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Triangle read by rows: T(n,k) is the number of length-n left factors of Dyck paths having length of the first ascent equal to k (1 <= k <= n).
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%I #11 Nov 03 2019 19:42:07

%S 1,1,1,1,1,1,2,2,1,1,3,3,2,1,1,6,6,4,2,1,1,10,10,7,4,2,1,1,20,20,14,8,

%T 4,2,1,1,35,35,25,15,8,4,2,1,1,70,70,50,30,16,8,4,2,1,1,126,126,91,56,

%U 31,16,8,4,2,1,1,252,252,182,112,62,32,16,8,4,2,1,1,462,462,336,210,119,63,32,16,8,4,2,1,1

%N Triangle read by rows: T(n,k) is the number of length-n left factors of Dyck paths having length of the first ascent equal to k (1 <= k <= n).

%C Sum of entries in row n is binomial(n, floor(n/2)) = A001405(n).

%C T(n,1) = A001405(n-2) (n>=1).

%C T(n,2) = A001405(n-2) (n>=2).

%C Sum_{k=1..n} k*T(n,k) = A191782(n).

%F G.f.: G(t,z) = (1 - z*c + t*z^3*c^2)/((1 - z*c)*(1 - t*z)*(1 - t*z^2)), where c = (1-sqrt(1 - 4*z^2))/(2*z^2).

%e T(5,2)=3 because we have UUDDU, UUDUD, and UUDUU, where U=(1,1) and D=(1,-1).

%e Triangle starts:

%e 1;

%e 1, 1;

%e 1, 1, 1;

%e 2, 2, 1, 1;

%e 3, 3, 2, 1, 1;

%e 6, 6, 4, 2, 1, 1;

%p c := ((1-sqrt(1-4*z^2))*1/2)/z^2: G := (1-z*c+t*z^3*c^2)/((1-z*c)*(1-t*z)*(1-t*z^2*c)): Gser := simplify(series(G, z = 0, 17)): for n to 13 do P[n] := sort(coeff(Gser, z, n)) end do: for n to 13 do seq(coeff(P[n], t, k), k = 1 .. n) end do; # yields sequence in triangular form

%Y Cf. A001405, A191782.

%K nonn,tabl

%O 1,7

%A _Emeric Deutsch_, Jun 18 2011