A058304 Continued fraction for Liouville's number (A012245).

0, 9, 11, 99, 1, 10, 9, 999999999999, 1, 8, 10, 1, 99, 11, 9, 999999999999999999999999999999999999999999999999999999999999999999999999, 1, 8, 11, 99, 1, 10, 8, 1, 999999999999, 9, 10, 1, 99, 11, 9

Offset

0,2

Comments

From A.H.M. Smeets, Jun 06 2018: (Start)

Except for the first term, the only values that occur in this sequence are 1,8,9,10,11,and values 10^((m-1)*m!)-1 for m > 1. The probability of occurrence P(a(n) = k) are given by:

P(a(n) = 1) = 1/4,

P(a(n) = 8) = 1/8,

P(a(n) = 9) = 1/8,

P(a(n) = 10) = 1/8,

P(a(n) = 11) = 1/8 and

P(a(n) = 10^((m-1)*m!)-1) = 2^-(m+1) for m > 1. (End)

References

Harold M. Stark, "An Introduction to Number Theory," The MIT Press, Cambridge, MA and London, England, Eighth Printing, 1994, pages 172 - 177.

Links

Muniru A Asiru, Table of n, a(n) for n = 0..62

J. O. Shallit, Simple Continued Fractions for Some Irrational Numbers II, J. Number Theory 14 (1982), 228-231.

Eric Weisstein's World of Mathematics, Liouville's Constant

G. Xiao, Contfrac

Index entries for continued fractions for constants

Formula

From A.H.M. Smeets, Jun 26 2018: (Start)

a(n) = 1 iff n in A317331,

a(n) = 8 iff n in A317332,

a(n) = 9 iff n in A317333,

a(n) = 10 iff n = 8*m - 6 + 3*(m mod 2) for m > 0,

a(n) = 11 iff n = 8*m - 3 - 3*(m mod 2) for m > 0,

a(n) = 10^((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. (End)

Example

0.1100010000000000000000010... = 0 + 1/(9 + 1/(11 + 1/(99 + 1/(1 + ...)))). - Harry J. Smith, May 15 2009

Maple

with(numtheory): cfrac(add(1/10^factorial(n), n=1..7), 62, 'quotients'); # Muniru A Asiru, Aug 08 2018

Mathematica

ContinuedFraction[ Sum[ 1 /10^(n!), {n, 1, 7} ], 40 ]

Prog

(PARI) { allocatemem(932245000); default(realprecision, 200000); x=contfrac(suminf(n=1, 1.0/10^n!)); for (n=1, 255, write("b058304.txt", n, " ", x[n])); } \\ Harry J. Smith, May 15 2009
(Python)
n, f, i, p, q, base = 1, 1, 0, 0, 1, 10
while i < 1000:
    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)
# A.H.M. Smeets, Aug 03 2018

Crossrefs

Cf. A012245.

Cf. A317413 (in base 2), A317414 (in base 3) A317661 (in base 4 and general).

Sequence in context: A137016 A053886 A137020 * A359457 A365221 A027727

Adjacent sequences: A058301 A058302 A058303 * A058305 A058306 A058307

Keyword

cofr,nonn

Author

Robert G. Wilson v, Dec 08 2000

Extensions

Offset changed to 0 on the advice of A.H.M. Smeets by Muniru A Asiru, Aug 11 2018

Status

approved