OFFSET
0,2
COMMENTS
(17*a(n))^2 - 293*b(n)^2 = -4 with b(n)=A098250(n) give all positive solutions of this Pell equation.
LINKS
Indranil Ghosh, Table of n, a(n) for n = 0..405
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 (291,-1).
FORMULA
a(n) = (-2/17)*i*((-1)^n)*T(2*n+1, 17*i/2) with the imaginary unit i and Chebyshev's polynomials of the first kind. See the T-triangle A053120.
G.f.: (1+x)/(1-291*x+x^2).
a(n) = S(n, 291) + S(n-1, 291) = S(2*n, sqrt(293)), 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) = 291*a(n-1) - a(n-2), n > 1; a(0)=1, a(1)=292. [Philippe Deléham, Nov 18 2008]
EXAMPLE
All positive solutions of Pell equation x^2 - 293*y^2 = -4 are (17=17*1,1), (4964=17*292,290), (1444507=17*84971,84389), (420346573=17*24726269,24556909), ...
MATHEMATICA
LinearRecurrence[{291, -1}, {1, 292}, 20] (* Harvey P. Dale, Jan 01 2020 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Sep 10 2004
STATUS
approved