A050132 a(n) = floor(a(n-1)/2) if this is not among 0,a(1),...,a(n-1); otherwise a(n) = 3*n.

1, 6, 3, 12, 15, 7, 21, 10, 5, 2, 33, 16, 8, 4, 45, 22, 11, 54, 27, 13, 63, 31, 69, 34, 17, 78, 39, 19, 9, 90, 93, 46, 23, 102, 51, 25, 111, 55, 117, 58, 29, 14, 129, 64, 32, 138, 141, 70, 35, 150, 75, 37, 18, 162, 81, 40, 20, 174, 87, 43

Offset: 1

Author

Clark Kimberling

Keywords

• nonn

Comments

This sequence is a permutation of the natural numbers. Sketch of proof: that it is one-to-one is trivial. Inductively, the halving operation can never happen more than 4 times in a row. There are at least 5 multiples of 3 amongst 16m .. 16m+15; by the induction, one of these will be a value a(n) = 3n and then 4 halving operations will get m (if it has not previously appeared). It follows that m will occur in the sequence no later than floor((16m+26)/3). Empirically, it appears that the 26 in this formula could be replaced by 21. The first occurrence of 4 consecutive halvings starts at n = 226, winding up with a(230)=42. - Franklin T. Adams-Watters, Mar 10 2006

Crossrefs

Cf. A050000.