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Search: id:A120982
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| A120982 |
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Triangle read by rows: T(n,k) is the number of ternary trees with n edges and having k vertices of outdegree 2 (n>=0, k>=0). 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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+0
5
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| 1, 3, 9, 3, 28, 27, 93, 162, 18, 333, 825, 270, 1272, 3915, 2430, 135, 5085, 18144, 17199, 2835, 20925, 84000, 106596, 34020, 1134, 87735, 391554, 612360, 308448, 30618, 372879, 1838295, 3369600, 2364390, 459270, 10206, 1602450, 8674380
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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Row n has 1+floor(n/2) terms. Row sums yield A001764. T(n,0)=A120985(n). Sum(k*T(n,k),k>=1)=3*binom(3n,n-2)=3*A003408(n-2).
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FORMULA
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T(n,k)=(1/(n+1))*binomial(n+1,k)*sum(3^(n-k-3j)*binomial(n+1-k,k+1+2j)*binomial(n-2k-2j,j), j=0..n/2-k). G.f.=G=G(t,z) satisfies G = 1+3zG+3tz^2*G^2+z^3*G^3.
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EXAMPLE
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T(2,1)=3 because we have (Q,L,M), (Q,L,R), and (Q,M,R), where Q denotes the root and L (M,R) denotes a left (middle, right) child of Q.
Triangle starts:
1;
3;
9,3;
28,27;
93,162,18;
333,825,270;
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MAPLE
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T:=(n, k)->(1/(n+1))*binomial(n+1, k)*sum(3^(n-k-3*q)*binomial(n+1-k, k+1+2*q)*binomial(n-2*k-2*q, q), q=0..n/2-k):for n from 0 to 12 do seq(T(n, k), k=0..floor(n/2)) od; # yields sequence in triangular form
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CROSSREFS
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Cf. A001764, A003408, A120429, A120981, A120983, A120985.
Sequence in context: A010259 A120429 A101431 this_sequence A125143 A130701 A050000
Adjacent sequences: A120979 A120980 A120981 this_sequence A120983 A120984 A120985
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KEYWORD
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nonn,tabf
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AUTHOR
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Emeric Deutsch (deutsch(AT)duke.poly.edu), Jul 21 2006
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