|
%I #24 Jan 23 2020 04:10:00
%S 1,628,393755,246883757,154795721884,97056670737511,60854377756697513,
%T 38155597796778603140,23923498964202427471267,
%U 14999995694957125245881269,9404973377239153326740084396
%N Chebyshev polynomials S(n,627) + S(n-1,627) with Diophantine property.
%C (25*a(n))^2 - 629*b(n)^2 = -4 with b(n)=A098262(n) give all positive solutions of this Pell equation.
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H Giovanni Lucca, <a href="http://forumgeom.fau.edu/FG2019volume19/FG201902index.html">Integer Sequences and Circle Chains Inside a Hyperbola</a>, Forum Geometricorum (2019) Vol. 19, 11-16.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (627,-1).
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%F 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).
%F 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.
%F G.f.: (1+x)/(1-627*x+x^2).
%F a(n) = 627*a(n-1) - a(n-2), n > 1; a(0)=1, a(1)=628. [_Philippe Deléham_, Nov 18 2008]
%e 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), ...
%K nonn,easy
%O 0,2
%A _Wolfdieter Lang_, Sep 10 2004
|
