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%I #13 Oct 05 2019 04:09:28
%S 1,1,1,1,2,3,6,6,2,17,15,10,46,51,30,5,128,175,91,35,372,568,336,140,
%T 14,1109,1827,1296,504,126,3349,5980,4785,2010,630,42,10221,19833,
%U 17215,8415,2640,462,31527,66078,61908,34210,11385,2772,132,98178,220649,223444,134706,50908,13299,1716
%N Triangle read by rows: T(n,k) is the number of Dyck paths of semilength n having k ascents of length 2.
%C T(n,k) is the number of Łukasiewicz paths of length n having k steps (1,1). A Łukasiewicz path of length n is a path in the first quadrant from (0,0) to (n,0) using rise steps (1,k) for any positive integer k, level steps (1,0) and fall steps (1,-1) (see R. P. Stanley, Enumerative Combinatorics, Vol. 2, Cambridge Univ. Press, Cambridge, 1999, p. 223, Exercise 6.19w; the integers are the slopes of the steps). Example: T(3,0)=2 because we have HHH and U(2)DD, where H=(1,0), U(2)=(1,2) and D=(1,-1). Row n has 1+floor(n/2) terms. Row sums yield the Catalan numbers (A000108). T(2n,n)=A000108(n). Column 0 is A102403
%H Alois P. Heinz, <a href="/A102402/b102402.txt">Rows n = 0..200, flattened</a>
%H <a href="/index/Lu#Lukasiewicz">Index entries for sequences related to Łukasiewicz</a>
%F G.f.: G=G(t,z) satisfies z^3*(1-t)G^3+z(1-z+tz)G^2-G+1=0.
%e T(4,2) = 2 because we have UUDDUUDD and UUDUUDDD, where U=(1,1) and D=(1,-1).
%e Triangle begins:
%e 1;
%e 1;
%e 1, 1;
%e 2, 3;
%e 6, 6, 2;
%e 17, 15, 10;
%t m = 14; G[_, _] = 0;
%t Do[G[t_, z_] = 1 + G[t, z]^2 z + G[t, z]^2 t z^2 - G[t, z]^2 z^2 + G[t, z]^3 z^3 - G[t, z]^3 t z^3 + O[t]^m + O[z]^m, {m}];
%t CoefficientList[#, t]& /@ Take[CoefficientList[G[t, z], z], m] // Flatten (* _Jean-François Alcover_, Oct 05 2019 *)
%Y Cf. A000108, A102403.
%K nonn,tabf
%O 0,5
%A _Emeric Deutsch_, Jan 06 2005
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