A097783 -id:A097783 - cp's OEIS frontend

3, 36, 393, 4287, 46764, 510117, 5564523, 60699636, 662131473, 7222746567, 78788080764, 859446141837, 9375119479443, 102266868132036, 1115560429972953, 12168897861570447, 132742316047301964, 1447996578658751157, 15795220049198960763, 172299423962529817236, 1879498443538629028833, 20502183454962389499927

Offset: 1

Derek Orr, Jun 18 2015

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

Index entries for linear recurrences with constant coefficients, signature (11,-1).

Cf. A002878, A077444.

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

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 *)

for(n=1,20,q=((3+sqrt(13))/2)^(2*n-1);print1(contfrac(q)[1],", "))

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

cp's OEIS Frontend

A259131 Numbers n such that 13*n^2 + 52 is a square.

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