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%I #39 Sep 08 2022 08:45:36
%S 1,4,9,40,89,396,881,3920,8721,38804,86329,384120,854569,3802396,
%T 8459361,37639840,83739041,372596004,828931049,3688320200,8205571449,
%U 36510605996,81226783441,361417739760,804062262961,3577666791604,7959395846169,35415250176280
%N Denominators of continued fraction convergents to sqrt(3/2).
%C sqrt(3/2) = 1.224744871... = 2/2 + 2/9 + 2/(9*89) + 2/(89*881) + 2/(881*8721) + 2/(8721*86329) + ... - _Gary W. Adamson_, Oct 08 2008
%C From _Charlie Marion_, Jan 07 2009: (Start)
%C In general, denominators, a(k,n) and numerators, b(k,n), of continued fraction convergents to sqrt((k+1)/k) may be found as follows:
%C a(k,0) = 1, a(k,1) = 2k;
%C for n > 0, a(k,2n) = 2*a(k,2n-1)+a(k,2n-2) and a(k,2n+1)=(2k)*a(k,2n)+a(k,2n-1);
%C b(k,0) = 1, b(k,1) = 2k+1;
%C for n > 0, b(k,2n) = 2*b(k,2n-1)+b(k,2n-2) and b(k,2n+1)=(2k)*b(k,2n)+b(k,2n-1).
%C For example, the convergents to sqrt(3/2) start 1/1, 5/4, 11/9, 49/40, 109/89.
%C In general, if a(k,n) and b(k,n) are the denominators and numerators, respectively, of continued fraction convergents to sqrt((k+1)/k) as defined above, then
%C k*a(k,2n)^2 - a(k,2n-1)*a(k,2n+1) = k = k*a(k,2n-2)*a(k,2n) - a(k,2n-1)^2 and
%C b(k,2n-1)*b(k,2n+1) - k*b(k,2n)^2 = k+1 = b(k,2n-1)^2 - k*b(k,2n-2)*b(k,2n);
%C for example, if k=2 and n=3, then a(2,n)=a(n) and
%C 2*a(2,6)^2 - a(2,5)*a(2,7) = 2*881^2 - 396*3920 = 2;
%C 2*a(2,4)*a(2,6) - a(2,5)^2 = 2*89*881 - 396^2 = 2;
%C b(2,5)*b(2,7) - 2*b(2,6)^2 = 485*4801 - 2*1079^2 = 3;
%C b(2,5)^2 - 2*b(2,4)*b(2,6) = 485^2 - 2*109*1079 = 3.
%C (End)
%C For n > 0, a(n) equals the permanent of the n X n tridiagonal matrix with the main diagonal alternating sequence [4, 2, 4, 2, 4, ...] and 1's along the superdiagonal and the subdiagonal. - _Rogério Serôdio_, Apr 01 2018
%H Vincenzo Librandi, <a href="/A142239/b142239.txt">Table of n, a(n) for n = 0..200</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,10,0,-1).
%F G.f.'s for numerators and denominators are -(1+5*x+x^2-x^3)/(-1-x^4+10*x^2) and -(1+4*x-x^2)/(-1-x^4+10*x^2).
%F a(n) = 10*a(n-2) - a(n-4) for n > 3. - _Vincenzo Librandi_, Feb 01 2014
%F From: _Rogério Serôdio_, Apr 02 2018: (Start)
%F Recurrence formula: a(n) = (3-(-1)^n)*a(n-1) + a(n-2), a(0) = 1, a(1) = 4;
%F Some properties:
%F (1) a(n)^2 - a(n-2)^2 = (3-(-1)^n)*a(2*n-1), for n > 1;
%F (2) a(2*n+1) = a(n)*(a(n+1) + a(n-1)), for n > 0;
%F (3) a(2*n) = A041007(2*n);
%F (4) a(2*n+1) = 2*A041007(2*n+1). (End)
%e The initial convergents are 1, 5/4, 11/9, 49/40, 109/89, 485/396, 1079/881, 4801/3920, 10681/8721, 47525/38804, 105731/86329, ...
%p with(numtheory): cf := cfrac (sqrt(3)/sqrt(2),100): [seq(nthnumer(cf,i), i=0..50)]; [seq(nthdenom(cf,i), i=0..50)]; [seq(nthconver(cf,i), i=0..50)];
%t Table[Denominator[FromContinuedFraction[ContinuedFraction[Sqrt[3/2], n]]], {n, 1, 50}] (* _Vladimir Joseph Stephan Orlovsky_, Jun 23 2011 *)
%t Denominator[Convergents[Sqrt[3/2], 30]] (* _Bruno Berselli_, Nov 11 2013 *)
%o (Magma) I:=[1,4,9,40]; [n le 4 select I[n] else 10*Self(n-2)-Self(n-4): n in [1..30]]; // _Vincenzo Librandi_, Feb 01 2014
%Y Cf. A115754, A142238.
%Y Cf. A000129, A001333, A142238, A153313-A153318.
%K nonn,frac,easy
%O 0,2
%A _N. J. A. Sloane_, Oct 05 2008, following a suggestion from Rob Miller (rmiller(AT)AmtechSoftware.net)
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