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

1,1

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]

Links

Table of n, a(n) for n=1..105.

Formula

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

Crossrefs

Cf. A000069 A001969 A178253

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

Sequence in context: A060586 A076662 A164359 * A327465 A079063 A031352

Adjacent sequences: A178304 A178305 A178306 * A178308 A178309 A178310

Keyword

nonn

Author

Vladimir Shevelev, May 24 2010, May 25 2010

Extensions

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

Extended by Jeremy Gardiner, May 26 2010

Status

approved