

A258123


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).


0



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,7


COMMENTS

Sum of entries in row n = A000055(n) = number of trees with n vertices.


LINKS

Table of n, a(n) for n=1..81.


FORMULA

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 Mindices 9,12,16 (the Mindex 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.


EXAMPLE

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).


CROSSREFS

Cf. A000055, A235111, A182907.
Sequence in context: A108643 A133838 A182138 * A121583 A228924 A246862
Adjacent sequences: A258120 A258121 A258122 * A258124 A258125 A258126


KEYWORD

nonn,tabf


AUTHOR

Emeric Deutsch, Aug 19 2015


STATUS

approved



