A227152 Nonnegative solutions of the Pell equation x^2 - 101*y^2 = +1. Solutions x = a(n).
1, 201, 80801, 32481801, 13057603201, 5249124005001, 2110134792407201, 848268937423689801, 341002002709530892801, 137081956820293995216201, 55106605639755476546020001, 22152718385224881277504824201
Offset: 0
Examples
n=0: 1^2 - 101*0^2 = +1 (a proper, but not a positive solution), n=1: 201^2 - 101*(20*1)^2 = +1, where 20 is the positive fundamental y-solution. n=2: 80801^2 - 101*(20*402)^2 = +1, where 80801 = 7^2*17*97 and 20*402 = 8040 = 2^3*3*5*67.
References
- T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964, ch. VI, 56., pp. 115-200.
- O. Perron, Die Lehre von den Kettenbruechen, Band I, Teubner, Stuttgart, 1954, Paragraph 27, pp. 92-95.
Links
Programs
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Mathematica
LinearRecurrence[{402,-1},{1,201},20] (* Harvey P. Dale, Jan 17 2020 *)
Formula
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.
a(n) = 2*201*a(n-1) - a(n-2), n >= 1, with input a(-1) = 201 and a(0) = 1.
O.g.f.: (1 - 201*x)/(1 - 2*201*x + x^2).
Comments