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%I #16 Apr 20 2023 14:56:39
%S 1,1,0,1,1,0,1,2,1,0,1,3,3,1,0,1,4,6,4,1,0,1,5,10,10,6,1,0,1,6,15,20,
%T 21,12,2,0,1,7,21,35,55,63,31,2,0,1,8,28,56,120,220,227,78,4,0,1,9,36,
%U 84,231,600,1040,891,234,6,0,1,10,45,120,406,1386,3530,5480,3876,722,11,0
%N Array read by antidiagonals: T(n,k) is the number of leaf colored trees with n leaves of k colors and all non-leaf nodes having degree 3.
%C Not all k colors need to be used. The total number of nodes will be 2n-1.
%C See table 4.1 in the Johnson reference.
%H Andrew Howroyd, <a href="/A339649/b339649.txt">Table of n, a(n) for n = 0..1325</a>
%H Virginia Perkins Johnson, <a href="https://people.math.sc.edu/czabarka/Theses/JohnsonThesis.pdf">Enumeration Results on Leaf Labeled Trees</a>, Ph. D. Dissertation, Univ. South Carolina, 2012.
%F G.f. of column k: 1 + R(x) + (R(x^3) - R(x)^3)/3 where R(x) is the g.f. of column k of A319539.
%e Array begins:
%e ======================================================
%e n\k| 0 1 2 3 4 5 6 7
%e ---+--------------------------------------------------
%e 0 | 1 1 1 1 1 1 1 1 ...
%e 1 | 0 1 2 3 4 5 6 7 ...
%e 2 | 0 1 3 6 10 15 21 28 ...
%e 3 | 0 1 4 10 20 35 56 84 ...
%e 4 | 0 1 6 21 55 120 231 406 ...
%e 5 | 0 1 12 63 220 600 1386 2842 ...
%e 6 | 0 2 31 227 1040 3530 9772 23366 ...
%e 7 | 0 2 78 891 5480 23250 77112 214718 ...
%e 8 | 0 4 234 3876 31420 165510 655599 2122099 ...
%e 9 | 0 6 722 17790 190360 1243825 5878446 22102577 ...
%e ...
%o (PARI) \\ here U(n,k) gives column k as a vector.
%o R(n, k)={my(v=vector(n)); v[1]=k; for(n=2, n, v[n]=sum(j=1, (n-1)\2, v[j]*v[n-j]) + if(n%2, 0, binomial(v[n/2]+1, 2))); v}
%o U(n, k)={my(g=x*Ser(R(n,k))); Vec(1 + g + (subst(g + O(x*x^(n\3)), x, x^3) - g^3)/3)}
%o {my(T=Mat(vector(8, k, U(8, k-1)~))); for(n=1, #T~, print(T[n,]))}
%Y Columns k=1..4 are A129860, A220826, A220827, A220828.
%Y Cf. A319539 (rooted), A339650, A339779.
%K nonn,tabl
%O 0,8
%A _Andrew Howroyd_, Dec 14 2020
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