A191523 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).

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, 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, 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

Offset

1,8

Comments

Row n contains n entries.

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

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

Links

Table of n, a(n) for n=1..91.

Formula

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).

Example

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).
Triangle starts:
  1;
  1, 1;
  1, 1, 1;
  1, 3, 1, 1;
  1, 3, 4, 1, 1;
  1, 6, 6, 5, 1, 1;

Maple

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

Crossrefs

Cf. A001405, A191524.

Sequence in context: A214635 A166030 A351149 * A132890 A370783 A295295

Adjacent sequences: A191520 A191521 A191522 * A191524 A191525 A191526

Keyword

nonn,tabl

Author

Emeric Deutsch, Jun 05 2011

Status

approved