A129167 Number of base pyramids in all skew Dyck paths of semilength n.

0, 1, 3, 9, 30, 109, 420, 1685, 6960, 29391, 126291, 550359, 2426502, 10803801, 48507843, 219377949, 998436792, 4569488371, 21016589073, 97090411019, 450314942682, 2096122733211, 9788916220518, 45850711498859, 215348942668680, 1013979873542689, 4785437476592805, 22633143884165985, 107258646298581390

Offset: 0

Author

Emeric Deutsch, Apr 04 2007

Keywords

• nonn

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 the path is defined to be the number of its steps. A pyramid in a skew Dyck word (path) is a factor of the form u^h d^h, h being the height of the pyramid. A base pyramid is a pyramid starting on the x-axis.

a(n) = |A091699(n+1)|. Partial sums of A033321(n), n = 1, 2, 3, ....

a(n+1) is the number of 3-colored Motzkin paths of length n with no peaks at level 1. - José Luis Ramírez Ramírez, Mar 31 2013

Examples

			a(2)=3 because in the paths (UD)(UD), (UUDD) and UUDL we have altogether 3 base pyramids (shown between parentheses).
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• E. Deutsch, E. Munarini, S. Rinaldi, Skew Dyck paths, J. Stat. Plann. Infer. 140 (8) (2010) 2191-2203

Crossrefs

Cf. A033321, A091699, A129165.

Programs

Maple

G:=(1-3*z-sqrt(1-6*z+5*z^2))/z/(3-3*z-sqrt(1-6*z+5*z^2)): Gser:=series(G,z=0,30): seq(coeff(Gser,z,n),n=0..27);

Mathematica

CoefficientList[Series[(1-3*x-Sqrt[1-6*x+5*x^2])/(x*(3-3*x-Sqrt[1-6*x+5*x^2])), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 20 2012 *)

PARI

z='z+O('z^66); concat([0], Vec((1-3*z-sqrt(1-6*z+5*z^2))/z/(3-3*z-sqrt(1-6*z+5*z^2)))) \\ Joerg Arndt, Aug 27 2014

Formula

a(n) = Sum_{k=0..n} k*A129165(n,k).

G.f.: (1 - 3*z - sqrt(1 - 6*z + 5*z^2))/(z*(3 - 3*z - sqrt(1 - 6*z + 5*z^2))).

Recurrence: 2*(n+1)*a(n) = (13*n-3)*a(n-1) - 4*(4*n-3)*a(n-2) + 5*(n-1)*a(n-3) . - Vaclav Kotesovec, Oct 20 2012

a(n) ~ 5^(n+5/2)/(72*sqrt(Pi)*n^(3/2)) . - Vaclav Kotesovec, Oct 20 2012