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A120985
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Number of ternary trees with n edges and having no vertices of degree 2. A ternary tree is a rooted tree in which each vertex has at most three children and each child of a vertex is designated as its left or middle or right child.
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4
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1, 3, 9, 28, 93, 333, 1272, 5085, 20925, 87735, 372879, 1602450, 6953824, 30438138, 134255403, 596154495, 2662813341, 11955684591, 53927330037, 244250703252, 1110401393067, 5065143385647, 23176155530394, 106344639962973
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OFFSET
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0,2
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COMMENTS
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LINKS
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FORMULA
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a(n) = (1/(n+1)) * Sum_{j..floor(n/3)} 3^(n-3*j) * binomial(n+1,2*j+1) * binomial(n-2*j,j).
G.f.=G(z) satisfies G=1+3zG + z^3*G^3.
Recurrence: 2*n*(2*n+3)*a(n) = 6*(6*n^2-1)*a(n-1) - 54*(n-1)*(2*n-1)*a(n-2) + 135*(n-2)*(n-1)*a(n-3). - Vaclav Kotesovec, Oct 19 2012
a(n) ~ (3+3/2^(2/3))^(n+3/2)/(2*sqrt(3*Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 19 2012
a(n) = (1/(n+1)) * Sum_{k=0..floor(n/2)} (-3)^k * binomial(n+1,k) * binomial(3*n-3*k+3,n-2*k). - Seiichi Manyama, Mar 23 2024
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EXAMPLE
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a(1)=3 because we have (Q,L), (Q,M) and (Q,R), where Q denotes the root and L (M,R) denotes a left (middle, right) child of Q.
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MAPLE
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a:=n->sum(3^(n-3*j)*binomial(n+1, 2*j+1)*binomial(n-2*j, j), j=0..n/2)/(n+1): seq(a(n), n=0..27);
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MATHEMATICA
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Table[1/(n+1)*Sum[3^(n-3*j)*Binomial[n+1, 2*j+1]*Binomial[n-2*j, j], {j, 0, n/2}], {n, 0, 20}] (* Vaclav Kotesovec, Oct 19 2012 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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