A293028 -id:A293028 - cp's OEIS frontend

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

Labos Elemer, Mar 01 2001

nonn

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

			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.

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

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

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

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

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

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

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

A294226 Length of period of continued fraction expansion of sqrt(3*2^n).

Original entry on oeis.org

2, 2, 2, 2, 2, 4, 4, 8, 8, 12, 16, 32, 36, 60, 72, 128, 136, 244, 292, 508, 576, 972, 1120, 1992, 2272, 3948, 4588, 7924, 9056, 15764, 18132, 31832, 36444, 63216, 72808, 126456, 145332, 253112, 290968, 507096, 581952, 1012312, 1163452, 2026504, 2327844, 4051424, 4656388

Offset: 0

A.H.M. Smeets, Oct 25 2017

nonn

Lim {n->inf} a(2n)/2^n = 0.555...
Lim {n->inf} a(2n+1)/2^n = 0.966...
It seems that Lim {n->inf} a(2n+1)/a(2n) = sqrt(3).
It seems that Lim {n->inf} a(n)/2^n = (Lim {n -> inf} A064932(n)/3^n)/2.

Chai Wah Wu, Table of n, a(n) for n = 0..80 (n = 0..46 from A.H.M. Smeets)

Cf. A003285, A007283, A059927, A064932, A293028.

Array[Length@ Last@ ContinuedFraction@ Sqrt[3*2^#] &, 47, 0] (* Michael De Vlieger, Oct 25 2017 *)

# for odd n
m, p, q = 0, 6, 2
tl, nl, tb, nb = 3, 1, 2, 1
while nl < 10**100000000:
    tl = tl * nb + tb * nl
    nl = 2 * nl * nb
    nb = tl
    tb = p * nl
tl = tl *nb + tb * nl
nl = 2 * nl * nb
tel, noe = tl, nl
while m >= 0:
    tl = tel*q**m
    nl = noe
    a0 = tl//nl
    t = 0
    an = a0
    while an != 2*a0:
        tl = tl - an*nl
        tl, nl = nl, tl
        an = tl//nl
        t = t + 1
    print(2*m+1, t)
    m = m+1

a(n) = A003285(A007283(n)). - Michel Marcus, Oct 02 2019

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

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