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A339649
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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.
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6
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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, 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, 84, 231, 600, 1040, 891, 234, 6, 0, 1, 10, 45, 120, 406, 1386, 3530, 5480, 3876, 722, 11, 0
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OFFSET
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0,8
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COMMENTS
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Not all k colors need to be used. The total number of nodes will be 2n-1.
See table 4.1 in the Johnson reference.
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LINKS
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FORMULA
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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.
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EXAMPLE
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Array begins:
======================================================
n\k| 0 1 2 3 4 5 6 7
---+--------------------------------------------------
0 | 1 1 1 1 1 1 1 1 ...
1 | 0 1 2 3 4 5 6 7 ...
2 | 0 1 3 6 10 15 21 28 ...
3 | 0 1 4 10 20 35 56 84 ...
4 | 0 1 6 21 55 120 231 406 ...
5 | 0 1 12 63 220 600 1386 2842 ...
6 | 0 2 31 227 1040 3530 9772 23366 ...
7 | 0 2 78 891 5480 23250 77112 214718 ...
8 | 0 4 234 3876 31420 165510 655599 2122099 ...
9 | 0 6 722 17790 190360 1243825 5878446 22102577 ...
...
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PROG
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(PARI) \\ here U(n, k) gives column k as a vector.
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}
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)}
{my(T=Mat(vector(8, k, U(8, k-1)~))); for(n=1, #T~, print(T[n, ]))}
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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