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.

A077587 a(n) = C(n+1) + n*C(n) where C = A000108 (Catalan numbers).

Original entry on oeis.org

1, 3, 9, 29, 98, 342, 1221, 4433, 16302, 60554, 226746, 854658, 3239044, 12332140, 47137005, 180780345, 695367510, 2681600130, 10364759790, 40142121030, 155748675420, 605274171060, 2355676013730, 9180275261274, 35819645937228
Offset: 0

Views

Author

Michael Somos, Nov 09 2002

Keywords

Comments

Number of ascents of length 2 starting at an even level in all Dyck paths of semilength n+2. Example: a(1)=3 because all Dyck paths of semilength 3 are UDUDUD, UD(UU)DD, (UU)DDUD, (UU)DUDD and UUUDDD, where U=(1,1), D=(1,-1), having altogether 3 ascents of length 2 that start at an even level (shown between parentheses). - Emeric Deutsch, Nov 29 2005
a(n) is the number of parking functions of size n+1 avoiding the patterns 132, 231, and 321. - Lara Pudwell, Apr 10 2023

Crossrefs

Programs

  • Maple
    egf := x -> exp(2*x)*(1+1/x)*BesselI(1, 2*x);
    seq(n!*coeff(series(egf(x), x, n+2), x, n), n=0..24); # Peter Luschny, Apr 14 2014
  • Mathematica
    Table[(CatalanNumber[n + 1] + n CatalanNumber[n]), {n, 0, 40}] (* Vincenzo Librandi, Apr 15 2014 *)
  • PARI
    a(n)=if(n<0,0,(n^2+6*n+2)*(2*n)!/n!/(n+2)!)
    
  • PARI
    a(n)=if(n<0,0,polcoeff((4+x+1/x-(x+1/x)^2)*(1+x)^(2*n),n)/2)

Formula

a(n) = binomial(2n+1, n+1) - binomial(2n, n+2).
a(n) = (3*(3*n+2)*a(n-1) - 2*(11*n-7)*a(n-2) + 4*(2*n-5)*a(n-3))/(n+2), n>2.
G.f.: A(x) = (1 - 3*x - (1-5*x+2*x^2)/sqrt(1-4*x) )/(2*x^2) satisfies 0 = (x^2+4*x-1) + (12*x^2-7*x+1)*A + (4*x^3-x^2)*A^2.
E.g.f.: A(x) = (1+x)B(x)' where B(x) = e.g.f. of A000108.
a(n) = Sum_{k=0..n} binomial(n,k)*A057977(k)*2^(n-k); here the A057977 are understood as the extended Catalan numbers (see also A063549). Related to Touchard's identity. - Peter Luschny, Jul 14 2016
a(n) ~ 4^n/sqrt(Pi*n). - Ilya Gutkovskiy, Jul 14 2016
Asymptotic starts a(n) ~ (4^n/sqrt(Pi*n))*(1 + (23/2^3)/n - (1199/2^7)/n^2 +(22685/2^10)/n^3 - (1562421/2^15)/n^4 + ... ). - Peter Luschny, Jul 14 2016