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%I #10 Nov 16 2019 10:32:26
%S 1,3,9,3,28,27,93,162,18,333,825,270,1272,3915,2430,135,5085,18144,
%T 17199,2835,20925,84000,106596,34020,1134,87735,391554,612360,308448,
%U 30618,372879,1838295,3369600,2364390,459270,10206,1602450,8674380
%N 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).
%C 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.
%F T(n,k) = (1/(n+1))*binomial(n+1,k)*Sum_{j=0..floor(n/2)-k} 3^(n-k-3j)*binomial(n+1-k, k+1+2j)*binomial(n-2k-2j, j).
%F G.f.: G = G(t,z) satisfies G = 1 + 3zG + 3tz^2*G^2 + z^3*G^3.
%F Row n has 1+floor(n/2) terms.
%F Row sums yield A001764.
%F T(n,0) = A120985(n).
%F Sum_{k>=1} k*T(n,k) = 3*binomial(3n,n-2) = 3*A003408(n-2).
%e 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.
%e Triangle starts:
%e 1;
%e 3;
%e 9, 3;
%e 28, 27;
%e 93, 162, 18;
%e 333, 825, 270;
%p 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
%Y Cf. A001764, A003408, A120429, A120981, A120983, A120985.
%K nonn,tabf
%O 0,2
%A _Emeric Deutsch_, Jul 21 2006
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