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.

A128750 Number of skew Dyck paths of semilength n having no ascents of length 1.

Original entry on oeis.org

1, 0, 2, 4, 14, 44, 150, 520, 1850, 6696, 24602, 91500, 343846, 1303572, 4979822, 19150352, 74075890, 288022160, 1125076210, 4413061972, 17375007294, 68641377980, 272014578822, 1081009104664, 4307221752874, 17203123381304
Offset: 0

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Author

Emeric Deutsch, Mar 31 2007

Keywords

Comments

A skew Dyck path is a path in the first quadrant which begins at the origin, ends on the x-axis, consists of steps U=(1,1)(up), D=(1,-1)(down) and L=(-1,-1)(left) so that up and left steps do not overlap. The length of a path is defined to be the number of its steps. An ascent in a path is a maximal sequence of consecutive U steps.
Hankel transform is 2^ceiling(n(n+1)/3). Binomial transform is A059278. - Paul Barry, Feb 11 2009

Examples

			a(3)=4 because we have UUUDDD, UUUDLD, UUUDDL and UUUDLL.
		

Crossrefs

Cf. A128749.

Programs

  • Maple
    G:=(1-z^2-sqrt((1-z^2)*(1-4*z-z^2)))/2/z/(1+z): Gser:=series(G,z=0,35): seq(coeff(Gser,z,n),n=0..30);
  • Mathematica
    CoefficientList[Series[(1-x^2-Sqrt[(1-x^2)*(1-4*x-x^2)])/2/x/(1+x), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 08 2014 *)

Formula

a(n) = A128749(n,0).
G.f.: G = G(z) satisfies z(1 + z)G^2 - (1 - z^2)G + 1 - z = 0.
G.f.: 1/(1+x-x/(1-x-x/(1+x-x/(1-x-x/(1+x-x/(1-... (continued fraction). - Paul Barry, Feb 11 2009
From Paul Barry, Feb 11 2009: (Start)
G.f.: (1/(1+x))c(x/(1-x^2)) where c(x) is the g.f. of A000108;
G.f.: 1/(1-2x^2/(1-2x-x^2/(1-2x-2x^2/(1-x-2x^2/(1-2x-x^2/(1-2x-2x^2/(1-x-2x^2/(1-.... (continued fraction);
a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(floor((n+k)/2),k)*A000108(k).
(End)
Conjecture: (n+1)*a(n) +(-4*n+3)*a(n-1) +(-2*n-1)*a(n-2) +(4*n-11)*a(n-3) +(n-4)*a(n-4)=0. - R. J. Mathar, Nov 15 2012
a(n) ~ sqrt(5+3*sqrt(5)) * (2+sqrt(5))^n / (4 * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Feb 08 2014