A307572 Number of Motzkin meanders of length n with an odd number of humps.

0, 0, 1, 5, 18, 56, 161, 443, 1196, 3228, 8823, 24579, 69810, 201380, 586843, 1719081, 5044584, 14800352, 43384747, 127076015, 372100654, 1089864344, 3194496987, 9372984609, 27532712140, 80966582548, 238342592353, 702222958797, 2070454005078, 6108341367004

Offset: 0

Author

Andrei Asinowski, Apr 15 2019

Keywords

• nonn

Comments

A Motzkin meander is a lattice path with steps from the set {D=-1, H=0, U=1} that starts at (0,0), and never goes below the x-axis.

A hump is an occurrence of the pattern UHH...HD (the number of Hs in the pattern is not fixed, and can be 0).

Examples

			For n = 3 the a(3) = 5 paths are UDH, HUD, UHD, UUD, UDU.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1000
• Andrei Asinowski, Axel Bacher, Cyril Banderier, Bernhard Gittenberger, Analytic combinatorics of lattice paths with forbidden patterns, the vectorial kernel method, and generating functions for pushdown automata, Algorithmica (2019).
• Tamás Szakács, Linear recursive sequences and factorials, Ph. D. Thesis, Univ. Debrecen (Hungary, 2024). See pp. 51, 58.

Crossrefs

Cf. A001006.

Programs

Maple

b:= proc(x, y, t, c) option remember; `if`(y<0, 0, `if`(x=0, c,
      b(x-1, y-1, 0, irem(c+t, 2))+b(x-1, y, t, c)+b(x-1, y+1, 1, c)))
    end:
a:= n-> b(n, 0$3):
seq(a(n), n=0..35);  # Alois P. Heinz, Apr 16 2019

Mathematica

b[x_, y_, t_, c_] := b[x, y, t, c] = If[y<0, 0, If[x==0, c, b[x-1, y-1, 0, Mod[c+t, 2]] + b[x-1, y, t, c] + b[x-1, y+1, 1, c]]];
a[n_] := b[n, 0, 0, 0];
Table[a[n], {n, 0, 35}] (* Jean-François Alcover, Apr 29 2019, after Alois P. Heinz *)

Formula

G.f.: (sqrt((1-t^2)/(1-4*t+3*t^2)) - sqrt((1+t^2)/(1-4*t+5*t^2))) / (4*t).

Conjecture: D-finite with recurrence -3*(n+1)*(n-2)*a(n) +12*(2*n^2-4*n-1)*a(n-1) +2*(-35*n^2+107*n-48)*a(n-2) +4*(21*n^2-89*n+80)*a(n-3) +4*(-5*n^2+32*n-43)*a(n-4) +4*(-8*n^2+62*n-115)*a(n-5) +2*(31*n^2-283*n+616)*a(n-6) -4*(23*n-97)*(n-6)*a(n-7) +15*(n-6)*(n-7)*a(n-8)=0. - R. J. Mathar, May 06 2020

a(n) ~ 3^(n + 1/2) / (2*sqrt(Pi*n)). - Vaclav Kotesovec, Mar 08 2023