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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).

%I A258123 #11 Aug 20 2015 13:01:34
%S A258123 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,
%T A258123 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,
%U A258123 227,124,69,11,0,0,0,0,0,1,2105,556,256,185,52,4,0,0,0,0,0,1
%N 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).
%C A258123 Sum of entries in row n = A000055(n) = number of trees with n vertices.
%F A258123 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.
%e A258123 Triangle starts:
%e A258123 1;
%e A258123 0,1;
%e A258123 1;
%e A258123 1,1;
%e A258123 2,0,1;
%e A258123 4,1,0,1;
%e A258123 8,2,0,0,1;
%e A258123 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).
%Y A258123 Cf. A000055, A235111, A182907.
%K A258123 nonn,tabf
%O A258123 1,7
%A A258123 _Emeric Deutsch_, Aug 19 2015