

A182880


Triangle read by rows: T(n,k) is the number of weighted lattice paths in L_n having k (1,1)steps. L_n is the set of lattice paths of weight n that start at (0,0) and end on the horizontal axis and whose steps are of the following four kinds: a (1,0)step with weight 1; a (1,0)step with weight 2; a (1,1)step with weight 2; a (1,1)step with weight 1. The weight of a path is the sum of the weights of its steps.


1



1, 1, 2, 3, 2, 5, 6, 8, 18, 13, 44, 6, 21, 102, 30, 34, 222, 120, 55, 466, 390, 20, 89, 948, 1140, 140, 144, 1884, 3066, 700, 233, 3672, 7770, 2800, 70, 377, 7044, 18780, 9800, 630, 610, 13332, 43710, 31080, 3780, 987, 24946, 98610, 91560, 17850, 252, 1597, 46218, 216732, 254400, 72450, 2772
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OFFSET

0,3


COMMENTS

Sum of entries in row n is A051286(n).


REFERENCES

M. Bona and A. Knopfmacher, On the probability that certain compositions have the same number of parts, Ann. Comb., 14 (2010), 291306.
E. Munarini, N. Zagaglia Salvi, On the rank polynomial of the lattice of order ideals of fences and crowns, Discrete Mathematics 259 (2002), 163177.


LINKS



FORMULA

T(n,0) = A000045(n+1) (the Fibonacci numbers).
Sum_{k=0..n} k*T(n,k) = A182881(n).
G.f.: G(t,z) = 1/sqrt(1  2*z  z^2 + 2*z^3 + z^4  4*t*z^3).
The g.f. of column k is binomial(2n,n)*z^(3n)/(1zz^2)^(2n+1).


EXAMPLE

T(3,1)=2. Indeed, denoting by h (H) the (1,0)step of weight 1 (2), and u=(1,1), d=(1,1), the five paths of weight 3 are ud, du, hH, Hh, and hhh; two of them, ud and du, have exactly one u step.
Triangle starts:
1;
1;
2;
3, 2;
5, 6;
8, 18;
13, 44, 6;


MAPLE

G:=1/sqrt(12*zz^2+2*z^3+z^44*t*z^3): Gser:=simplify(series(G, z=0, 18)): for n from 0 to 16 do P[n]:=sort(coeff(Gser, z, n)) od: for n from 0 to 16 do seq(coeff(P[n], t, k), k=0..floor(n/3)) od; # yields sequence in triangular form


CROSSREFS



KEYWORD

nonn,tabf


AUTHOR



STATUS

approved



