A199405 y-values in the solution to 13*x^2 - 12 = y^2.

1, 14, 25, 155, 274, 1691, 2989, 18446, 32605, 201215, 355666, 2194919, 3879721, 23942894, 42321265, 261176915, 461654194, 2849003171, 5035874869, 31077857966, 54932969365, 339007434455, 599226788146, 3698003921039, 6536561700241, 40339035696974

Offset

1,2

Comments

When are both n+1 and 13*n+1 perfect squares? This problem gives the equation 13*x^2-12=y^2.

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..200

Index entries for linear recurrences with constant coefficients, signature (0,11,0,-1).

Formula

a(n+4) = 11*a(n+2)-a(n) with a(1)=1, a(2)=14, a(3)=25, a(4)=155.

G.f.: x*(1+x)*(1+13*x+x^2)/((1+3*x-x^2)*(1-3*x-x^2)). - Bruno Berselli, Nov 08 2011

a(n) = 2^(-1-n)*(2*(3-sqrt(13))^n+(-3-sqrt(13))^n*(-3+sqrt(13))-3*(-3+sqrt(13))^n-sqrt(13)*(-3+sqrt(13))^n+2*(3+sqrt(13))^n). - Colin Barker, Mar 27 2016

Mathematica

LinearRecurrence[{0, 11, 0, -1}, {1, 14, 25, 155}, 50]  (* Bruno Berselli, Nov 08 2011 *)

Prog

(Magma) m:=27; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!(x*(1+x)*(1+13*x+x^2)/((1+3*x-x^2)*(1-3*x-x^2)))); // Bruno Berselli, Nov 08 2011
(PARI) Vec(x*(1+x)*(1+13*x+x^2)/((1+3*x-x^2)*(1-3*x-x^2)) + O(x^50)) \\ Colin Barker, Mar 27 2016

Crossrefs

Cf. A199404.

Sequence in context: A174519 A079505 A039604 * A256573 A030786 A094163

Adjacent sequences: A199402 A199403 A199404 * A199406 A199407 A199408

Keyword

nonn,easy

Author

Sture Sjöstedt, Nov 05 2011

Extensions

More terms from Bruno Berselli, Nov 08 2011

Status

approved