A217975 Integers k such that 2*k^2 - 7 is a square.

2, 4, 8, 22, 46, 128, 268, 746, 1562, 4348, 9104, 25342, 53062, 147704, 309268, 860882, 1802546, 5017588, 10506008, 29244646, 61233502, 170450288, 356895004, 993457082, 2080136522, 5790292204, 12123924128, 33748296142, 70663408246, 196699484648

Offset

1,1

Comments

a(n) gives y-values solving the Diophantine equation x^2 + 7 = 2*y^2. A077446(n) gives the x-values. - Sture Sjöstedt, Oct 16 2012

Positive values of x (or y) satisfying x^2 - 6xy + y^2 + 28 = 0. - Colin Barker, Feb 08 2014

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Formula

a(n) = 6*a(n - 2) - a(n - 4) with a(1)=2, a(2)=4, a(3)=8, a(4)=22. - Sture Sjöstedt, Oct 16 2012

a(n)*a(n+3)-a(n+1)*a(n+2) = 10-2*(-1)^n. - Bruno Berselli, Oct 25 2012

a(n) = 2*A006452(n). - R. J. Mathar, Oct 17 2012

G.f.: -2*x*(x - 1)*(x^2 + 3*x + 1)/((x^2 - 2*x - 1)*(x^2 + 2*x - 1)). - Colin Barker, Oct 24 2012

a(n) = a(-n+1) = ((4+sqrt(2))*(1-(-1)^n*sqrt(2))^(2*floor(n/2))+(4-sqrt(2))*(1+(-1)^n*sqrt(2))^(2*floor(n/2)))/4. - Bruno Berselli, Oct 25 2012

a(2n-1) = A078343(2n-1), a(2n) = A100525(n-1). - Bruno Berselli, Oct 25 2012

Example

Since 2(4^2) - 7 = 25 = 5^2, and 4 is the second number with this property, a(2) = 4.

Mathematica

LinearRecurrence[{0, 6, 0, -1}, {2, 4, 8, 22}, 50] (* Sture Sjöstedt, Oct 16 2012 *)

Prog

(Magma) I:=[2, 4, 8, 22]; [n le 4 select I[n] else 6*Self(n-2)-Self(n-4): n in [1..31]]; // Vincenzo Librandi, Oct 16 2012
(PARI) Vec(2*x*(1-x)*(x^2+3*x+1)/(x^2-2*x-1)/(x^2+2*x-1)+O(x^99)) \\ Charles R Greathouse IV, Oct 24 2012

Crossrefs

Cf. A077442 (2*n^2 + 7 is a square).

Sequence in context: A295628 A027713 A155765 * A027385 A158324 A002075

Adjacent sequences: A217972 A217973 A217974 * A217976 A217977 A217978

Keyword

nonn,easy

Author

Sture Sjöstedt, Oct 16 2012

Status

approved