OFFSET
1,2
COMMENTS
As n increases, the ratio of consecutive terms settles into an approximate 2-cycle with the ratio a(n)/a(n-1) bounded above and below by 2024 + 765*sqrt(7) and 8 + 3*sqrt(7) respectively. - Ant King, Dec 29 2011
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..200
Eric Weisstein's World of Mathematics, Nonagonal Hexagonal Number.
Index entries for linear recurrences with constant coefficients, signature (1,64514,-64514,-1,1).
FORMULA
G.f.: x*(-1 - 12*x + 12902*x^2 + 3036*x^3 + 203*x^4) / ( (x-1)*(x^2 - 254*x + 1)*(x^2 + 254*x + 1) ). - R. J. Mathar, Dec 21 2011
From Ant King, Dec 29 2011: (Start)
a(n) = 64514*a(n-2) - a(n-4) - 16128.
a(n) = (1/56)*sqrt(7)*(3*((3 - sqrt(7)*(-1)^n)*(8 + 3*sqrt(7))^(2*n-2) - (3 + sqrt(7)*(-1)^n)*(8 - 3*sqrt(7))^(2*n-2)) + 2*sqrt(7)).
a(n) = ceiling((3/56)*sqrt(7)*(3 - sqrt(7)*(-1)^n)*(8 + 3*sqrt(7))^(2*n-2)).
(End)
MATHEMATICA
LinearRecurrence[{1, 64514, -64514, -1, 1}, {1, 13, 51625, 822757, 3330519121}, 210] (* Vincenzo Librandi, Dec 27 2011 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved