A080806 Positive integer values of n such that 6*n^2-5 is a square.

1, 3, 7, 29, 69, 287, 683, 2841, 6761, 28123, 66927, 278389, 662509, 2755767, 6558163, 27279281, 64919121, 270037043, 642633047, 2673091149, 6361411349, 26460874447, 62971480443, 261935653321, 623353393081, 2592895658763

Offset: 1

Author

John W. Layman, Mar 24 2003

Keywords

• nonn
• easy

Comments

The corresponding sequence for which 6n^2-6 is a square is A001079.

Positive values of x (or y) satisfying x^2 - 10xy + y^2 + 20 = 0. - Colin Barker, Feb 09 2014

Examples

			29 is a term of the sequence since 6*29^2 - 5 = 5041 = 71^2.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..200
• Eric Weisstein's World of Mathematics, Pell Equation.
• Index entries for linear recurrences with constant coefficients, signature (0,10,0,-1).

Crossrefs

Cf. A001079, A077288.

Programs

Magma

I:=[1,3,7,29]; [n le 4 select I[n] else 10*Self(n-2)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Feb 10 2014

Mathematica

Do[ If[ IntegerQ[ Sqrt[6n^2 - 5]], Print[n]], {n, 1, 3*10^7}]
a[1]=1; a[2]=3; a[3]=7; a[4]=29; a[n_] := a[n]=10a[n-2]-a[n-4]
CoefficientList[Series[(1 - x) (1 + 4 x + x^2)/(1 - 10 x^2 + x^4), {x, 0, 30}], x] (* Vincenzo Librandi, Feb 10 2014 *)

Formula

a(n) = 10*a(n-2)-a(n-4).

G.f.: x*(1-x)*(1+4*x+x^2)/(1-10*x^2+x^4). - Colin Barker, Jun 13 2012

a(2*n+1) = ((6+r)*(5+2*r)^n+(6-r)*(5+2*r)^n)/12, a(2*n+2) = ((18+7*r)*(5+2*r)^n+(18-7*r)*(5-2*r)^n)/12, where r=sqrt(6) and n>=0. - Paul Weisenhorn, Sep 01 2012

Extensions

Extended by Robert G. Wilson v, Apr 14 2003