%I #31 Feb 02 2020 07:56:08
%S 1,1,13,14,181,195,2521,2716,35113,37829,489061,526890,6811741,
%T 7338631,94875313,102213944,1321442641,1423656585,18405321661,
%U 19828978246,256353060613,276182038859,3570537526921,3846719565780,49731172316281,53577891882061
%N Denominators of continued fraction convergents to sqrt(48).
%C The following remarks assume an offset of 1. This is the sequence of Lehmer numbers U_n(sqrt(R),Q) for the parameters R = 12 and Q = -1; it is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for all positive integers n and m. Consequently, this is a divisibility sequence: if n divides m then a(n) divides a(m). - _Peter Bala_, May 28 2014
%H Vincenzo Librandi, <a href="/A041083/b041083.txt">Table of n, a(n) for n = 0..200</a>
%H Eric W. Weisstein, <a href="http://mathworld.wolfram.com/LehmerNumber.html">MathWorld: Lehmer Number</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,14,0,-1).
%F From _Colin Barker_, Jul 15 2012: (Start)
%F a(n) = 14*a(n-2) - a(n-4).
%F G.f.: (1+x-x^2)/((1-4*x+x^2)*(1+4*x+x^2)). (End)
%F From _Peter Bala_, May 28 2014: (Start)
%F The following remarks assume an offset of 1.
%F Let alpha = sqrt(3) + 2 and beta = sqrt(3) - 2 be the roots of the equation x^2 - sqrt(12)*x - 1 = 0. Then a(n) = (alpha^n - beta^n)/(alpha - beta) for n odd, while a(n) = (alpha^n - beta^n)/(alpha^2 - beta^2) for n even.
%F a(n) = Product_{k = 1..floor((n-1)/2)} ( 12 + 4*cos^2(k*Pi/n) ).
%F Recurrence equations: a(0) = 0, a(1) = 1 and for n >= 1, a(2*n) = a(2*n - 1) + a(2*n - 2) and a(2*n + 1) = 12*a(2*n) + a(2*n - 1). (End)
%t Table[Denominator[FromContinuedFraction[ContinuedFraction[Sqrt[48], n]]], {n, 1, 50}] (* _Vladimir Joseph Stephan Orlovsky_, Jun 23 2011 *)
%t Denominator[Convergents[Sqrt[48], 40]] (* _Vincenzo Librandi_, Oct 24 2013 *)
%t LinearRecurrence[{0,14,0,-1},{1,1,13,14},30] (* _Harvey P. Dale_, Mar 15 2015 *)
%Y Cf. A010502, A041082, A002530.
%K nonn,cofr,frac,easy
%O 0,3
%A _N. J. A. Sloane_