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%I #18 Apr 20 2023 14:56:34
%S 1,0,1,0,1,1,0,1,2,1,0,1,4,6,3,0,1,10,30,36,15,0,2,27,140,310,300,105,
%T 0,2,74,663,2376,3990,3150,945,0,4,226,3186,17304,44850,59805,39690,
%U 10395,0,6,710,15642,123508,462735,925890,1018710,582120,135135
%N Triangle read by rows: T(n,k) is the number of trees with n leaves of exactly k colors and all non-leaf nodes having degree 3.
%C See table 4.2 in the Johnson reference.
%H Andrew Howroyd, <a href="/A339650/b339650.txt">Table of n, a(n) for n = 0..1325</a> (rows 0..50)
%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 T(n,k) = Sum_{i=0..k} (-1)^(k-i)*binomial(k,i)*A339649(n,i).
%e Triangle begins:
%e 1;
%e 0, 1;
%e 0, 1, 1;
%e 0, 1, 2, 1;
%e 0, 1, 4, 6, 3;
%e 0, 1, 10, 30, 36, 15;
%e 0, 2, 27, 140, 310, 300, 105;
%e 0, 2, 74, 663, 2376, 3990, 3150, 945;
%e 0, 4, 226, 3186, 17304, 44850, 59805, 39690, 10395;
%e ...
%o (PARI) \\ here U(n,k) is column k of A339649 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 M(n, m=n)={my(v=vector(m+1, k, U(n, k-1)~)); Mat(vector(m+1, k, k--; sum(i=0, k, (-1)^(k-i)*binomial(k, i)*v[1+i])))}
%o {my(T=M(10)); for(n=1, #T~, print(T[n, ][1..n]))}
%Y Columns k=1..4 are A129860, A220829, A220830, A220831.
%Y Main diagonal is A001147(n-2) for n >= 2.
%Y Row sums are A339651.
%Y Cf. A319541 (rooted), A339649, A339780.
%K nonn,tabl
%O 0,9
%A _Andrew Howroyd_, Dec 14 2020
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