A359838 Continued fraction for binary expansion of A359456 interpreted in base 2.

0, 1, 3, 3, 1, 2, 1, 262143, 3, 1, 3, 3, 1, 1532495540865888858358347027150309183618739122183602175, 4, 3, 1, 3, 262143, 1, 2, 1, 3, 3, 1

Offset: 0

Author

A.H.M. Smeets, Jan 14 2023

Keywords

• nonn
• base
• cofr

Comments

The continued fraction of the number obtained by reading A359456 as a binary fraction.

Except for the first term, the only values that occur in this sequence are 1, 2, 3, 4 and values 2^A359458(m) - 1 for m > 2. The probabilities of occurrence P(a(n) = k) are given by:

P(a(n) = 1) = 1/3,

P(a(n) = 2) = 1/12,

P(a(n) = 3) = 1/3,

P(a(n) = 4) = 1/12 and

P(a(n) = 2^A359458(m)-1) = 1/(3*2^m) for m > 1.

Crossrefs

Cf. A014710, A317538, A317539, A359456, A359458.

Cf. A359457 (in base 10).

Formula

a(n) = 1 if and only if n in A317538.

a(n) = 2 if and only if n in {24*m - 19 | m > 0} union {24*m - 4 | m > 0}.

a(n) = 3 if and only if n in A317539.

a(n) = 4 if and only if n in {12*m - 3*A014710(m-1) + 5 | m > 0}

a(n) = 2^A359458(m)-1 if and only if n in {3*2^(m-1)*(1+k*4) + 1 | k >= 0} union {3*2^(m-1)*(3+k*4) | k >= 0} for m > 1.