A191523 - OEIS

%I #13 Jul 17 2017 02:16:54

%S 1,1,1,1,1,1,1,3,1,1,1,3,4,1,1,1,6,6,5,1,1,1,6,12,8,6,1,1,1,10,20,20,

%T 10,7,1,1,1,10,30,35,28,12,8,1,1,1,15,50,70,54,37,14,9,1,1,1,15,65,

%U 115,116,75,47,16,10,1,1,1,21,105,210,224,175,99,58,18,11,1,1,1,21,126,315,420,357,246,126,70,20,12,1,1

%N Triangle read by rows: T(n,k) is the number of left factors of Dyck paths of length n and having k double rises, i.e., two consecutive (1,1)-steps (n>=1, 0<=k<=n-1).

%C Row n contains n entries.

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

%C Sum_{k>=0} k*T(n,k) = A191524(n).

%F G.f.: G(t,z)=(z+r+r*z)/(1-t*z*(1+r)) where r=r(t,z) is a solution of z^2*(1+r)*(1+t*r) (the Narayana function with argument z^2).

%e T(5,2)=4 because we have UD(U[U)U], (UU)D(UU), (U[U)U]DD, and (U[U)U]DU, where U=(1,1) and D=(1,-1) (the double rises are shown between parentheses).

%e Triangle starts:

%e   1;

%e   1, 1;

%e   1, 1, 1;

%e   1, 3, 1, 1;

%e   1, 3, 4, 1, 1;

%e   1, 6, 6, 5, 1, 1;

%p eqr := R = z^2*(1+R)*(1+t*R): r := RootOf(eqr, R): G := (z+r+r*z)/(1-t*z*(1+r)): 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 = 0 .. n-1) end do; # yields sequence in triangular form

%Y Cf. A001405, A191524.

%K nonn,tabl

%O 1,8

%A _Emeric Deutsch_, Jun 05 2011