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%I #24 Oct 16 2015 23:00:31
%S 1,1,2,4,1,10,3,1,26,12,3,1,72,41,15,3,1,206,143,58,18,3,1,606,492,
%T 231,76,21,3,1,1820,1693,891,335,95,24,3,1,5558,5823,3403,1411,455,
%U 115,27,3,1,17206,20040,12870,5848,2061,591,136,30,3,1,53872,69033,48318,23858,9143,2850,743,158,33,3,1
%N Triangle read by rows: T(n,k) is number of Dyck paths of semilength n having k ascents of length 1 that start at an odd level.
%C T(n,k) is number of Łukasiewicz paths of length n having k level steps at an odd 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(3,1)=1 because we have only UHD with exactly one level step at an odd level; here U=(1,1), H=(1,0) and D=(1,-1). Row n has n-1 terms (n>=2). Row sums are the Catalan numbers (A000108). Column 0 yields A102407.
%C T(n,k) is the number of Dyck paths of semilength n with k DUDU's. - I. Tasoulas (jtas(AT)unipi.gr), Feb 19 2006
%H Alois P. Heinz, <a href="/A102405/b102405.txt">Rows n = 0..150, flattened</a>
%H A. Sapounakis, I. Tasoulas and P. Tsikouras, <a href="http://dx.doi.org/10.1016/j.disc.2007.03.005">Counting strings in Dyck paths</a>, Discrete Math., 307 (2007), 2909-2924.
%H <a href="/index/Lu#Lukasiewicz">Index entries for sequences related to Łukasiewicz</a>
%F G.f.: G=G(t, z) satisfies zG^2-(1+z-z^2-tz+tz^2)G+1+z-tz=0.
%e T(4,1) = 3 because we have UDUUD(U)DD, UUD(U)DDUD and UUUDD(U)DD, where U=(1,1), D=(1,-1) and the ascents of length 1 that start at an odd level are shown between parentheses.
%e Triangle starts:
%e 00 : 1;
%e 01 : 1;
%e 02 : 2;
%e 03 : 4, 1;
%e 04 : 10, 3, 1;
%e 05 : 26, 12, 3, 1;
%e 06 : 72, 41, 15, 3, 1;
%e 07 : 206, 143, 58, 18, 3, 1;
%e 08 : 606, 492, 231, 76, 21, 3, 1;
%e 09 : 1820, 1693, 891, 335, 95, 24, 3, 1;
%e 10 : 5558, 5823, 3403, 1411, 455, 115, 27, 3, 1;
%p b:= proc(x, y, t) option remember; `if`(y<0 or y>x, 0,
%p `if`(x=0, 1, expand(b(x-1, y-1, [2, 2, 4, 2][t])
%p +b(x-1, y+1, [1, 3, 1, 3][t])*`if`(t=4, z, 1))))
%p end:
%p T:= n-> (p-> seq(coeff(p, z, i), i=0..degree(p)))(b(2*n, 0, 1)):
%p seq(T(n), n=0..15); # _Alois P. Heinz_, Jun 02 2014
%t b[x_, y_, t_] := b[x, y, t] = If[y<0 || y>x, 0, If[x == 0, 1, Expand[b[x-1, y-1, {2, 2, 4, 2}[[t]]] + b[x-1, y+1, {1, 3, 1, 3}[[t]]]*If[t == 4, z, 1]]]]; T[n_] := Function[{p}, Table[Coefficient[p, z, i], {i, 0, Exponent[p, z]}]][b[2*n, 0, 1]]; Table[T[n], {n, 0, 15}] // Flatten (* _Jean-François Alcover_, May 20 2015, after _Alois P. Heinz_ *)
%Y Cf. A000108, A094507 (the same for UDUD), A102404, A102407, A263173.
%K nonn,tabf
%O 0,3
%A _Emeric Deutsch_, Jan 06 2005
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