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

Author

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

Keywords

• nonn
• nice

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

Crossrefs

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

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

Programs

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 *)

Extensions

Title corrected by Charlie Neder, Feb 06 2019