%I #40 Jun 28 2024 09:29:28
%S 1,2,8,56,551,6930,105937,1905632,39424240,922080050,24057287759,
%T 692686638072,21817946138353,746243766783074,27543862067299424,
%U 1091228270370045824,46187969968474139807,2080128468827570457762,99318726126650358502921,5011361251329169946919800
%N Denominators of the continued fraction n-1/(n-1/...) [n times].
%C The (n-1)-th term of the Lucas sequence U(n,1). The numerator is the n-th term. Adjacent terms of the sequence U(n,1) are relatively prime.
%H Alois P. Heinz, <a href="/A097691/b097691.txt">Table of n, a(n) for n = 1..387</a>
%H Spencer Daugherty, Pamela E. Harris, Ian Klein, and Matt McClinton, <a href="https://arxiv.org/abs/2406.12941">Metered Parking Functions</a>, arXiv:2406.12941 [math.CO], 2024. See pp. 8, 22.
%H Pascual Jara and Miguel L. RodrÃguez, <a href="https://amj-math.com/wp-content/uploads/2022/01/AMJ2020-vol7iss2.pdf#page=6">Solving quadratic congruences</a>, Arhimede Math. J. (2020) Vol. 7, No. 2, 105-120.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/LucasSequence.html">Lucas Sequence</a>
%F a(n) = ChebyshevU(n-1,n/2). - _Gary Detlefs_, Oct 15 2011
%F a(n) = abs((2^(-n) * (sqrt(4 - n^2) + i*n)^n - 2^n*(-sqrt(4 - n^2) - i*n)^(-n))/sqrt(4 - n^2)), where i is the imaginary unit, for n > 2. - _Daniel Suteu_, May 31 2017
%F a(n) ~ n^(n-1). - _Vaclav Kotesovec_, Jun 03 2017
%e a(4) = 56 because 4-1/(4-1/(4-1/4)) = 209/56.
%t Table[s=n; Do[s=n-1/s, {n-1}]; Denominator[s], {n, 20}]
%t Table[Abs[Fibonacci[n, I n]], {n, 20}] (* _Vladimir Reshetnikov_, Oct 16 2018 *)
%o (Sage) [lucas_number1(n,n,1) for n in range(1,19)] # _Zerinvary Lajos_, Jul 16 2008
%Y Cf. A084844, A084845, A097690 (numerators).
%K easy,frac,nonn
%O 1,2
%A _T. D. Noe_, Aug 19 2004