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Array counting nodes in rooted trees of height n in which the root and internal nodes have valency k (and the leaf nodes have valency one).
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%I #10 Feb 23 2014 12:39:14

%S 1,1,1,1,2,1,1,3,2,1,1,4,5,2,1,1,5,10,7,2,1,1,6,17,22,9,2,1,1,7,26,53,

%T 46,11,2,1,1,8,37,106,161,94,13,2,1,1,9,50,187,426,485,190,15,2,1,1,

%U 10,65,302,937,1706,1457,382,17,2,1,1,11,82,457,1814,4687,6826,4373,766,19

%N Array counting nodes in rooted trees of height n in which the root and internal nodes have valency k (and the leaf nodes have valency one).

%C Rows of the square array have g.f. (1+x)/((1-x)(1-kx)). They are the partial sums of the coordination sequences for the infinite tree of valency k. Row sums are A112740.

%C Rows of the square array are successively: A000012, A040000, A005408, A033484, A048473, A020989, A057651, A061801, A238275, A238276, A138894, A090843, A199023. - _Philippe Deléham_, Feb 22 2014

%D L. He, X. Liu and G. Strang, (2003) Trees with Cantor Eigenvalue Distribution. Studies in Applied Mathematics 110 (2), 123-138.

%D L. He, X. Liu and G. Strang, Laplacian eigenvalues of growing trees, Proc. Conf. on Math. Theory of Networks and Systems, Perpignan (2000).

%F As a square array read by antidiagonals, T(n, k)=sum{j=0..k, (2-0^j)*(n-1)^(k-j)}; T(n, k)=(n(n-1)^k-2)/(n-2), n<>2, T(2, n)=2n+1; T(n, k)=sum{j=0..k, (n(n-1)^j-0^j)/(n-1)}, j<>1. As a triangle read by rows, T(n, k)=if(k<=n, sum{j=0..k, (2-0^j)*(n-k-1)^(k-j)}, 0).

%e As a square array, rows begin

%e 1,1,1,1,1,1,... (A000012)

%e 1,2,2,2,2,2,... (A040000)

%e 1,3,5,7,9,11,... (A005408)

%e 1,4,10,22,46,94,... (A033484)

%e 1,5,17,53,161,485,... (A048473)

%e 1,6,26,106,426,1706,... (A020989)

%e 1,7,37,187,937,4687,... (A057651)

%e 1,8,50,302,1814,10886,... (A061801)

%e As a number triangle, rows start

%e 1;

%e 1,1;

%e 1,2,1;

%e 1,3,2,1;

%e 1,4,5,2,1;

%e 1,5,10,7,2,1;

%Y Cf. A112468, A000012, A040000, A005408, A033484, A048473, A020989, A057651, A061801, A238275, A238276, A138894, A090843, A199023.

%K easy,nonn,tabl

%O 0,5

%A _Paul Barry_, Sep 16 2005