A325921 Number of Motzkin meanders of length n with an even number of humps and an even number of peaks.
1, 2, 4, 8, 17, 38, 92, 239, 653, 1832, 5192, 14726, 41683, 117822, 333312, 945952, 2698117, 7740920, 22337788, 64788768, 188683267, 551179370, 1613612996, 4731245903, 13888157307, 40804653640, 119984904744, 353085202434, 1039830559085, 3064566227434
Offset: 0
Keywords
Examples
For n=0, 1, 2, 3 there are 2^n paths: all the paths without D (0 humps, 0 peaks). For example, for n=3: UUU, UUH, UHU, UHH, HUU, HUH, HHU, HHH. For n=4, the "extra" path is UDUD (2 humps, 2 peaks). The smallest example with #(humps) <> #(peaks) is UHDUHD (2 humps, 0 peaks).
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..2100
- 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).
Programs
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Maple
b:= proc(x, y, t, p, h) option remember; `if`(x=0, `if`(p+h=0, 1, 0), `if`(y>0, b(x-1, y-1, 0, irem(p+`if`(t=1, 1, 0), 2), irem(h+ `if`(t=2, 1, 0), 2)), 0)+b(x-1, y, `if`(t>0, 2, 0), p, h)+ b(x-1, y+1, 1, p, h)) end: a:= n-> b(n, 0$4): seq(a(n), n=0..35); # Alois P. Heinz, Jul 03 2019
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Mathematica
CoefficientList[Series[(1/(8*x))*((-1 + 4*x - 3*x^2 + Sqrt[(-(-1 + x)^2)* (-1 + 2*x + 3*x^2)])/ (1 - 4*x + 3*x^2) - (-1 + 4*x - 5*x^2 + 2*x^3 + Sqrt[(-1 + x)^3*(-1 + x + 4*x^3)])/((-1 + x)^2* (-1 + 2*x)) + (-1 + 4*x - 5*x^2 + Sqrt[1 - 4*x + 6*x^2 - 4*x^3 + 5*x^4])/ (1 - 4*x + 5*x^2) + (-1 + 4*x - 3*x^2 - 2*x^3 + Sqrt[1 - 4*x + 2*x^2 + 8*x^3 - 11*x^4 + 4*x^5 + 4*x^6])/(1 - 4*x + 3*x^2 + 2*x^3)), {x, 0, 30}], x] (* Vaclav Kotesovec, Jul 03 2019 *)
Formula
G.f.: ( (-1+4*t-3*t^2+sqrt(-3*t^4+4*t^3+2*t^2-4*t+1))/(3*t^2-4*t+1) + (-1+4*t-5*t^2+2*t^3+sqrt(4*t^6-12*t^5+13*t^4-8*t^3+6*t^2-4*t+1))/(-2*t^3+5*t^2-4*t+1) + (-1+4*t-5*t^2+sqrt(5*t^4-4*t^3+6*t^2-4*t+1))/(5*t^2-4*t+1) + (-1+4*t-3*t^2-2*t^3+sqrt(4*t^6+4*t^5-11*t^4+8*t^3+2*t^2-4*t+1))/(2*t^3+3*t^2-4*t+1) ) / (8*t).
a(n) ~ 3^(n + 1/2) / (4*sqrt(Pi*n)). - Vaclav Kotesovec, Jul 03 2019
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