cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A112739 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).

Original entry on oeis.org

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, 46, 11, 2, 1, 1, 8, 37, 106, 161, 94, 13, 2, 1, 1, 9, 50, 187, 426, 485, 190, 15, 2, 1, 1, 10, 65, 302, 937, 1706, 1457, 382, 17, 2, 1, 1, 11, 82, 457, 1814, 4687, 6826, 4373, 766, 19
Offset: 0

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Author

Paul Barry, Sep 16 2005

Keywords

Comments

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.
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

Examples

			As a square array, rows begin
1,1,1,1,1,1,... (A000012)
1,2,2,2,2,2,... (A040000)
1,3,5,7,9,11,... (A005408)
1,4,10,22,46,94,... (A033484)
1,5,17,53,161,485,... (A048473)
1,6,26,106,426,1706,... (A020989)
1,7,37,187,937,4687,... (A057651)
1,8,50,302,1814,10886,... (A061801)
As a number triangle, rows start
1;
1,1;
1,2,1;
1,3,2,1;
1,4,5,2,1;
1,5,10,7,2,1;
		

References

  • L. He, X. Liu and G. Strang, (2003) Trees with Cantor Eigenvalue Distribution. Studies in Applied Mathematics 110 (2), 123-138.
  • L. He, X. Liu and G. Strang, Laplacian eigenvalues of growing trees, Proc. Conf. on Math. Theory of Networks and Systems, Perpignan (2000).

Crossrefs

Formula

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).