A307723 Naturally ordered prime factorization of n as a quasi-logarithmic word over the binary alphabet {1,0}.

10, 1100, 1010, 110100, 101100, 11011000, 101010, 11001100, 10110100, 1101101000, 10110010, 1101100100, 1011011000, 1100110100, 10101010, 1101010100, 1011001100, 110110011000, 1010110100, 110011011000

Offset

2,1

Comments

Let m(n) be the number of digits (letters) in a(n).

m(n) = 2*A064097(n) = 2*(A073933(n)-1).

Split the word a(n) into two parts of equal length. The number of 1's in the left part equals the number of 0's in the right part and vice versa.

Links

I. V. Serov, Table of n, a(n) for n = 2..10000

Formula

a(1) is empty.

a(n) = concatenation(1, a(n-1), 0) if n is prime.

a(n) = concatenation_{k=1..A001222(n)} a(A307746(n,k)) if n is composite.

a(n) = concatenation(a(n/A088387(n)), a(A088387(n))) if n is composite.

Example

The sequence begins:
   n a(n)
  -- -----------
   1
   2 10
   3 1100
   4 1010
   5 110100
   6 101100
   7 11011000
   8 101010
   9 11001100
  10 10110100
  11 1101101000
  12 10110010
  ...

Crossrefs

Cf. A010051, A088387, A307641, A307746, A064097, A073933.

Sequence in context: A069886 A133383 A268229 * A071672 A290155 A265849

Adjacent sequences: A307720 A307721 A307722 * A307724 A307725 A307726

Keyword

nonn,base

Author

I. V. Serov, Apr 24 2019

Status

approved