=0..30); # G. C. Greubel, May 03 2021
  • Mathematica
    Table[(27n^3 +81n^2 +108n +24)*n!*Binomial[2n, n]/(n+4)!, {n, 0, 40}] (* Vincenzo Librandi, Feb 05 2014 *)
    Differences[CatalanNumber[Range[0,40]],3] (* Harvey P. Dale, Jul 05 2020 *)
  • PARI
    a(n)=( 27*n^3 + 81*n^2 + 108*n + 24)*n!*binomial(2*n,n)/(n+4)!;
    
  • Sage
    [sum((-1)^j*binomial(3,j)*catalan_number(n-j+3) for j in (0..3)) for n in (0..40)] # G. C. Greubel, May 03 2021

Formula

a(n) = ( 27*n^3 + 81*n^2 + 108*n + 24 )*binomial(2*n, n)/( (n+1)*(n+2)*(n+3)*(n+4) ). - Benoit Cloitre, Jun 11 2004
a(n) = -binomial(2*n,n)/(n+1)*hypergeom([-3,n+1/2],[n+2],4). - Peter Luschny, Aug 15 2012
G.f.: (1 + x + x^2*C(x)^3)*C(x)^3 where C(x) is the g.f. of A000108. - Philippe Deléham, Feb 04 2014
From G. C. Greubel, May 03 2021: (Start)
G.f.: (x + (1-x)*C(x))*C(x)^3, where C(x) is the g.f. of A000108.
E.g.f.: exp(2*x)*(BesselI(0, 2*x) +2*BesselI(1, 2*x) -BesselI(2, 2*x) -BesselI(3, 2*x) - BesselI(4, 2*x)).
a(n) = Sum_{k=0..3} (-1)^k*binomial(3,k)*C(n-k+3), where C(n) = A000108(n). (End)
Sum_{n>=0} a(n)/4^n = 14. - Amiram Eldar, Jul 10 2023