A098246 Chebyshev polynomials S(n,227) + S(n-1,227) with Diophantine property.

1, 228, 51755, 11748157, 2666779884, 605347285511, 137411167031113, 31191729568777140, 7080385200945379667, 1607216248885032407269, 364831008111701411070396, 82815031625107335280572623

Offset: 0

Author

Wolfdieter Lang, Sep 10 2004

Keywords

• nonn
• easy

Comments

(15*a(n))^2 - 229*b(n)^2 = -4 with b(n)=A098247(n) give all positive solutions of this Pell equation.

Examples

			All positive solutions of Pell equation x^2 - 229*y^2 = -4 are (15=15*1,1), (3420=15*228,226), (776325=15*51755,51301), (176222355=15*11748157,11645101), ...
		

Links

• Indranil Ghosh, Table of n, a(n) for n = 0..423
• Tanya Khovanova, Recursive Sequences
• Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
• Index entries for sequences related to Chebyshev polynomials.
• Index entries for linear recurrences with constant coefficients, signature (227, -1).

Programs

Mathematica

LinearRecurrence[{227,-1},{1,228},20] (* Harvey P. Dale, May 29 2014 *)

Formula

a(n) = S(n, 227) + S(n-1, 227) = S(2*n, sqrt(229)), with S(n, x)=U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x) = 0 = U(-1, x). S(n, 227)=A098245(n).

a(n) = (-2/15)*i*((-1)^n)*T(2*n+1, 15*i/2) with the imaginary unit i and Chebyshev's polynomials of the first kind. See the T-triangle A053120.

G.f.: (1+x)/(1-227*x+x^2).

a(n) = 227*a(n-1) - a(n-2), n > 1; a(0)=1, a(1)=228. [Philippe Deléham, Nov 18 2008]