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%I #25 Sep 08 2022 08:44:54
%S 1,1,23,24,551,575,13201,13776,316273,330049,7577351,7907400,
%T 181540151,189447551,4349386273,4538833824,104203730401,108742564225,
%U 2496540143351,2605282707576,59812759710023,62418042417599,1433009692897201,1495427735314800
%N Denominators of continued fraction convergents to sqrt(143).
%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 = 22 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="/A041263/b041263.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 (0,24,0,-1).
%F G.f.: (1 + x - x^2)/(x^4 - 24*x^2 + 1). - _Vincenzo Librandi_, Dec 14 2013
%F a(n) = 24*a(n-2) - a(n-4). - _Vincenzo Librandi_, Dec 14 2013
%F From _Peter Bala_, May 28 2014: (Start)
%F The following remarks assume an offset of 1.
%F Let alpha = ( sqrt(22) + sqrt(26) )/2 and beta = ( sqrt(22) - sqrt(26) )/2 be the roots of the equation x^2 - sqrt(22)*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)} ( 22 + 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) = 22*a(2*n) + a(2*n - 1). (End)
%t Denominator[Convergents[Sqrt[143], 30]] (* or *) CoefficientList[Series[(1 + x - x^2)/(x^4 - 24 x^2 + 1), {x, 0, 30}], x] (* _Vincenzo Librandi_, Dec 14 2013 *)
%o (Magma) I:=[1,1,23,24]; [n le 4 select I[n] else 24*Self(n-2)-Self(n-4): n in [1..40]]; // _Vincenzo Librandi_, Dec 14 2013
%Y Cf. A041262, A002530.
%K nonn,frac,easy,less
%O 0,3
%A _N. J. A. Sloane_
%E More terms from _Vincenzo Librandi_, Dec 14 2013
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