A143359 Triangle read by rows, T(n,k) = number of symmetric ordered trees with n edges and root degree k (1 <= k <= n).

1, 1, 1, 2, 0, 1, 3, 1, 1, 1, 6, 0, 3, 0, 1, 10, 2, 4, 2, 1, 1, 20, 0, 10, 0, 4, 0, 1, 35, 5, 15, 5, 5, 3, 1, 1, 70, 0, 35, 0, 15, 0, 5, 0, 1, 126, 14, 56, 14, 21, 9, 6, 4, 1, 1, 252, 0, 126, 0, 56, 0, 21, 0, 6, 0, 1, 462, 42, 210, 42, 84, 28, 28, 14, 7, 5, 1, 1, 924, 0, 462, 0, 210, 0, 84, 0, 28, 0, 7, 0, 1

Offset: 1

Author

Emeric Deutsch, Aug 15 2008

Keywords

• nonn
• tabl

Comments

Number of symmetric Dyck n-paths with k returns to the x-axis. - David Scambler, Aug 16 2012

Examples

			Triangle starts:
   1;
   1,  1;
   2,  0,  1;
   3,  1,  1,  1;
   6,  0,  3,  0,  1;
  10,  2,  4,  2,  1,  1;
  20,  0, 10,  0,  4,  0,  1;
  35,  5, 15,  5,  5,  3,  1,  1;
		

Crossrefs

Cf. A001405, A000108 (column 2), A143360, A037952 (column 3).

Programs

Maple

C:=proc(z) options operator, arrow: (1/2-(1/2)*sqrt(1-4*z))/z end proc: S:=1/(1-z-z^2*C(z^2)): G:=(1+t*z*S)/(1-t^2*z^2*C(z^2))-1: Gser:=simplify(series(G, z=0,15)): for n to 13 do P[n]:=coeff(Gser,z,n) end do: for n to 13 do seq(coeff(P[n],t,j),j=1..n) end do; # yields sequence in triangular form

Mathematica

Module[{nmax = 13, G, C, S},
   G = (1 + t*z*S[z])/(1 - t^2*z^2*C[z^2]) - 1;
   S[z_] = 1/(1 - z - z^2*C[z^2]) ;
   C[z_] = (1 - Sqrt[1 - 4 z])/(2 z);
   CoefficientList[#/t + O[t]^nmax, t]& /@
   CoefficientList[G/z + O[z]^nmax, z]
] // Flatten (* Jean-François Alcover, Apr 09 2024 *)

Formula

G.f. = (1+t*z*S)/(1-t^2*z^2*C(z^2))-1, where S = 1/(1-z-z^2*C(z^2)) is the g.f. of the sequence binomial(n, floor(n/2)) (A001405) and C(z) = (1-sqrt(1-4z))/(2z) is the generating function of the Catalan numbers (A000108).

Sum_{k=1..n} T(n,k) = A001405(n).

T(n,1) = A001405(n-1).

Sum_{k=1..n} k*T(n,k) = A143360(n).

Sum_{k=2..n} T(n,k) = A037952(n). - R. J. Mathar, Sep 24 2021