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A213657 Irregular triangle read by rows: T(n,k) is the number of dominating subsets with k vertices of the graph G(n) consisting of an edge ab, n vertices c_1, c_2, ..., c_n, and 2n edges ac_i, bc_i (i=1..n). (n triangles with a common edge).

%I A213657 #15 Dec 09 2017 12:20:15
%S A213657 3,3,1,2,6,4,1,2,7,10,5,1,2,9,16,15,6,1,2,11,25,30,21,7,1,2,13,36,55,
%T A213657 50,28,8,1,2,15,49,91,105,77,36,9,1,2,17,64,140,196,182,112,45,10,1,2,
%U A213657 19,81,204,336,378,294,156,55,11,1
%N A213657 Irregular triangle read by rows: T(n,k) is the number of dominating subsets with k vertices of the graph G(n) consisting of an edge ab, n vertices c_1, c_2, ..., c_n, and 2n edges ac_i, bc_i (i=1..n). (n triangles with a common edge).
%C A213657 Row n contain n + 2 entries.
%C A213657 Sum of entries in row n = 1 + 3*2^n = A181565(n).
%H A213657 S. Alikhani and E. Deutsch, <a href="http://arxiv.org/abs/1305.3734">Graphs with domination roots in the right half-plane</a>, arXiv preprint arXiv:1305.3734, 2013
%H A213657 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.
%H A213657 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.
%F A213657 Generating polynomial of row n is x^n + x*(2+x)*(1+x)^n; this is the domination polynomial of the graph G(n).
%F A213657 T(n,n) = (n+1)*(n+3)/2; T(n,k) = 2*binomial(n, k-1) + binomial(n, k-2) if k != n.
%e A213657 Row 1 is 3,3,1 because the graph G(1) is the triangle abc; there are 3 dominating subsets of size 1 ({a}, {b}, {c}), 3 dominating subsets of size 2 ({a,b}, {a,c}, {b,c}), and 1 dominating subset of size 3 ({a,b,c}).
%e A213657 T(n,1)=2 for n >= 2 because {a} and {b} are the only dominating subsets of size k=1.
%e A213657 Triangle starts:
%e A213657   3, 3,  1;
%e A213657   2, 6,  4,  1;
%e A213657   2, 7, 10,  5, 1;
%e A213657   2, 9, 16, 15, 6, 1;
%p A213657 T := proc (n, k) if k = n then (1/2)*(n+1)*(n+2) else 2*binomial(n, k-1)+binomial(n, k-2) end if end proc: for n to 12 do seq(T(n, k), k = 1 .. n+2) end do; # yields sequence in triangular form
%t A213657 T[n_, k_] := If[k==n, (n+1)*(n+2)/2, 2*Binomial[n, k-1]+Binomial[n, k-2]];
%t A213657 Table[T[n, k], {n, 1, 10}, {k, 1, n+2}] // Flatten (* _Jean-François Alcover_, Dec 09 2017 *)
%Y A213657 Cf. A181565
%K A213657 nonn,tabf
%O A213657 1,1
%A A213657 _Emeric Deutsch_, Jun 29 2012