A268642 Seelmann's sequence: a(1) = 1; thereafter a(n + 1) = ceiling(a(n)/2) unless this is already in the sequence, in which case a(n + 1) = 3*a(n).

1, 3, 2, 6, 18, 9, 5, 15, 8, 4, 12, 36, 108, 54, 27, 14, 7, 21, 11, 33, 17, 51, 26, 13, 39, 20, 10, 30, 90, 45, 23, 69, 35, 105, 53, 159, 80, 40, 120, 60, 180, 540, 270, 135, 68, 34, 102, 306, 153, 77, 231, 116, 58, 29, 87, 44, 22, 66, 198, 99, 50, 25, 75, 38

Offset

1,2

Comments

It is conjectured that this is a permutation of the positive integers, along with any Seelmann sequence in which a(n+1) = M*a(n) if the divide by 2 rule cannot be applied, for any integer M>1 and not of the form M = 2^N. [Corrected by Charlie Neder, Feb 06 2019]

Reminiscent of the 3x+1 or Collatz problem, cf. A006577. - N. J. A. Sloane, Feb 09 2016

The Reddit link contains what is claimed to be a proof that this sequence is a permutation. I don't know if it has been checked. - N. J. A. Sloane, Feb 11 2016

Links

Peter Kagey, Table of n, a(n) for n = 1..10000

David Seelmann, Proving a sequence of integers reaches every integer, Posting to Reddit Web Site, Jan 09 2016

Mathematica

a[1]=1; a[n_] := a[n] = Module[{an1, an}, an1 = a[n-1]; an = If[EvenQ[an1], an1/2, (an1+1)/2]; If[FreeQ[Array[a, n-1], an], an, 3*a[n-1]]]; Array[a, 100] (* Jean-François Alcover, Feb 27 2016 *)
Fold[Append[#1, If[FreeQ[#1, #3], #3, 3 #1[[-1]]]] & @@ {#1, #2, Ceiling[#1[[-1]]/2]} &, {1}, Range@ 63] (* Michael De Vlieger, Jan 13 2018 *)

Crossrefs

Cf. A006577, A050000 (with floor instead of ceiling).

For records see A268529, A268530. For inverse see A268531.

Sequence in context: A007812 A276225 A082561 * A110768 A140230 A356563

Adjacent sequences: A268639 A268640 A268641 * A268643 A268644 A268645

Keyword

nonn,nice

Author

Peter Kagey, Feb 09 2016, based on a posting by David Seelmann to the Reddit web site.

Extensions

Title corrected by Charlie Neder, Feb 06 2019

Status

approved