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

1, 2, 4, 4, 12, 24, 48, 96, 196, 368, 760, 1524, 3064, 6068, 12168, 24360, 48668, 97160, 194952, 389416, 778832, 1557780, 3116216, 6229836, 12462296, 24923320, 49849604, 99694536, 199394616, 398783628, 797556364, 1595117676, 3190297400, 6380517544, 12761088588, 25522110948, 51044281208, 102088450460, 204177067944, 408353857832, 816708255152

Offset: 0

Author

Labos Elemer, Mar 01 2001

Keywords

• nonn

Comments

K. R. Matthews (Feb 2007) showed that lim_{n -> oo} a(n)/2^n = 0.7427.... - A.H.M. Smeets, Nov 14 2017

Examples

			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.
		

Links

• K. R. Matthews, On the continued fraction expansion of sqrt(2^(2n+1)), Feb 2007.

Crossrefs

Cf. A003285, A004171, A059866 (for sqrt(2^n-1)).

Cf. A064932 (for sqrt(3^(2n+1))), A293028 (for sqrt(5^(2n+1))).

Programs

Maple

with(numtheory): [seq(nops(cfrac(sqrt(2^(2*k-1)),'periodic','quotients')[2]),k=1..15)];

Mathematica

Array[Length@ ContinuedFraction[Sqrt[2^(2 # + 1)]][[-1]] &, 15, 0] (* Michael De Vlieger, Oct 09 2017 *)

Formula

a(n) = A003285(A004171(n)). - Michel Marcus, Sep 27 2019

Extensions

More terms from Don Reble, Oct 31 2001

a(32) = 3190297400 from Don Reble, Feb 10 2007

a(33)-a(35) from Keith Matthews (keithmatt(AT)gmail.com), Feb 16 2007, Feb 28 2007

Name clarified by Joerg Arndt, Oct 09 2017

a(36)-a(37) from Chai Wah Wu, Sep 26 2019

a(38)-a(40) from Chai Wah Wu, Sep 30 2019