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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.

A101275 Triangle read by rows: T(n,k) is the number of Schroeder paths of length 2n having exactly k down steps hitting the x-axis.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 13, 7, 1, 1, 44, 34, 10, 1, 1, 165, 150, 64, 13, 1, 1, 680, 659, 346, 103, 16, 1, 1, 3001, 2973, 1753, 659, 151, 19, 1, 1, 13880, 13844, 8716, 3798, 1116, 208, 22, 1, 1, 66345, 66300, 43384, 20798, 7226, 1744, 274, 25, 1, 1, 324908, 324853, 217804
Offset: 0

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Author

Emeric Deutsch, Dec 20 2004

Keywords

Comments

A Schroeder path of length 2n is a lattice path starting from (0,0), ending at (2n,0), consisting only of steps U=(1,1) (up steps), D=(1,-1) (down steps) and H=(2,0) (level steps) and never going below the x-axis. Schroeder paths are counted by the large Schroeder numbers (A006318).

Examples

			T(2,1)=4 because we have UHD, UUDD, HUD and UDH.
Triangle begins:
  1;
  1,  1;
  1,  4,  1;
  1, 13,  7,  1;
  1, 44, 34, 10, 1;
		

Crossrefs

Cf. A006318.

Programs

  • Maple
    G:=2/(2-2*z-t+t*z+t*sqrt(1-6*z+z^2)): Gser:=simplify(series(G,z=0,12)): P[0]:=1: for n from 1 to 10 do P[n]:=coeff(Gser,z^n) od: for n from 0 to 10 do seq(coeff(t*P[n],t^k),k=1..n+1) od; # yields the sequence in triangular form
  • Maxima
    T(n,k):=if k=0 then 1 else k*sum(((sum(binomial(m+k,i)*binomial(2*m+k-i-1,m+k-1),i,0,m))*binomial(n-m,k))/(m+k),m,0,n-k); /* Vladimir Kruchinin, Apr 20 2015 */
    
  • PARI
    T(n, k)= if (k==0, 1, k*sum(m=0,n-k,sum(i=0,m, binomial(m+k, i)*binomial(2*m+k-i-1, m+k-1)*binomial(n-m, k))/(m+k)));
    tabl(nn) = {for (n=0, nn, for (k=0, n, print1(T(n, k), ", ");); print(););} \\ Michel Marcus, Apr 21 2015

Formula

G.f.: 2/(2-2*z-t+t*z+t*sqrt(1-6*z+z^2)).
1/(1-x-xy/(1-x-x/(1-x-x/(1-x-x/(1-x-x/(1-.... (continued fraction). - Paul Barry, Feb 01 2009
T(n,k) = k*Sum_{m=0..n-k} (C(n-m,k)/(m+k))*Sum_{i=0..m} C(m+k,i)*C(2*m+k-i-1,m+k-1), T(n,0) = 1. - Vladimir Kruchinin, Apr 20 2015