A099250 Bisection of Motzkin numbers A001006.

1, 4, 21, 127, 835, 5798, 41835, 310572, 2356779, 18199284, 142547559, 1129760415, 9043402501, 73007772802, 593742784829, 4859761676391, 40002464776083, 330931069469828, 2750016719520991, 22944749046030949, 192137918101841817, 1614282136160911722

Offset: 0

Author

N. J. A. Sloane, Nov 16 2004

Keywords

• nonn
• easy

Comments

a(n) is the number of grand Motzkin paths from (0,0) to (2n+2,0) that avoid vertices (k,0) for all odd k and end on a down step. - Alexander Burstein, May 11 2021

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• Veronika Irvine, Lace Tessellations: A mathematical model for bobbin lace and an exhaustive combinatorial search for patterns, PhD Dissertation, University of Victoria, 2016.
• Veronika Irvine, Stephen Melczer, and Frank Ruskey, Vertically constrained Motzkin-like paths inspired by bobbin lace, arXiv:1804.08725 [math.CO], 2018.

Crossrefs

Cf. A001006, A026945, A048990.

Programs

Maple

G:=(1-x-(1-2*x-3*x^2)^(1/2))/(2*x^2): GG:=series(G,x=0,60): seq(coeff(GG,x^(2*n-1)),n=1..24);  # Emeric Deutsch
M := proc(n) option remember; `if`(n<2,1,(3*(n-1)*M(n-2)+(2*n+1)*M(n-1))/(n+2)) end: A099250 := n -> M(2*n+1):
seq(A099250(i),i=0..20); # Peter Luschny, Sep 11 2011

Mathematica

Take[CoefficientList[Series[(1-x-(1-2x-3x^2)^(1/2))/(2x^2), {x,0,60}], x], {2,-1,2}] (* Harvey P. Dale, Sep 11 2011 *)
Table[Hypergeometric2F1[-1/2-n, -n, 2, 4], {n, 0, 30}] (* Jean-François Alcover, Apr 03 2015 *)
MotzkinNumber = DifferenceRoot[Function[{y, n}, {(-3n-3)*y[n] + (-2n-5)*y[n+1] + (n+4)*y[n+2] == 0, y[0] == 1, y[1] == 1}]];
Table[MotzkinNumber[2n+1], {n, 0, 20}] (* Jean-François Alcover, Oct 27 2021 *)

PARI

my(x='x+O('x^66)); v=Vec((1-x-(1-2*x-3*x^2)^(1/2))/(2*x^2)); vector(#v\2,n,v[2*n]) \\ Joerg Arndt, May 12 2013

PARI

{a(n)=polcoeff(1/x*serreverse( x*(1+x)/((1+2*x)^2*(1+x+x^2) +x^2*O(x^n)) ),n)}
for(n=0,30,print1(a(n),", ")) \\ Paul D. Hanna, Oct 03 2014

Formula

a(n) = (2/Pi)*Integral_{x=-1..1} (1+2*x)^(2*n+1)*sqrt(1-x^2). [Peter Luschny, Sep 11 2011]

Recurrence: (n+1)*(2*n+3)*a(n) = (14*n^2+23*n+6)*a(n-1) + 3*(14*n^2-37*n+21)*a(n-2) - 27*(n-2)*(2*n-3)*a(n-3). - Vaclav Kotesovec, Oct 17 2012

a(n) ~ 3^(2*n+5/2)/(4*sqrt(2*Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 17 2012

G.f.: (1/x) * Series_Reversion( x*(1+x) / ( (1+2*x)^2 * (1+x+x^2) ) ). - Paul D. Hanna, Oct 03 2014

From Peter Bala, Apr 20 2024: (Start)

a(n) = Sum_{k = 0..2*n+1} (-1)^(k+1) * binomial(2*n+1, k)*Catalan(k+1).

a(n) = Sum_{k = 0..2*n+1} (-1)^k * binomial(2*n+1, k)*Catalan(k+1)*3^(2*n-k+1).

(4*n - 1)*(2*n + 3)*(n + 1)*a(n) = 2*(4*n + 1)*(10*n^2 + 5*n - 3)*a(n-1) - 9*(4*n + 3)*(2*n - 1)*(n - 1)*a(n-2) with a(0) = 1 and a(1) = 4. (End)

Extensions

More terms from Emeric Deutsch, Nov 17 2004