A104550 Number of horizontal segments in all Schroeder paths of length 2n (a horizontal segment is a maximal string of horizontal steps).
1, 4, 20, 104, 552, 2972, 16172, 88720, 489872, 2719028, 15157188, 84799992, 475894200, 2677788492, 15102309468, 85347160608, 483183316512, 2739851422820, 15558315261812, 88462135512712, 503569008273992, 2869602773253884, 16368396446913420, 93449566652932784, 533954950648248752
Offset: 1
Keywords
Examples
a(2)=4 because we have (HH),(H)UD,UD(H),U(H)D,UDUD and UUDD; the 4 horizontal segments are shown between parentheses.
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..200
Programs
-
Maple
G:=(1-z)*(1-z-sqrt(1-6*z+z^2))/2/sqrt(1-6*z+z^2): Gser:=series(G,z=0,28): seq(coeff(Gser,z^n),n=1..24); a := n -> hypergeom([-n, n-1], [1], -1); seq(round(evalf(a(n),36)),n=1..23); # Peter Luschny, Aug 02 2014
-
Mathematica
Rest[CoefficientList[Series[(1-x)*(1-x-Sqrt[1-6*x+x^2])/ (2*Sqrt[1 -6*x+x^2]), {x, 0, 20}], x]] (* Vaclav Kotesovec, Oct 17 2012 *) -
Maxima
a(n):=sum(binomial(n+1,k)*binomial(n+k-1,k),k,0,n+1); /* Vladimir Kruchinin, Jun 15 2020 */
-
PARI
x='x+O('x^66); Vec((1-x)*(1-x-sqrt(1-6*x+x^2))/(2*sqrt(1-6*x+x^2))) \\ Joerg Arndt, May 13 2013
Formula
G.f.: (1-x)*(1-x-sqrt(1-6*x+x^2))/(2*sqrt(1-6*x+x^2)).
a(n) = Jacobi_P(n+1,-1,-2,3). [Paul Barry, Sep 27 2009]
Recurrence: n*a(n) = (7*n-6)*a(n-1) - (7*n-22)*a(n-2) + (n-4)*a(n-3). - Vaclav Kotesovec, Oct 17 2012
a(n) ~ sqrt(6*sqrt(2)-8)*(3+2*sqrt(2))^n/(2*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 17 2012
a(n) = Hyper2F1([-n, n-1], [1], -1). - Peter Luschny, Aug 02 2014
a(n) = Sum_{k=0..n+1} C(n+1,k)*C(n+k-1,k). - Vladimir Kruchinin, Jun 15 2020
Comments