A090731 a(n) = 23a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 23.
2, 23, 527, 12098, 277727, 6375623, 146361602, 3359941223, 77132286527, 1770682648898, 40648568638127, 933146396028023, 21421718540006402, 491766380024119223, 11289205022014735727, 259159949126314802498
Offset: 0
Examples
(x;y) = (0;2), (23;1), (527;23), (12098;528), ... give the nonnegative integer solutions to x^2 - 21*(5*y)^2 = 4.
References
- O. Perron, "Die Lehre von den Kettenbruechen, Bd.I", Teubner, 1954, 1957 (Sec. 30, Satz 3.35, p. 109 and table p. 108).
Links
Crossrefs
Programs
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Mathematica
a[0] = 2; a[1] = 23; a[n_] := 23a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 15}] (* Robert G. Wilson v, Jan 30 2004 *) LinearRecurrence[{23,-1},{2,23},30] (* Harvey P. Dale, Feb 20 2012 *) -
Sage
[lucas_number2(n,23,1) for n in range(0,20)] # Zerinvary Lajos, Jun 26 2008
Formula
a(n) = S(n, 23) - S(n-2, 23) = 2*T(n, 23/2) with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. S(n, 23)=A097778(n). U-, resp. T-, are Chebyshev's polynomials of the second, resp. first, case. See A049310 and A053120.
a(n) = ap^n + am^n, with ap := (23+5*sqrt(21))/2 and am := (23-5*sqrt(21))/2.
G.f.: (2-23*x)/(1-23*x+x^2).
Extensions
Chebyshev and Pell comments from Wolfdieter Lang, Sep 10 2004
Comments