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+9*x-25499*x^2+2295*x^3+154*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)-23040.
a(n) = 1/28*(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))+10).
a(n) = ceiling(3/28*(3-sqrt(7)*(-1)^n)*(8+3*sqrt(7))^(2*n-2)). (End)
MATHEMATICA
LinearRecurrence[{1, 64514, -64514, -1, 1}, {1, 10, 39025, 621946, 2517635809}, 210] (* Vincenzo Librandi, Dec 28 2011 *)
CROSSREFS
KEYWORD
nonn,easy,changed
AUTHOR
STATUS
approved