OFFSET
0,2
COMMENTS
(15*a(n))^2 - 229*b(n)^2 = -4 with b(n)=A098247(n) give all positive solutions of this Pell equation.
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 linear recurrences with constant coefficients, signature (227, -1).
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]
EXAMPLE
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), ...
MATHEMATICA
LinearRecurrence[{227, -1}, {1, 228}, 20] (* Harvey P. Dale, May 29 2014 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Sep 10 2004
STATUS
approved