OFFSET
0,2
COMMENTS
(25*a(n))^2 - 629*b(n)^2 = -4 with b(n)=A098262(n) give all positive solutions of this Pell equation.
LINKS
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 (627,-1).
FORMULA
a(n) = S(n, 627) + S(n-1, 627) = S(2*n, sqrt(629)), 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, 531)=A098260(n).
a(n) = (-2/25)*i*((-1)^n)*T(2*n+1, 25*i/2) with the imaginary unit i and Chebyshev's polynomials of the first kind. See the T-triangle A053120.
G.f.: (1+x)/(1-627*x+x^2).
a(n) = 627*a(n-1) - a(n-2), n > 1; a(0)=1, a(1)=628. [Philippe Deléham, Nov 18 2008]
EXAMPLE
All positive solutions of Pell equation x^2 - 629*y^2 = -4 are (25=25*1,1), (15700=25*628,626), (9843875=25*393755,392501), (6172093925=25*246883757,246097501), ...
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Sep 10 2004
STATUS
approved