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A126177
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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)).
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1
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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
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OFFSET
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1,1
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COMMENTS
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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).
Sum_{k=1..1+floor(n/2)} k*T(n,k) = A026375(n).
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LINKS
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FORMULA
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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.
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EXAMPLE
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Triangle starts:
3;
9, 1;
27, 9;
81, 54, 2;
243, 270, 30;
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MAPLE
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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
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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