A114583 Triangle read by rows: T(n,k) is the number of Motzkin paths of length n and having k UHD's, where U=(1,1),H=(1,0),D=(1,-1) (0<=k<=floor(n/3)).
1, 1, 2, 3, 1, 7, 2, 15, 6, 36, 14, 1, 85, 39, 3, 209, 102, 12, 517, 280, 37, 1, 1303, 758, 123, 4, 3312, 2085, 381, 20, 8510, 5730, 1194, 76, 1, 22029, 15849, 3657, 295, 5, 57447, 43914, 11187, 1056, 30, 150709, 122090, 33903, 3734, 135, 1, 397569, 340104
Offset: 0
Examples
T(5,1)=6 because we have HH(UHD), UD(UHD), (UHD)HH, (UHD)UD, H(UHD)H and U(UHD)D, where U=(1,1),H=(1,0),D=(1,-1) (the UHD's are shown between parentheses). Triangle begins: 1; 1; 2; 3, 1; 7, 2; 15, 6; 36, 14, 1; ...
Links
- Alois P. Heinz, Rows n = 0..300, flattened
- Marilena Barnabei, Flavio Bonetti, and Niccolò Castronuovo, Motzkin and Catalan Tunnel Polynomials, J. Int. Seq., Vol. 21 (2018), Article 18.8.8.
- Zhuang, Yan. A generalized Goulden-Jackson cluster method and lattice path enumeration, Discrete Mathematics 341.2 (2018): 358-379. Also arXiv: 1508.02793v2.
Programs
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Maple
G:=(1-z-t*z^3+z^3-sqrt((1-3*z+z^3-t*z^3)*(1+z+z^3-t*z^3)))/2/z^2: Gser:=simplify(series(G,z=0,20)): P[0]:=1: for n from 1 to 17 do P[n]:=sort(coeff(Gser,z^n)) od: for n from 0 to 17 do seq(coeff(t*P[n],t^j),j=1..1+floor(n/3)) od; # yields sequence in triangular form # second Maple program: b:= proc(x, y, t) option remember; expand(`if`(y<0 or y>x, 0, `if`(x=0, 1, b(x-1, y, `if`(t=1, 2, 0))+b(x-1, y-1, 0)* `if`(t=2, z, 1)+b(x-1, y+1, 1)))) end: T:= n-> (p-> seq(coeff(p, z, i), i=0..degree(p)))(b(n, 0$2)): seq(T(n), n=0..15); # Alois P. Heinz, Feb 01 2019 -
Mathematica
CoefficientList[#, t]& /@ CoefficientList[(1 - z - t z^3 + z^3 - Sqrt[(1 - 3z + z^3 - t z^3)(1 + z + z^3 - t z^3)])/2/z^2 + O[z]^17, z] // Flatten (* Jean-François Alcover, Aug 07 2018 *)
Formula
G.f.=G=G(t, z) satisfies G=1+zG+z^2*G(tz-z+G).
Comments