

A058879


Triangle read by rows: T(n,k) = number of connected graphs with one cycle of length m = nk+1 and n nodes (n>=3, 1<=k<=n2).


1



1, 1, 1, 1, 1, 3, 1, 1, 4, 7, 1, 1, 4, 9, 18, 1, 1, 5, 10, 28, 44, 1, 1, 5, 13, 32, 71, 117, 1, 1, 6, 14, 45, 89, 202, 299, 1, 1, 6, 17, 52, 130, 264, 542, 793, 1, 1, 7, 19, 69, 163, 413, 751, 1507, 2095, 1, 1, 7, 22, 79, 224, 544, 1221, 2179, 4114, 5607
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OFFSET

3,6


COMMENTS

Diagonals give A000226, A000368. Row sums give A001429.
Comments from Washington Bomfim, Jul 06 2008 (Start): T(n,3) = 2 + floor(m/2). When k = 3, n = m+2, so we have unicyclic graphs of order m+2 with a cycle of length m. Only two nodes of those graphs belong to the rooted trees attached to the cycle, so the orders of those trees can be only 1, 2, or 3.
We can have only one tree of order 3 in those graphs. So the two different rooted trees of order 3 correspond to two unicycles.
We can have two trees of order 2 in those graphs. Those trees can be rooted at two points r_1, r_2 of the cycle in h = floor(m/2) ways. They can be neighbors, i.e., we have an edge of the cycle (r_1, r_2). They can be 2, 3,..., h edges apart, but they cannot be h+1 edges away from each other. This is true because we obtain an isomorphic graph if r_1 and r_2 are h+1 (or more) edges apart, since there are also n  (h+1) edges between r_1 and r_2 and nh1 <= h. Note that there is only one rooted tree of order two.
The five unicyclic graphs of order 9 with a cycle of length 7 are depicted in the picture corresponding to the link.
T(n,4) = 4 + 2floor(m/2) + nearest integer to m^2/12.
We have unicyclic graphs of order m+3 with a cycle of length m. Only three nodes of those graphs belong to the rooted trees attached to the cycle, so the orders of those trees can be only 1, 2, 3, or 4. The set of unicycles can be divided in graphs with trees of orders
4,1,1,...,1
3,2,1,...,1
2,2,2,1,...,1.
Since there are 4 rooted trees of order 4, the orders 4,1,1,...,1 correspond to 4 unicycles.
The orders 3,2,1,...,1 correspond to 2floor(m/2) unicycles. For each one of the two rooted trees of order 3, we see above that there are floor(m/2) possibilities to choose a root for the tree of order 2.
The orders 2,2,2,1,...,1 correspond to i unicycles, i = nearest integer to m^2/12. This follows from the number of necklaces with n+3 beads 3 of which are red, that is equal to the nearest integer to (n+3)^2/12. See A001399. In our case we have necklaces with m beads. The 3 red beads are the roots of the trees of order 2. (End)


REFERENCES

F. Harary and E. M. Palmer, Graphical Enumeration, Academic Press, NY, 1973, p. 69, (3.4.1).
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 150, Table 9.


LINKS

Table of n, a(n) for n=3..68.
Washington Bomfim, The five unicyclic graphs of order 9 with a cycle of length 7: T(9,3)=5.


EXAMPLE

1; 1,1; 1,1,3; 1,1,4,7; 1,1,4,9,18; ...


MATHEMATICA

Needs["NumericalDifferentialEquationAnalysis`"];
t[n_, k_] := Block[{x}, Coefficient[CycleIndexPolynomial[DihedralGroup[n + 1  k], Table[ButcherTreeCount[n].x^(p Range[n]), {p, n + 1  k}]], x, n]];
Table[t[n, k], {n, 13}, {k, 1, n  2}] // Flatten
(* requires Mathematica 9+; Andrey Zabolotskiy, May 12 2017 *)


CROSSREFS

Cf. A217781.
Sequence in context: A133380 A105687 A209415 * A208344 A209172 A263950
Adjacent sequences: A058876 A058877 A058878 * A058880 A058881 A058882


KEYWORD

nonn,easy,tabl,nice


AUTHOR

N. J. A. Sloane, Jan 07 2001


EXTENSIONS

More terms from Washington Bomfim, May 12 2008
More terms from Washington Bomfim, Jul 06 2008
Rows n=11 to 13 added, name and offset corrected by Andrey Zabolotskiy, May 12 2017


STATUS

approved



