A213669 - OEIS

%I #19 Dec 06 2017 03:43:21

%S 4,4,1,1,6,11,6,1,1,6,17,26,22,8,1,1,8,28,58,78,68,37,10,1,1,10,45,

%T 120,212,262,230,140,56,12,1,1,12,66,220,495,794,936,822,535,250,79,

%U 14,1,1,14,91,364,1001,2002,3005,3446,3045,2072,1071,406,106,16,1

%N Irregular triangle read by rows: T(n,k) is the number of dominating subsets with k vertices of the double star graph G(n) obtained by joining with an edge the centers of two star trees each having n+1 vertices (n>=1, k>=2).

%C Number of entries in row n is 2n+1.

%C Sum of entries in row n is (2^n +1)^2 = A028400(n).

%C The Matula-Goebel number of the rooted tree obtained from G(n), by selecting the center of one of the trees as the root, is 2^n*(2^n-th prime); (knowing this, see A212630 for another approach to find this sequence).

%C Closely related to the connected domination polynomial of the n-book graph (divided by x^2), which is 1 less in the 3rd-to-last term of each row. - _Eric W. Weisstein_, May 12 2017

%H S. Alikhani and Y. H. Peng, <a href="http://arxiv.org/abs/0905.2251">Introduction to domination polynomial of a graph</a>, arXiv:0905.2251 [math.CO], 2009.

%H S. Akbari, S. Alikhani, and Y. H. Peng, <a href="https://doi.org/10.1016/j.ejc.2010.03.007">Characterization of graphs using domination polynomials</a>, European J. Comb., 31, 2010, 1714-1724.

%H T. Kotek, J. Preen, F. Simon, P. Tittmann, and M. Trinks, <a href="http://arxiv.org/abs/1206.5926">Recurrence relations and splitting formulas for the domination polynomial</a>, arXiv:1206.5926 [math.CO], 2012.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/BookGraph.html">Book Graph</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/ConnectedDominatingSet.html">Connected Dominating Set</a>

%F The generating polynomial of row n is (x^n + x(1+x)^n)^2; this is the domination polynomial of the graph G(n).

%F The domination polynomial of the double star graph obtained by joining with an edge the center of a star tree having m+1 vertices and the center of a star tree having n+1 vertices is (x^m+x(1+x)^m)*(x^n + x(1+x)^n) (m,n >=1).

%e Row 1 is 4,4,1 because the graph G(1) is the path abcd; there are 4 dominating subsets of size 2 (ac,ad,bc,bd), 4 dominating subsets of size 3 (abc,abd,acd,bcd) and 1 dominating subset of size 4 (abcd).

%e Triangle starts:

%e   4,  4,  1;

%e   1,  6, 11,  6,  1;

%e   1,  6, 17, 26, 22,  8,  1;

%e   1,  8, 28, 58, 78, 68, 37, 10,  1;

%p P := proc (n) options operator, arrow: (x^n+x*(1+x)^n)^2 end proc: for n to 9 do seq(coeff(P(n), x, k), k = 2 .. 2*n+2) end do; # yields sequence in triangular form

%t T[n_, k_] := SeriesCoefficient[(x^n + x (1 + x)^n)^2, {x, 0, k}];

%t Table[T[n, k], {n, 1, 9}, {k, 2, 2 n + 2}] // Flatten (* _Jean-François Alcover_, Dec 06 2017 *)

%Y Cf. A028400, A212630.

%K nonn,tabf

%O 1,1

%A _Emeric Deutsch_, Jul 10 2012