cp's OEIS Frontend

A059927 Period of the continued fraction for sqrt(2^(2n+1)).

%I A059927 #48 Nov 20 2023 08:14:19
%S A059927 1,2,4,4,12,24,48,96,196,368,760,1524,3064,6068,12168,24360,48668,
%T A059927 97160,194952,389416,778832,1557780,3116216,6229836,12462296,24923320,
%U A059927 49849604,99694536,199394616,398783628,797556364,1595117676,3190297400,6380517544,12761088588,25522110948,51044281208,102088450460,204177067944,408353857832,816708255152
%N A059927 Period of the continued fraction for sqrt(2^(2n+1)).
%C A059927 K. R. Matthews (Feb 2007) showed that lim_{n -> oo} a(n)/2^n = 0.7427.... - _A.H.M. Smeets_, Nov 14 2017
%H A059927 K. R. Matthews, <a href="http://www.numbertheory.org/pdfs/period.pdf">On the continued fraction expansion of sqrt(2^(2n+1))</a>, Feb 2007.
%F A059927 a(n) = A003285(A004171(n)). - _Michel Marcus_, Sep 27 2019
%e A059927 For n=5 we look at the square root of 2^11 = 2048, and find that the cycle has length 24. Here is Maple's calculation:  cfrac(sqrt(2048),'periodic','quotients') = [[45],[3,1,12,5,1,1,2,1,2,4,1,21,1,4,2,1,2,1,1,5,12,1,3,90]], the periodic part having length 24.
%p A059927 with(numtheory): [seq(nops(cfrac(sqrt(2^(2*k-1)),'periodic','quotients')[2]),k=1..15)];
%t A059927 Array[Length@ ContinuedFraction[Sqrt[2^(2 # + 1)]][[-1]] &, 15, 0] (* _Michael De Vlieger_, Oct 09 2017 *)
%Y A059927 Cf. A003285, A004171, A059866 (for sqrt(2^n-1)).
%Y A059927 Cf. A064932 (for sqrt(3^(2n+1))), A293028 (for sqrt(5^(2n+1))).
%K A059927 nonn
%O A059927 0,2
%A A059927 _Labos Elemer_, Mar 01 2001
%E A059927 More terms from _Don Reble_, Oct 31 2001
%E A059927 a(32) = 3190297400 from _Don Reble_, Feb 10 2007
%E A059927 a(33)-a(35) from Keith Matthews (keithmatt(AT)gmail.com), Feb 16 2007, Feb 28 2007
%E A059927 Name clarified by _Joerg Arndt_, Oct 09 2017
%E A059927 a(36)-a(37) from _Chai Wah Wu_, Sep 26 2019
%E A059927 a(38)-a(40) from _Chai Wah Wu_, Sep 30 2019