OFFSET
1,1
COMMENTS
The corresponding values of y are given by a(n+2).
Positive values of x (or y) satisfying x^2 - 18xy + y^2 + 1216 = 0.
LINKS
Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-1).
FORMULA
a(n) = 3*a(n-2)-a(n-4).
G.f.: -x*(x-1)*(4*x^2+9*x+4) / ((x^2-x-1)*(x^2+x-1)).
a(n) = (1/2) * (F(n+4) + (-1)^n*F(n-5)), n>4, with F the Fibonacci numbers (A000045). - Ralf Stephan, Feb 05 2014
EXAMPLE
11 is in the sequence because (x, y) = (11, 28) is a solution to x^2 - 3xy + y^2 + 19 = 0.
MATHEMATICA
LinearRecurrence[{0, 3, 0, -1}, {4, 5, 7, 11}, 40] (* Harvey P. Dale, Dec 15 2014 *)
PROG
(PARI) Vec(-x*(x-1)*(4*x^2+9*x+4)/((x^2-x-1)*(x^2+x-1)) + O(x^100))
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Colin Barker, Feb 04 2014
STATUS
approved