A126177 Triangle read by rows: T(n,k) is number of hex trees with n edges and k leaves (n >= 1, 1 <= k <= 1 + floor(n/2)).

3, 9, 1, 27, 9, 81, 54, 2, 243, 270, 30, 729, 1215, 270, 5, 2187, 5103, 1890, 105, 6561, 20412, 11340, 1260, 14, 19683, 78732, 61236, 11340, 378, 59049, 295245, 306180, 85050, 5670, 42, 177147, 1082565, 1443420, 561330, 62370, 1386, 531441, 3897234

Offset: 1

Author

Emeric Deutsch, Dec 19 2006

Keywords

• nonn
• tabf

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 middle child, or a right child (name due to an obvious bijection with certain tree-like polyhexes; see the Harary-Read paper).

Also number of hex trees with n edges and k-1 nodes of outdegree 2.

Row n has 1 + floor(n/2) terms.

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

T(n,1) = 3^n (A000244).

T(n,2) = A027472(n+1).

Sum_{k=1..1+floor(n/2)} k*T(n,k) = A026375(n).

Examples

			Triangle starts:
    3;
    9,   1;
   27,   9;
   81,  54,   2;
  243, 270,  30;
		

Links

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

Crossrefs

Cf. A002212, A000244, A027472, A026375.

Programs

Maple

T:=(n,k)->3^(n-2*k+2)*binomial(2*k-2,k-1)*binomial(n,2*k-2)/k: for n from 1 to 13 do seq(T(n,k),k=1..1+floor(n/2)) od; # yields sequence in triangular form

Formula

T(n,k) = 3^(n-2k+2)*binomial(2k-2,k-1)*binomial(n,2k-2)/k. Proof: There are Catalan(k-1) full binary trees with k leaves. Each of them has 2k-2 edges. Additional n-2k+2 edges can be inserted as paths at the existing 2k-1 vertices in 3^(n-2k+2)*binomial(n,2k-2) ways.

G.f.: G=G(t,z) satisfies z^2*G^2-(1-3z-2tz^2)G+tz(3+tz)=0.