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Revision History for A112739 (Bold, blue-underlined text is an addition; faded, red-underlined text is a deletion.)

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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).
(history; published version)
#10 by Joerg Arndt at Sun Feb 23 12:39:14 EST 2014
STATUS

proposed

approved

#9 by Philippe Deléham at Sun Feb 23 12:29:47 EST 2014
STATUS

editing

proposed

#8 by Philippe Deléham at Sun Feb 23 12:29:36 EST 2014
KEYWORD

easy,nonn,changed,tabl

STATUS

approved

editing

#7 by N. J. A. Sloane at Fri Feb 21 22:06:48 EST 2014
STATUS

editing

approved

#6 by N. J. A. Sloane at Fri Feb 21 22:06:45 EST 2014
STATUS

proposed

editing

#5 by Philippe Deléham at Fri Feb 21 20:21:04 EST 2014
STATUS

editing

proposed

#4 by Philippe Deléham at Fri Feb 21 20:20:54 EST 2014
COMMENTS

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

EXAMPLE

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)

CROSSREFS

Cf. A112468

STATUS

approved

editing

#3 by Russ Cox at Fri Mar 30 18:59:12 EDT 2012
AUTHOR

_Paul Barry (pbarry(AT)wit.ie), _, Sep 16 2005

Discussion
Fri Mar 30
18:59
OEIS Server: https://oeis.org/edit/global/287
#2 by N. J. A. Sloane at Fri Feb 24 03:00:00 EST 2006
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).

KEYWORD

easy,nonn,new

#1 by N. J. A. Sloane at Wed Sep 21 03:00:00 EDT 2005
NAME

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).

DATA

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

0,5

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.

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).

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).

EXAMPLE

As a square array, rows begin

1,1,1,1,1,1,...

1,2,2,2,2,2,...

1,3,5,7,9,11,...

1,4,10,22,46,94,...

1,5,17,53,161,485,...

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;

KEYWORD

easy,nonn,new

AUTHOR

Paul Barry (pbarry(AT)wit.ie), Sep 16 2005

STATUS

approved