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A258123
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Irregular triangle read by rows: T(n,k) is the number of trees with n vertices that have k vertices of maximum degree (n, k>=1).
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0
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1, 0, 1, 1, 1, 1, 2, 0, 1, 4, 1, 0, 1, 8, 2, 0, 0, 1, 15, 6, 1, 0, 0, 1, 32, 11, 3, 0, 0, 0, 1, 68, 25, 10, 2, 0, 0, 0, 1, 156, 47, 25, 6, 0, 0, 0, 0, 1, 361, 105, 58, 24, 2, 0, 0, 0, 0, 1, 869, 227, 124, 69, 11, 0, 0, 0, 0, 0, 1, 2105, 556, 256, 185, 52, 4, 0, 0, 0, 0, 0, 1
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OFFSET
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1,7
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COMMENTS
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Sum of entries in row n = A000055(n) = number of trees with n vertices.
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LINKS
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FORMULA
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No formula for T(n,k) or generating function is known to the author. Row 5, for example, has been obtained in the following manner. The 3 trees with 5 vertices have M-indices 9,12,16 (the M-index of a tree T is the smallest of the Matula numbers of the rooted trees isomorphic (as a tree) to T; see A235111). In A182907 one finds the degree sequence (and the degree sequence polynomial) of a rooted tree with known Matula number. To the Matula numbers 9, 12, 16, there correspond the degree sequence polynomials 3x^2 + 2x, x^3 + x^2 + 3x, x^4 + 4x, respectively. From here, number of vertices of maximum degree are 3, 1, and 1, respectively. In other words, 2 trees have 1 vertex of maximum degree, 0 trees have 2 vertices of maximum degree, and 1 tree has 3 vertices of maximum degree; this leads us to row 5: 2, 0, 1.
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EXAMPLE
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Triangle starts:
1;
0,1;
1;
1,1;
2,0,1;
4,1,0,1;
8,2,0,0,1;
Row 4 is 1,1; indeed, the star S(4) has degree sequence (0,0,0,1) and the path P(4) has degree sequence (1,1,2,2).
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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