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A317414 Continued fraction for ternary expansion of Liouville's number interpreted in base 3 (A012245). 6
0, 2, 4, 8, 1, 3, 2, 531440, 1, 1, 3, 1, 8, 4, 2, 22528399544939174411840147874772640, 1, 1, 4, 8, 1, 3, 1, 1, 531440, 2, 3, 1, 8, 4, 2 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
The continued fraction of the number obtained by reading A012245 as a ternary fraction.
Except for the first term, the only values that occur in this sequence are 1,2,3,4, and values 3^((m-1)*m!)-1 for m > 1. The probability of occurrence P(a(n) = k) are given by:
P(a(n) = 1) = 3/8,
P(a(n) = 2) = 1/8,
P(a(n) = 3) = 1/8,
P(a(n) = 4) = 1/8 and
P(a(n) = 3^((m-1)*m!)-1) = 2^-(m+1) for m > 1.
More generally it seems that for any base > 2, P(a(n) <= base+1) = 3/4, P(a(n) > base+1) = 1/4, and P(a(n) = base^((m-1)*m!)-1) = 2^-(m+1) for m > 1.
LINKS
FORMULA
a(n) = 1 if and only if n in {floor(8*n/3) + A317627(n) | n > 0}.
a(n) = 2 if and only if n in {8*n - 10 + 3*A089013(n-1) | n > 0}.
a(n) = 3 if and only if n in {16*n - 11 | n > 0} union {16*n - 6 | n > 0}.
a(n) = 4 if and only if n in {16*n - 14 | n > 0} union {16*n - 3 | n > 0}.
a(n) = 3^((m-1)*m!)-1 iff n in {2^m*(1+k*4) - 1 | k >= 0} union {2^m*(3+k*4) | k >= 0} for m > 1.
MAPLE
with(numtheory): cfrac(add(1/3^factorial(n), n=1..7), 30, 'quotients'); # Muniru A Asiru, Aug 11 2018
MATHEMATICA
ContinuedFraction[ FromDigits[ RealDigits[ Sum[1/10^n!, {n, 8}], 10, 10000], 3], 60] (* Robert G. Wilson v, Aug 09 2018 *)
PROG
(Python)
n, f, i, p, q, base = 1, 1, 0, 0, 1, 3
while i < 100000:
....i, p, q = i+1, p*base, q*base
....if i == f:
........p, n = p+1, n+1
........f = f*n
n, a, j = 0, 0, 0
while p%q > 0:
....a, f, p, q = a+1, p//q, q, p%q
....print(a-1, f)
CROSSREFS
Cf. A058304 (in base 10), A317413 (in base 2), A317661 (in base 4).
Sequence in context: A023104 A133145 A350208 * A008952 A268516 A021407
KEYWORD
nonn
AUTHOR
A.H.M. Smeets, Jul 27 2018
STATUS
approved

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Last modified May 14 21:33 EDT 2024. Contains 372533 sequences. (Running on oeis4.)