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%I #51 Feb 07 2019 04:24:15
%S 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,
%T 20,10,30,90,45,23,69,35,105,53,159,80,40,120,60,180,540,270,135,68,
%U 34,102,306,153,77,231,116,58,29,87,44,22,66,198,99,50,25,75,38
%N 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).
%C 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]
%C Reminiscent of the 3x+1 or Collatz problem, cf. A006577. - _N. J. A. Sloane_, Feb 09 2016
%C 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
%H Peter Kagey, <a href="/A268642/b268642.txt">Table of n, a(n) for n = 1..10000</a>
%H David Seelmann, <a href="https://www.reddit.com/r/math/comments/44vvkb/proving_a_sequence_of_integers_reaches_every/">Proving a sequence of integers reaches every integer</a>, Posting to Reddit Web Site, Jan 09 2016
%t 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 *)
%t 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 *)
%Y Cf. A006577, A050000 (with floor instead of ceiling).
%Y For records see A268529, A268530. For inverse see A268531.
%K nonn,nice
%O 1,2
%A _Peter Kagey_, Feb 09 2016, based on a posting by _David Seelmann_ to the Reddit web site.
%E Title corrected by _Charlie Neder_, Feb 06 2019
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