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%I #12 Jul 17 2017 02:18:01
%S 1,1,2,3,5,1,8,2,13,6,1,21,12,2,34,26,9,1,55,50,19,2,89,97,52,13,1,
%T 144,180,108,28,2,233,332,243,97,18,1,377,600,488,210,39,2,610,1076,
%U 999,552,170,24,1,987,1908,1942,1168,376,52,2,1597,3361,3773,2663,1162,282,31,1,2584,5878,7140,5466,2543,630,67,2
%N Triangle read by rows: T(n,k) is the number of dispersed Dyck paths (i.e., Motzkin paths with no (1,0) steps at positive heights) of length n having k peaks at height >=2.
%C Also number of dispersed Dyck paths of length n having k valleys (i.e., DU's). Example: T(7,2)=2 because we have HUDUDUD and UDUDUDH, where U=(1,1), D=(1,-1), H=(1,0).
%C Also number of dispersed Dyck paths of length n having k doublerises (i.e., UU's). Example: T(7,2)=2 because we have HUUUDDD and UUUDDDH, where U=(1,1), D=(1,-1), H=(1,0).
%C Row n>=2 has floor(n/2) entries.
%C Sum of entries in row n is binomial(n, floor(n/2)) = A001405(n).
%C T(n,0) = A000045(n+1) (the Fibonacci numbers).
%C T(n,1) = A054454(n-4).
%C Sum_{k>=0} k*T(n,k) = A191309(n).
%F G.f.: G=G(t,z) satisfies G = 1+z*G + z^2*G(1+t*r), where r=r(t,z) is given by r=z^2*(1+r)*(1+t*r) (the Narayana function).
%e T(7,2)=2 because we have HUUDUDD and UUDUDDH, where U=(1,1), D=(1,-1), H=(1,0).
%e Triangle starts:
%e 1;
%e 1;
%e 2;
%e 3;
%e 5, 1;
%e 8, 2;
%e 13, 6, 1;
%e 21, 12, 2;
%p r := ((1-z^2-t*z^2-sqrt(((1-z)^2-t*z^2)*((1+z)^2-t*z^2)))*1/2)/(t*z^2): G := 1/(1-z-z^2*(1+t*r)): Gser := simplify(series(G, z = 0, 20)): for n from 0 to 17 do P[n] := sort(coeff(Gser, z, n)) end do: 1; 1; for n from 2 to 17 do seq(coeff(P[n], t, k), k = 0 .. floor((1/2)*n)-1) end do; # yields sequence in triangular form
%Y Cf. A000045, A001405, A054454, A191309.
%K nonn,tabf
%O 0,3
%A _Emeric Deutsch_, May 30 2011
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