OFFSET
1,1
COMMENTS
The independence polynomial is given by I(n,x) = 1+(3*n-1)*x*(x+1)^(n-1).
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..200
Eric Weisstein's World of Mathematics, Andrásfai Graph
Eric Weisstein's World of Mathematics, Independent Vertex Set
Index entries for linear recurrences with constant coefficients, signature (5,-8,4).
FORMULA
a(n) = 1 + (3*n-1)*2^(n-1).
From Colin Barker, Jun 05 2017: (Start)
G.f.: x*(3 - 4*x + 2*x^2) / ((1 - x)*(1 - 2*x)^2).
a(n) = 5*a(n-1) - 8*a(n-2) + 4*a(n-3) for n>3.
(End)
MATHEMATICA
Table[(3 n - 1) 2^(n - 1) + 1, {n, 20}]
LinearRecurrence[{5, -8, 4}, {3, 11, 33}, 20]
CoefficientList[Series[(-3 + 4 x - 2 x^2)/((-1 + x) (-1 + 2 x)^2), {x, 0, 20}], x]
PROG
(PARI) Vec(x*(3 - 4*x + 2*x^2) / ((1 - x)*(1 - 2*x)^2) + O(x^30)) \\ Colin Barker, Jun 05 2017
(PARI) a(n) = 1 + (3*n-1)*2^(n-1); \\ Michel Marcus, Jun 05 2017
CROSSREFS
KEYWORD
nonn,easy,changed
AUTHOR
Andrew Howroyd, Jun 04 2017
STATUS
approved