A178307 Remove powers of 2 from A000069. Let b(n) be n-th term of the remaining sequence. Then a(n) is the least number m such that (b(n))^m is in A001969.

3, 3, 2, 3, 3, 2, 3, 4, 2, 3, 3, 4, 2, 3, 3, 2, 3, 2, 4, 4, 2, 3, 3, 5, 2, 3, 4, 2, 4, 2, 2, 3, 4, 2, 3, 2, 3, 3, 3, 2, 2, 4, 4, 4, 9, 2, 2, 2, 3, 3, 2, 4, 5, 3, 2, 3, 3, 4, 2, 4, 2, 2, 4, 3, 3, 2, 2, 2, 3, 2, 4, 4, 2, 2, 3, 4, 2, 2, 4, 3, 3, 4, 5, 3, 5, 2, 2, 2, 6, 4, 4, 2, 4, 2, 2, 9, 2, 2, 2, 2, 3, 2, 3, 3, 3

Offset: 1

Author

Vladimir Shevelev, May 24 2010, May 25 2010

Keywords

• nonn

Comments

The sequence {b(n)} coincides with A075930. Conjecture. For every n>=1, a(n) does exist.

The sequence b(n) is A075930 (Positions of check bits in code in A075928); see comment in that sequence. [From Jeremy Gardiner, May 26 2010]

Crossrefs

Cf. A000069 A001969 A178253

Cf. A075930. [From Jeremy Gardiner, May 26 2010]

Formula

If k=b(n)=2^m*b(s), where b(s) is odd, then a(n)=a(s).

Extensions

Edited by N. J. A. Sloane, May 29 2010

Extended by Jeremy Gardiner, May 26 2010