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

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

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

Table of n, a(n) for n=1..22.

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

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

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

Cf. A002878, A077444.

Sequence in context: A068177 A249894 A099670 * A199258 A262465 A322491

Adjacent sequences: A259128 A259129 A259130 * A259132 A259133 A259134

Keyword

nonn,easy

Author

Derek Orr, Jun 18 2015

Status

approved