A157762 a(n) = 15780962*n^2 - 5618000*n + 500001.

10662963, 52387849, 125674659, 230523393, 366934051, 534906633, 734441139, 965537569, 1228195923, 1522416201, 1848198403, 2205542529, 2594448579, 3014916553, 3466946451, 3950538273, 4465692019, 5012407689, 5590685283

Offset

1,1

Comments

The identity (15780962*n^2 - 5618000*n + 500001)^2 - (2809*n^2 - 1000*n + 89)*(297754*n - 53000)^2 = 1 can be written as a(n)^2 -A157760(n)*A157761(n)^2 = 1.

This is the case s=53 and r=500 of the identity (2*(s^2*n-r)^2+1)^2 - (((s^2*n-r)^2+1)/s^2)*(2*s*(s^2*n-r))^2 = 1, where ((s^2*n-r)^2+1)/s^2 is an integer if r^2 == -1 (mod s^2). Therefore, for s=53, nonnegative r values are: 500, 2309, 3309, 5118, 6118, 7927, 8927, 10736, 11736, ... - Bruno Berselli, Apr 24 2018

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..10000

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Formula

G.f: x*(10662963 + 20398960*x + 500001*x^2)/(1 - x)^3.

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).

Mathematica

LinearRecurrence[{3, -3, 1}, {10662963, 52387849, 125674659}, 30]

Prog

(Magma) I:=[10662963, 52387849, 125674659]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..40]];
(PARI) a(n) = 15780962*n^2 - 5618000*n + 500001;

Crossrefs

Cf. A157760, A157761.

Sequence in context: A250963 A233500 A195224 * A234390 A237149 A344634

Adjacent sequences: A157759 A157760 A157761 * A157763 A157764 A157765

Keyword

nonn,easy

Author

Vincenzo Librandi, Mar 06 2009

Status

approved