A033434 Third differences of Catalan numbers A000108.
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
Keywords
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
- Jocelyn Quaintance and Harris Kwong, A combinatorial interpretation of the Catalan and Bell number difference tables, Integers, 13 (2013), #A29.
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