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A059927
Period of the continued fraction for sqrt(2^(2n+1)).
5
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
0,2
COMMENTS
K. R. Matthews (Feb 2007) showed that lim_{n -> oo} a(n)/2^n = 0.7427.... - A.H.M. Smeets, Nov 14 2017
FORMULA
a(n) = A003285(A004171(n)). - Michel Marcus, Sep 27 2019
EXAMPLE
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.
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 *)
CROSSREFS
Cf. A003285, A004171, A059866 (for sqrt(2^n-1)).
Cf. A064932 (for sqrt(3^(2n+1))), A293028 (for sqrt(5^(2n+1))).
Sequence in context: A129882 A129017 A086915 * A290437 A154987 A089419
KEYWORD
nonn
AUTHOR
Labos Elemer, Mar 01 2001
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
STATUS
approved