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%I #9 Jul 21 2019 13:26:17
%S 1,1,2,3,2,5,6,8,18,13,44,6,21,102,30,34,222,120,55,466,390,20,89,948,
%T 1140,140,144,1884,3066,700,233,3672,7770,2800,70,377,7044,18780,9800,
%U 630,610,13332,43710,31080,3780,987,24946,98610,91560,17850,252,1597,46218,216732,254400,72450,2772
%N 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.
%C Sum of entries in row n is A051286(n).
%D M. Bona and A. Knopfmacher, On the probability that certain compositions have the same number of parts, Ann. Comb., 14 (2010), 291-306.
%D E. Munarini, N. Zagaglia Salvi, On the rank polynomial of the lattice of order ideals of fences and crowns, Discrete Mathematics 259 (2002), 163-177.
%F T(n,0) = A000045(n+1) (the Fibonacci numbers).
%F Sum_{k=0..n} k*T(n,k) = A182881(n).
%F G.f.: G(t,z) = 1/sqrt(1 - 2*z - z^2 + 2*z^3 + z^4 - 4*t*z^3).
%F The g.f. of column k is binomial(2n,n)*z^(3n)/(1-z-z^2)^(2n+1).
%F Apparently, T(n,1) = 2*A001628(n-3), T(n,2) = 6*A001873(n-6), T(n,3) = 20*A001875(n-9). - _R. J. Mathar_, Dec 11 2010
%e 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.
%e Triangle starts:
%e 1;
%e 1;
%e 2;
%e 3, 2;
%e 5, 6;
%e 8, 18;
%e 13, 44, 6;
%p G:=1/sqrt(1-2*z-z^2+2*z^3+z^4-4*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
%Y Cf. A051286, A000045, A182881.
%K nonn,tabf
%O 0,3
%A _Emeric Deutsch_, Dec 11 2010
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