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A213659
a(n) = 3^n + 2^(2*n + 1).
2
11, 41, 155, 593, 2291, 8921, 34955, 137633, 543971, 2156201, 8565755, 34085873, 135812051, 541653881, 2161832555, 8632981313, 34488878531, 137826373961, 550918075355, 2202510039953, 8806553375411, 35215753148441, 140831631534155
OFFSET
1,1
COMMENTS
Number of dominating subsets of the graph G(n) consisting of an edge ab and n triangles, each having one vertex identified with the vertex b.
Resultant of polynomial 2*x^n+1 with the polynomial 3*x^(n+1)+2. - Philippe Deléham, Jan 19 2024
LINKS
S. Alikhani and Y. H. Peng, Introduction to domination polynomial of a graph, arXiv:0905.2251 [math.CO], 2009.
T. Kotek, J. Preen, F. Simon, P. Tittmann, and M. Trinks, Recurrence relations and splitting formulas for the domination polynomial, arXiv:1206.5926 [math.CO], 2012.
FORMULA
a(n) = Sum_{k=1..2*n+2} A213658(n,k).
a(n) = 3^n + 2^(2*n+1).
G.f.: -x*(-11+36*x) / ( (4*x-1)*(3*x-1) ). - R. J. Mathar, Jul 03 2012
a(n) = 7*a(n-1) - 12*a(n-2). - Philippe Deléham, Jan 19 2024
EXAMPLE
a(1)=11 because the graph G(1) is abcd with edges ab, bc, bd, and cd; there is 1 dominating subset of size 1 ({b}), all binomial(4,2)=6 subsets of size 2 of {a,b,c,d} with the exception of {c,d} are dominating; all binomial(4,3)=4 subsets of size 3 of {a,b,c,d} are dominating; obviously, {a,b,c,d} is dominating; 1+5+4+1 = 11.
a(1) = Det [2, 1, 0; 0, 2, 1; 3, 0, 2] = 11; a(2) = Det [2, 0, 1, 0, 0; 0, 2, 0, 1, 0; 0, 0, 2, 0, 1; 3, 0, 0, 2, 0; 0, 3, 0, 0, 2] = 41; a(3) = Det [2, 0, 0, 1, 0, 0, 0; 0, 2, 0, 0, 1, 0, 0; 0, 0, 2, 0, 0, 1, 0; 0, 0, 0, 2, 0, 0, 1; 3, 0, 0, 0, 2, 0, 0; 0, 3, 0, 0, 0, 2, 0; 0, 0, 3, 0, 0, 0, 2] = 155. - Philippe Deléham, Jan 19 2024
MAPLE
seq(3^n+2^(2*n+1), n=1..30);
MATHEMATICA
Table[3^n + 2^(2 n + 1), {n, 40}] (* Vincenzo Librandi, Jun 30 2017 *)
LinearRecurrence[{7, -12}, {11, 41}, 30] (* Harvey P. Dale, Sep 10 2017 *)
PROG
(Magma) [3^n+2^(2*n+1): n in [1..30]]; // Vincenzo Librandi, Jun 30 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Emeric Deutsch, Jun 29 2012
STATUS
approved