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Length of period of continued fraction expansion of sqrt(3*2^n).
1

%I #36 Oct 02 2019 01:54:15

%S 2,2,2,2,2,4,4,8,8,12,16,32,36,60,72,128,136,244,292,508,576,972,1120,

%T 1992,2272,3948,4588,7924,9056,15764,18132,31832,36444,63216,72808,

%U 126456,145332,253112,290968,507096,581952,1012312,1163452,2026504,2327844,4051424,4656388

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

%C Lim {n->inf} a(2n)/2^n = 0.555...

%C Lim {n->inf} a(2n+1)/2^n = 0.966...

%C It seems that Lim {n->inf} a(2n+1)/a(2n) = sqrt(3).

%C It seems that Lim {n->inf} a(n)/2^n = (Lim {n -> inf} A064932(n)/3^n)/2.

%H Chai Wah Wu, <a href="/A294226/b294226.txt">Table of n, a(n) for n = 0..80</a> (n = 0..46 from A.H.M. Smeets)

%F a(n) = A003285(A007283(n)). - _Michel Marcus_, Oct 02 2019

%t Array[Length@ Last@ ContinuedFraction@ Sqrt[3*2^#] &, 47, 0] (* _Michael De Vlieger_, Oct 25 2017 *)

%o (Python, for odd n)

%o m,p,q = 0,6,2

%o tl,nl,tb,nb = 3,1,2,1

%o while nl < 10**100000000:

%o ....tl = tl * nb + tb * nl

%o ....nl = 2 * nl * nb

%o ....nb = tl

%o ....tb = p * nl

%o tl = tl *nb + tb * nl

%o nl = 2 * nl * nb

%o tel,noe = tl,nl

%o while m >= 0:

%o ....tl = tel*q**m

%o ....nl = noe

%o ....a0 = tl//nl

%o ....t = 0

%o ....an = a0

%o ....while an != 2*a0:

%o ........tl = tl - an*nl

%o ........tl, nl = nl, tl

%o ........an = tl//nl

%o ........t = t + 1

%o ....print(2*m+1,t)

%o ....m = m+1

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

%K nonn

%O 0,1

%A _A.H.M. Smeets_, Oct 25 2017