A042103 - OEIS

%I #22 Sep 08 2022 08:44:55

%S 1,1,47,48,2255,2303,108193,110496,5191009,5301505,249060239,

%T 254361744,11949700463,12204062207,573336561985,585540624192,

%U 27508205274817,28093745899009,1319820516629231,1347914262528240,63323876592928271,64671790855456511

%N Denominators of continued fraction convergents to sqrt(575).

%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 = 46 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 27 2014

%H Vincenzo Librandi, <a href="/A042103/b042103.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 (0,48,0,-1).

%F G.f.: -(x^2-x-1) / (x^4-48*x^2+1). - _Colin Barker_, Dec 01 2013

%F a(n) = 48*a(n-2) - a(n-4) for n > 3. - _Vincenzo Librandi_, Jan 14 2014

%F From _Peter Bala_, May 27 2014: (Start)

%F The following remarks assume an offset of 1.

%F Let alpha = ( sqrt(46) + sqrt(50) )/2 and beta = ( sqrt(46) - sqrt(50) )/2 be the roots of the equation x^2 - sqrt(46)*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)} ( 46 + 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) = 46*a(2*n) + a(2*n - 1). (End)

%t Denominator[Convergents[Sqrt[575], 30]] (* _Vincenzo Librandi_, Jan 14 2014 *)

%o (Magma) I:=[1,1,47,48]; [n le 4 select I[n] else 48*Self(n-2)-Self(n-4): n in [1..30]]; // _Vincenzo Librandi_, Jan 14 2014

%Y Cf. A042102, A040551, A002530.

%K nonn,frac,easy

%O 0,3

%A _N. J. A. Sloane_

%E More terms from _Colin Barker_, Dec 01 2013