%I #13 Nov 15 2019 21:33:38
%S 1,0,1,1,0,1,2,2,0,1,5,5,3,0,1,14,14,9,4,0,1,39,46,27,14,5,0,1,114,
%T 143,101,44,20,6,0,1,339,466,341,184,65,27,7,0,1,1028,1524,1212,664,
%U 300,90,35,8,0,1,3163,5043,4279,2539,1145,454,119,44,9,0,1,9852,16812,15206,9564,4665,1819,651,152,54,10,0,1
%N Triangle read by rows: T(n,k) is number of Dyck paths of semilength n having k ascents of length 1 starting at an even level.
%C T(n,k) is number of Łukasiewicz paths of length n having k level steps at an even level. 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(4,1)=5 because we have (H)UHD, (H)U(2)DD, UHD(H), U(2)DD(H) and U(2)(H)DD, where H=(1,0), U(2)=(1,2) and D=(1,-1) and the level steps at even level are shown between parentheses. Row n contains n+1 terms. Row sums yield the Catalan numbers (A000108). Column 0 is A102406.
%H <a href="/index/Lu#Lukasiewicz">Index entries for sequences related to Łukasiewicz</a>
%F G.f.: G=G(t, z) satisfies z*(1+z-tz)^2*G^2 - (1 + z + z^2 - tz - tz^2)*G + 1 = 0.
%e T(3,1)=2 because we have (U)DUUDD and UUDD(U)D, where U=(1,1), D=(1,-1) and the ascents of length 1 starting at an even level are shown between parentheses.
%t m = 12; G[_] = 0;
%t Do[G[z_] = -((G[z]^2 z ((t-1)z - 1)^2 + 1)/((t-1)z^2 + (t-1)z - 1)) + O[z]^m, {m}];
%t CoefficientList[#, t]& /@ CoefficientList[G[z], z] // Flatten (* _Jean-François Alcover_, Nov 15 2019 *)
%Y Cf. A000108, A102405, A102406.
%K nonn,tabl
%O 0,7
%A _Emeric Deutsch_, Jan 06 2005