A317414 Continued fraction for ternary expansion of Liouville's number interpreted in base 3 (A012245).

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

Offset: 0

Author

A.H.M. Smeets, Jul 27 2018

Keywords

• nonn
• cofr

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

• A.H.M. Smeets, Table of n, a(n) for n = 0..62

Crossrefs

Cf. A012245, A089013, A317627.

Cf. A058304 (in base 10), A317413 (in base 2), A317661 (in base 4).

Programs

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 *)

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)

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.