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))^2 = 71 + 12*sqrt(35). - Ant King, Dec 31 2011
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..200
Eric Weisstein's World of Mathematics, Nonagonal Heptagonal Number.
Index entries for linear recurrences with constant coefficients, signature (143,-143,1).
FORMULA
From Bruno Berselli, Dec 20 2011: (Start)
G.f.: x*(1 - 39*x - 4*x^2)/((1-x)*(1 - 142*x + x^2)).
a(n) = (42 + (-21+5r)*(6+r)^(2n-1) - (21+5r)*(6-r)^(2n-1))/140, where r=sqrt(35). (End)
From Ant King, Dec 31 2011: (Start)
a(n) = 142*a(n-1) - a(n-2) - 42.
a(n) = ceiling(1/140*(49+9*sqrt(35))*(6+sqrt(35))^(2*n-2)).
(End)
MATHEMATICA
LinearRecurrence[{143, -143, 1}, {1, 104, 14725}, 30] (* Vincenzo Librandi, Dec 21 2011 *)
PROG
(Maxima) makelist(expand((42+(-21+5*sqrt(35))*(6+sqrt(35))^(2*n-1)-(21+5*sqrt(35))*(6-sqrt(35))^(2*n-1))/140), n, 1, 12); /* Bruno Berselli, Dec 20 2011 */
(Magma) I:=[1, 104, 14725]; [n le 3 select I[n] else 143*Self(n-1)-143*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Dec 21 2011
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved