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%I #19 Jan 26 2020 17:45:41
%S 1,1,1,3,4,1,11,19,9,1,45,96,66,16,1,197,501,450,170,25,1,903,2668,
%T 2955,1520,365,36,1,4279,14407,18963,12355,4165,693,49,1,20793,78592,
%U 119812,94528,41230,9856,1204,64,1,103049,432073,748548,693588,372078,117054
%N Triangle read by rows: T(n,k) is the number of bicolored Dyck paths of semilength n and having k peaks of the form ud (0 <= k <= n). A bicolored Dyck path is a Dyck path in which each up-step is of two kinds: u and U.
%C Row sums yield A052701. Column 0 yields the little Schroeder numbers (A001003). Sum_{k=0..n} k*T(n,k) = A069720(n).
%C Triangle T(n,k), 0 <= k <= n, read by rows; given by [1, 2, 1, 2, 1, 2, 1, 2, 1, 2, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...] where DELTA is the operator defined in A084938. - _Philippe Deléham_, Dec 23 2005
%H Paul Barry, <a href="https://arxiv.org/abs/1803.10297">Generalized Eulerian Triangles and Some Special Production Matrices</a>, arXiv:1803.10297 [math.CO], 2018.
%F T(n,k) = (1/n)*binomial(n,k)*Sum_{j=0..n-k} 2^j*binomial(n, j+1)*binomial(n-k, j) (k <= n-1); T(n, n)=1.
%F G.f. = G = G(t, z) satisfies G = 1 + z*(G-1+t)*G + z*G^2.
%e T(3,2)=9 because we have (ud)(ud)Ud, (ud)Ud(ud), Ud(ud)(ud), (ud)u(ud)d,
%e (ud)U(ud)d, u(ud)d(ud), U(ud)d(ud), u(ud)(ud)d and U(ud)(ud)d (the ud peaks are shown between parentheses).
%e Triangle starts:
%e 1;
%e 1, 1;
%e 3, 4, 1;
%e 11, 19, 9, 1;
%e 45, 96, 66, 16, 1;
%p T:=proc(n,k) if k<=n-1 then (1/n)*binomial(n,k)*sum(2^j*binomial(n,j+1)*binomial(n-k,j),j=0..n-k) elif k=n then 1 else 0 fi end: for n from 0 to 10 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form
%t T[n_, k_] := If[k <= n-1, (1/n)*Binomial[n, k]*Sum[2^j*Binomial[n, j+1]* Binomial[n-k, j], {j, 0, n-k}], If[k == n, 1, 0]];
%t Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Jul 11 2018, from Maple *)
%Y Cf. A001003, A052701, A069720.
%K nonn,tabl
%O 0,4
%A _Emeric Deutsch_, Dec 15 2005
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