OFFSET
1,1
COMMENTS
The limit of a(n)/a(n-1) approaches (11+sqrt(117))/2 as n -> infinity.
The continued fraction [a(n); a(n), a(n), ...] = ((3+sqrt(13))/2)^(2*n-1).
Equivalently, numbers n such that (n^2+4)/13 is a square.
Sequence of all positive integers k such that continued fraction [k,k,k,k,k,k,...] belongs to Q(sqrt(13)). - Greg Dresden, Jul 22 2019
As 13*n^2 + 52 = 13 * (n^2 + 4), n == 3 (mod 13) or n == 10 (mod 13) alternately. - Bernard Schott, Jul 23 2019
LINKS
FORMULA
G.f.: 3*x*(1+x)/(1-11*x+x^2).
a(n) = 11*a(n-1) - a(n-2); a(0) = 3, a(1) = 36.
a(n) = floor(((3+sqrt(13))/2)^(2*n+1)+((3+sqrt(13))/2)^(1-2*n)).
a(n) = 3*A097783(n-1). - R. J. Mathar, Jun 07 2016
E.g.f.: (1/4)*(sqrt(13) - 3)*exp((1/2)*(11-3*sqrt(13)*x)*(11 + 3*sqrt(13) - 2*exp(3*sqrt(13)*x)). - Stefano Spezia, Jul 24 2019
MATHEMATICA
Table[Floor[((3 + Sqrt[13])/2)^(2*n + 1) + ((3 + Sqrt[13])/2)^(1 - 2 n)], {n, 21}] (* Michael De Vlieger, Jun 20 2015 *)
LinearRecurrence[{11, -1}, {3, 36}, 25] (* Vincenzo Librandi, Jul 23 2019 *)
PROG
(PARI) for(n=1, 20, q=((3+sqrt(13))/2)^(2*n-1); print1(contfrac(q)[1], ", "))
(MAGMA) I:=[3, 36]; [n le 2 select I[n] else 11*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jul 23 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Derek Orr, Jun 18 2015
STATUS
proposed