|
Search: id:A126178
|
|
|
| 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). 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). |
|
+0
1
|
|
| 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
(list; table; graph; listen)
|
|
|
OFFSET
|
0,3
|
|
|
COMMENT
|
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 (A000244). sum(kT(n,k), k=0..n)=3*A026376(n) (n>=1).
|
|
REFERENCES
|
F. Harary and R. C. Read, The enumeration of tree-like polyhexes, Proc. Edinburgh Math. Soc. (2) 17 (1970), 1-13.
|
|
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.
|
|
EXAMPLE
|
Triangle starts:
1;
0,3;
1,0,9;
0,9,0,27;
2,0,54,0,81;
|
|
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
|
|
CROSSREFS
|
Cf. A002212, A000244, A026376.
Sequence in context: A078521 A137432 A135871 this_sequence A094753 A143398 A067176
Adjacent sequences: A126175 A126176 A126177 this_sequence A126179 A126180 A126181
|
|
KEYWORD
|
nonn,tabl
|
|
AUTHOR
|
Emeric Deutsch (deutsch(AT)duke.poly.edu), Dec 19 2006
|
|
|
Search completed in 0.002 seconds
|
