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A126178
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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).
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0
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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; refs; listen; history; internal format)
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OFFSET
| 0,3
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COMMENTS
| 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).
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REFERENCES
| F. Harary and R. C. Read, The enumeration of tree-like polyhexes, Proc. Edinburgh Math. Soc. (2) 17 (1970), 1-13.
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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.
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EXAMPLE
| Triangle starts:
1;
0,3;
1,0,9;
0,9,0,27;
2,0,54,0,81;
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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
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CROSSREFS
| Cf. A002212, A000244, A026376.
Sequence in context: A078521 A194938 A135871 * A094753 A143398 A202995
Adjacent sequences: A126175 A126176 A126177 * A126179 A126180 A126181
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KEYWORD
| nonn,tabl
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AUTHOR
| Emeric Deutsch (deutsch(AT)duke.poly.edu), Dec 19 2006
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