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

Author

I. V. Serov, Apr 24 2019

Keywords

• nonn
• base

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.

Examples

			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
  ...
		

Links

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

Crossrefs

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

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.