A157757 a(n) = 2809*n^2 - 4618*n + 1898.

89, 3898, 13325, 28370, 49033, 75314, 107213, 144730, 187865, 236618, 290989, 350978, 416585, 487810, 564653, 647114, 735193, 828890, 928205, 1033138, 1143689, 1259858, 1381645, 1509050, 1642073, 1780714, 1924973, 2074850

Offset

1,1

Comments

The identity (15780962*n^2-25943924*n+10662963)^2-(2809*n^2-4618*n+1898)*(297754*n-244754)^2=1 can be written as A157759(n)^2-a(n)*A157758(n)^2=1.

From Klaus Purath, Mar 31 2025: (Start)

Numbers k such that k*53^2-1 is a square, and k is the sum of two squares (see FORMULA).

All a(n) = D satisfy the Pell equation (k*x)^2 - D*(53*y)^2 = -1 for any integer n where a(1-n) = A157760(n). The values for k and the solutions x, y can be calculated using the following algorithm: k = sqrt(D*53^2 - 1), x(0) = 1, x(1) = 4*D*53^2 - 1, y(0) = 1, y(1) = 4*D*53^2 - 3. The two recurrences are of the form (4*D*53^2 - 2, -1).

It follows from the above that this sequence and A157760 are subsequences of A031396. (End)

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

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

G.f.: x*(-89-3631*x-1898*x^2)/(x-1)^3.

a(n) = (28*n - 23)^2 + (45*n - 37)^2. - Klaus Purath, Mar 31 2025

53^2*a(n) - 1 = (2809*n-2309)^2. - Klaus Purath, Mar 31 2025

Mathematica

LinearRecurrence[{3, -3, 1}, {89, 3898, 13325}, 40]
Table[2809n^2-4618n+1898, {n, 40}] (* Harvey P. Dale, Aug 02 2024 *)

Prog

(Magma) I:=[89, 3898, 13325]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]];
(PARI) a(n) = 2809*n^2 - 4618*n + 1898;

Crossrefs

Subsequence of A031396.

Cf. A157758, A157759.

Sequence in context: A264068 A189020 A322503 * A017805 A017752 A282478

Adjacent sequences: A157754 A157755 A157756 * A157758 A157759 A157760

Keyword

nonn,easy

Author

Vincenzo Librandi, Mar 06 2009

Status

approved