A126178 Triangle read by rows: T(n,k) is number of hex trees with n edges and k vertices of outdegree 1 (0<=k<=n).

1, 0, 3, 1, 0, 9, 0, 9, 0, 27, 2, 0, 54, 0, 81, 0, 30, 0, 270, 0, 243, 5, 0, 270, 0, 1215, 0, 729, 0, 105, 0, 1890, 0, 5103, 0, 2187, 14, 0, 1260, 0, 11340, 0, 20412, 0, 6561, 0, 378, 0, 11340, 0, 61236, 0, 78732, 0, 19683, 42, 0, 5670, 0, 85050, 0, 306180, 0, 295245, 0

Offset: 0

Author

Emeric Deutsch, Dec 19 2006

Keywords

• nonn
• tabl

Comments

A hex tree is a rooted tree where each vertex has 0, 1, or 2 children and, when only one child is present, it is either a left child, or a median child, or a right child (name due to an obvious bijection with certain tree-like polyhexes; see the Harary-Read paper).

Sum of terms in row n = A002212(n+1).

Column 0 yields the aerated Catalan numbers (1,0,1,0,2,0,5,0,14,...).

T(n,n) = 3^n (see A000244).

Sum_{k=0..n} k*T(n,k) = 3*A026376(n) (n>=1).

Examples

			Triangle starts:
  1;
  0,  3;
  1,  0,  9;
  0,  9,  0, 27;
  2,  0, 54,  0, 81;
		

Links

• F. Harary and R. C. Read, The enumeration of tree-like polyhexes, Proc. Edinburgh Math. Soc. (2) 17 (1970), 1-13.

Crossrefs

Cf. A000244, A002212, A026376.

Programs

Maple

T:=proc(n,k) if n-k mod 2 = 0 then 3^k*binomial(n+1,k)*binomial(n+1-k,(n-k)/2)/(n+1) else 0 fi end: for n from 0 to 11 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

Formula

T(n,k) = [3^k/(n+1)]binomial(n+1,k)*binomial(n+1-k,(n-k)/2) (0<=k<=n).

G.f.: G=G(t,z) satisfies G=1+3tzG+z^2*G^2.