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.

Showing 1-1 of 1 results.

A365327 Triangle read by rows: T(n,k) is the number of spanning subgraphs of the n-cycle graph with domination number k.

Original entry on oeis.org

2, 3, 1, 4, 3, 1, 0, 11, 4, 1, 0, 11, 15, 5, 1, 0, 10, 26, 21, 6, 1, 0, 0, 43, 49, 28, 7, 1, 0, 0, 33, 98, 80, 36, 8, 1, 0, 0, 22, 126, 189, 120, 45, 9, 1, 0, 0, 0, 141, 322, 325, 170, 55, 10, 1, 0, 0, 0, 89, 462, 671, 517, 231, 66, 11, 1, 0, 0, 0, 46, 480, 1162, 1236, 777, 304, 78, 12, 1, 0, 0, 0, 0, 417, 1586, 2483, 2093, 1118, 390, 91, 13, 1
Offset: 1

Views

Author

Roman Hros, Sep 01 2023

Keywords

Comments

For n >= 3 the n-cycle graph is a simple graph. In the case of n=1 the graph is a loop and for n=2 a double edge.
The number of spanning subgraphs of the n-cycle graph is given by 2^n which is also the sum of the n-th row Sum_{k=1..n} T(n,k).
The average domination number is given by (Sum_{k=1..n} k*T(n,k))/2^n.
The relative average domination number is given by ((Sum_{k=1..n} k*T(n,k))/2^n)/n.

Examples

			Example of spanning subgraphs of cycle with 2 vertices:
Domination number: 2      1      1      1
                          /\            /\
                  .  .   .  .   .  .   .  .
                                 \/     \/
The triangle T(n,k) begins:
n\k 1   2   3    4    5     6     7    8    9  10  11  12 ...
1:  2
2:  3   1
3:  4   3   1
4:  0  11   4    1
5:  0  11  15    5    1
6:  0  10  26   21    6     1
7:  0   0  43   49   28     7     1
8:  0   0  33   98   80    36     8    1
9:  0   0  22  126  189   120    45    9    1
10: 0   0   0  141  322   325   170   55   10   1
11: 0   0   0   89  462   671   517  231   66  11   1
12: 0   0   0   46  480  1162  1236  777  304  78  12   1
		

Crossrefs

Row sums are A000079.
Column sums are A002063(k-1).
Cf. A373436.

Formula

T(n,n) = 1 for n > 1.
T(n,n-1) = T(n-1, n-2) + 1 for n > 3.
T(n,n-2) = T(n-1, n-3) + T(n, n-1) for n > 5.
T(n,n-3) = T(n-1, n-4) + T(n, n-2) - 5 for n > 6.
T(n,n-4) = T(n-1, n-5) + T(n-1, n-4) + 11 + Sum_{i=1..n-9} (i+4) for n > 8.
G.f.:
For n > 3; G(n) = x*(G(n-1) + G(n-2) + 2*G(n-3)) + g(n); where
2*(1-x)*x^(n/3) for n mod 3 = 0.
g(n) = { 0 for n mod 3 = 1.
(1-x)*x^((n+1)/3) for n mod 3 = 2.
For n mod 3 = 0:
T(n,k) = 2*T(n-3,k-1) + T(n-2,k-1) + T(n-1,k-1) + 2 for k = n/3.
T(n,k) = 2*T(n-3,k-1) + T(n-2,k-1) + T(n-1,k-1) - 2 for k = n/3 + 1.
T(n,k) = 2*T(n-3,k-1) + T(n-2,k-1) + T(n-1,k-1) for k >= n/3 + 2.
For n mod 3 = 1:
T(n,k) = 2*T(n-3,k-1) + T(n-2,k-1) + T(n-1,k-1) for k >= (n+2)/3.
For n mod 3 = 2:
T(n,k) = 2*T(n-3,k-1) + T(n-2,k-1) + T(n-1,k-1) + 1 for k = (n+1)/3.
T(n,k) = 2*T(n-3,k-1) + T(n-2,k-1) + T(n-1,k-1) - 1 for k = (n+1)/3 + 1.
T(n,k) = 2*T(n-3,k-1) + T(n-2,k-1) + T(n-1,k-1) for k >= (n+1)/3 + 2.
Showing 1-1 of 1 results.