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

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

0,1

Comments

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.

Links

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

Formula

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

Mathematica

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

Prog

(Python, 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

Crossrefs

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

Sequence in context: A029104 A332886 A085543 * A083499 A029103 A175732

Adjacent sequences: A294223 A294224 A294225 * A294227 A294228 A294229

Keyword

nonn

Author

A.H.M. Smeets, Oct 25 2017

Status

approved