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%I #14 Jan 17 2020 13:11:45
%S 1,201,80801,32481801,13057603201,5249124005001,2110134792407201,
%T 848268937423689801,341002002709530892801,137081956820293995216201,
%U 55106605639755476546020001,22152718385224881277504824201
%N Nonnegative solutions of the Pell equation x^2 - 101*y^2 = +1. Solutions x = a(n).
%C The Pell equation x^2 - 101*y^2 = +1 has only proper solutions, namely x(n) = a(n) and y(n) = 20*A097740(n), n>= 0.
%D T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964, ch. VI, 56., pp. 115-200.
%D O. Perron, Die Lehre von den Kettenbruechen, Band I, Teubner, Stuttgart, 1954, Paragraph 27, pp. 92-95.
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (402,-1).
%F a(n) = (S(n, 2*201) - S(n-2, 2*201))/2 = T(n, 201) with the Chebyshev S- and T-polynomials (see A049310 and A053120, respectively). S(n, -2) = -1, S(n, -1) = 0. For S(n, 2*201) see A097740.
%F a(n) = 2*201*a(n-1) - a(n-2), n >= 1, with input a(-1) = 201 and a(0) = 1.
%F O.g.f.: (1 - 201*x)/(1 - 2*201*x + x^2).
%e n=0: 1^2 - 101*0^2 = +1 (a proper, but not a positive solution),
%e n=1: 201^2 - 101*(20*1)^2 = +1, where 20 is the positive fundamental y-solution.
%e n=2: 80801^2 - 101*(20*402)^2 = +1, where 80801 = 7^2*17*97 and 20*402 = 8040 = 2^3*3*5*67.
%t LinearRecurrence[{402,-1},{1,201},20] (* _Harvey P. Dale_, Jan 17 2020 *)
%Y Cf. A097740 (y/20 solutions and S(n,402)), A049310, A053120.
%K nonn,easy
%O 0,2
%A _Wolfdieter Lang_, Jul 05 2013
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