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.

A126223 Number of level steps in all 2-Motzkin paths (i.e., Motzkin paths with blue and red level steps) of length n, without red level steps on the x-axis.

Original entry on oeis.org

0, 1, 2, 7, 26, 98, 372, 1419, 5434, 20878, 80444, 310726, 1202852, 4665412, 18126760, 70538355, 274877370, 1072515990, 4189573740, 16383007410, 64126407180, 251226790620, 985033185240, 3865138313790, 15176957307876, 59633260964748, 234453859803352
Offset: 0

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Author

Emeric Deutsch, Dec 28 2006

Keywords

Comments

a(n) is the number of increasing strict binary trees with 2n-1 nodes that avoid 213 and 321 in the classical sense. For more information about increasing strict binary trees with an associated permutation, see A245894. - Manda Riehl, Aug 07 2014

Examples

			a(3) = 7 because the 2-Motzkin paths without red level steps on the x-axis are BBB, BUD, UBD, URD and UDB, where U=(1,1), D=(1,-1), B=blue (1,0), R=red (1,0); they have a total of 3+1+1+1+1 = 7 level steps.
		

Crossrefs

Cf. A126222.

Programs

  • Maple
    G:=(1-2*z)*(1-2*z-sqrt(1-4*z))/2/z/sqrt(1-4*z): Gser:=series(G,z=0,32): seq(coeff(Gser,z,n),n=0..28);
    # second Maple program:
    a:= proc(n) option remember; `if`(n<2, n,
          2*(2*n-3)*(n^2-n+1)*a(n-1)/((n+1)*(n^2-3*n+3)))
        end:
    seq(a(n), n=0..30);  # Alois P. Heinz, May 20 2014
  • Mathematica
    CoefficientList[Series[(1-2*x)*(1-2*x-Sqrt[1-4*x])/(2*x*Sqrt[1-4*x]), {x, 0, 20}], x] (* Vaclav Kotesovec, Sep 08 2014 *)
    Flatten[{0, Table[2*(1-n+n^2) * Binomial[2*n-2, n-1]/(n*(n+1)), {n, 1, 25}]}] (* Vaclav Kotesovec, Sep 08 2014 *)

Formula

a(n) = Sum_{k=0..n} k*A126222(n,k).
G.f.: (1-2z)(1-2z-sqrt(1-4*z))/(2z*sqrt(1-4z)).
a(n) = 2*(2*n-3)*(n^2-n+1)*a(n-1)/((n+1)*(n^2-3*n+3)) for n>1. - Alois P. Heinz, May 20 2014
a(n) = 2*(1-n+n^2) * C(2*n-2, n-1) / (n*(n+1)). - Vaclav Kotesovec, Sep 08 2014