A246182 Triangle read by rows: T(n,k) is the number of weighted lattice paths B(n) having k hh's. B(n) is the set of lattice paths of weight n that start in (0,0), end on the horizontal axis and never go below this axis, whose steps are of the following four kinds: a (1,0)-step h of weight 1; a (1,0)-step H of weight 2; a (1,1)-step u of weight 2; a (1,-1)-step d of weight 1. The weight of a path is the sum of the weights of its steps.
1, 1, 1, 1, 3, 0, 1, 5, 2, 0, 1, 9, 5, 2, 0, 1, 19, 9, 6, 2, 0, 1, 39, 21, 12, 7, 2, 0, 1, 79, 53, 27, 15, 8, 2, 0, 1, 167, 118, 74, 34, 18, 9, 2, 0, 1, 357, 269, 180, 96, 42, 21, 10, 2, 0, 1, 763, 639, 419, 254, 119, 51, 24, 11, 2, 0, 1, 1651, 1486, 1045, 605, 340, 143, 61, 27, 12, 2, 0, 1
Offset: 0
Examples
Row 3 is 3,0,1. Indeed, the four paths of weight 3 are: ud, hH, Hh, and hhh, having 0, 0, 0, and 2 hh's, respectively. Triangle starts: 1; 1; 1,1; 3,0,1; 5,2,0,1; 9,5,2,0,1;
Links
- Alois P. Heinz, Rows n = 0..141, flattened
- M. Bona and A. Knopfmacher, On the probability that certain compositions have the same number of parts, Ann. Comb., 14 (2010), 291-306.
Programs
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Maple
eq := z^3*(1+z-t*z)*G^2-(-z^3+1-z^2-t*z+t*z^3)*G+1+z-t*z = 0: g := RootOf(eq, G): gser := simplify(series(g, z = 0, 18)): for j from 0 to 15 do P[j] := coeff(gser, z, j) end do: 1; for j to 13 do seq(coeff(P[j], t, q), q = 0 .. j-1) end do; # yields sequence in triangular form # second Maple program: b:= proc(n, y, t) option remember; `if`(y<0 or y>n, 0, `if`(n=0, 1, expand(b(n-1, y, 1)*`if`(t=1, x, 1)+ `if`(n>1, b(n-2, y, 0)+ b(n-2, y+1, 0), 0) +b(n-1, y-1, 0)))) end: T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0$2)): seq(T(n), n=0..12); # Alois P. Heinz, Aug 24 2014
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Mathematica
b[n_, y_, t_] := b[n, y, t] = If[y<0 || y>n, 0, If[n==0, 1, Expand[b[n-1, y, 1] * If[t==1, x, 1] + If[n>1, b[n-2, y, 0] + b[n-2, y+1, 0], 0] + b[n-1, y-1, 0]]]]; T[n_] := Function[{p}, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][b[n, 0, 0]]; Table[T[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Jun 29 2015, after Alois P. Heinz *)
Formula
G.f. G=G(t,z) satisfies z^3*(1+z-t*z)*G^2 - (1-t*z-z^2+t*z^3-z^3)*G+1+z-t*z=0.
Comments