OFFSET
1,2
COMMENTS
As n increases, this sequence is approximately geometric with common ratio r = lim(n->Infinity,a(n)/a(n-1)) = (6+sqrt(35))^4 = 10081+1704*sqrt(35). - Ant King, Dec 31 2011
LINKS
Colin Barker, Table of n, a(n) for n = 1..233
Eric Weisstein's World of Mathematics, Nonagonal Heptagonal Number.
Index entries for linear recurrences with constant coefficients, signature (20163,-20163,1).
FORMULA
From Ant King, Dec 31 2011: (Start)
a(n) = 20163*a(n-1)-20163*a(n-2)+a(n-3).
a(n) = 20162*a(n-1)-a(n-2)+6768.
a(n) = 1/560*((39+4*sqrt(35))*(6+sqrt(35))^(4*n-3)+(39-4*sqrt(35))*(6-sqrt(35))^(4*n-3)-188).
a(n) = floor(1/560*(39+4*sqrt(35))*(6+sqrt(35))^(4*n-3)).
G.f.: x(1+6721*x+46*x^2) / ((1-x)(1-20162*x+x^2)).
(End)
MATHEMATICA
LinearRecurrence[{20163, -20163, 1}, {1, 26884, 542041975}, 9]; (* Ant King, Dec 31 2011 *)
PROG
(PARI) Vec(x*(1+6721*x+46*x^2)/((1-x)*(1-20162*x+x^2)) + O(x^20)) \\ Colin Barker, Jun 22 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved