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A322469 Irregular table: row i = 1, 2, 3, ... starts with 4*i - 1; then, as long as the number is divisible by 3, the next two terms are the result of dividing it by 3, then multiplying it by 2.

%I A322469 #42 Oct 16 2020 18:50:32
%S A322469 3,1,2,7,11,15,5,10,19,23,27,9,18,6,12,4,8,31,35,39,13,26,43,47,51,17,
%T A322469 34,55,59,63,21,42,14,28,67,71,75,25,50,79,83,87,29,58,91,95,99,33,66,
%U A322469 22,44,103,107,111,37,74
%N A322469 Irregular table: row i = 1, 2, 3, ... starts with 4*i - 1; then, as long as the number is divisible by 3, the next two terms are the result of dividing it by 3, then multiplying it by 2.
%C A322469 The sequence is the flattened form of an irregular table T(i, j) (see the example below) which has rows i >= 1 consisting of subsequences of varying length as defined by the following algorithm:
%C A322469     j := 1; T(i, j) := 4 * i - 1;
%C A322469     while T(i, j) is divisible by 3 do
%C A322469         T(i, j + 1) := T(i, j) / 3;
%C A322469         T(i, j + 2) := T(i, j + 1) * 2;
%C A322469         j := j + 2;
%C A322469     end while
%C A322469 The algorithm always stops.
%C A322469 The first rows which are longer than any previous row are 1, 7, 61, 547, 4921 ... (A066443).
%C A322469 Property: The sequence is a permutation of the natural numbers > 0.
%C A322469 Proof: (Start)
%C A322469 The values in the columns j of T for row indexes i of the form i = e * k + f,
%C A322469 k >= 0, if such columns are present, have the following residues modulo some power of 2:
%C A322469 j | Op.  | Form of i    |  T(i, j)     |  Residues  | Residues not yet covered
%C A322469 --+------+ -------------+--------------+------------+-------------------------
%C A322469 1 |      |  1 * k +  1  |   4 * k +  3 |   3 mod  4 |   0,  1,  2     mod  4
%C A322469 2 | / 3  |  3 * k +  1  |   4 * k +  1 |   1 mod  4 |   0,  2,  4,  6 mod  8
%C A322469 3 | * 2  |  3 * k +  1  |   8 * k +  2 |   2 mod  8 |   0,  4,  6     mod  8
%C A322469 4 | / 3  |  9 * k +  7  |   8 * k +  6 |   6 mod  8 |   0,  4,  8, 12 mod 16
%C A322469 5 | * 2  |  9 * k +  7  |  16 * k + 12 |  12 mod 16 |   0,  4,  8     mod 16
%C A322469 6 | / 3  | 27 * k +  7  |  16 * k +  4 |   4 mod 16 |   0,  8, 16, 24 mod 32
%C A322469 7 | * 2  | 27 * k +  7  |  32 * k +  8 |   8 mod 32 |   0, 16, 24     mod 32
%C A322469 8 | / 3  | 81 * k + 61  |  32 * k + 24 |  24 mod 32 |   0, 16, 32, 48 mod 64
%C A322469 9 | * 2  | 81 * k + 61  |  64 * k + 48 |  48 mod 64 |   0, 16, 32     mod 64
%C A322469 ..| ...  |  e * k +  f  |   g * k +  m |   m mod  g |   0, ...
%C A322469 The variables in the last, general line can be computed from the operations in the algorithm. They are the following:
%C A322469   e = 3^floor(j / 2)
%C A322469   f = A066443(floor(j / 4)) with A066443(n) = (3^(2*n+1)+1)/4
%C A322469   g = 2^floor((j + 3) / 2)
%C A322469   m = 2^floor((j - 1) / 4) * A084101(j + 1 mod 4) with A084101(0..3) = (1, 3, 3, 1)
%C A322469 The residues m in each column and therefore the T(i, j) are all disjoint. For numbers which contain a sufficiently high power of 3, the length of the rows in T grows beyond any limit, and the numbers containing any power of 2 will finally be covered.
%C A322469 (End)
%C A322469 All numbers > 0 up to and including 2^(2*j + 1) appear in the rows in T up to and including A066443(j). For example, 4096 and 8192 are the trailing elements in row 398581 = A066443(6).
%C A322469 Length of row n = 1, 2, ... is 1+2*A007949(A004767(n-1)). - _M. F. Hasler_, Dec 10 2018
%C A322469 From _Georg Fischer_, Oct 16 2020: (Start)
%C A322469 Whenever a row of T is longer than any previous rows, it defines the start values of the arithmetic progressions in the additional columns. These start values form the sequence A308709.
%C A322469 There is a hierarchy of such permutations of the positive integers derived by selecting and mapping the terms of the form 6*k - 2 to k:
%C A322469   Level 0: A307407, nodes in the graph of the "3x+1" or Collatz problem
%C A322469   Level 1: A322469 (this sequence), inverse is A338208
%C A322469   Level 2: A307048, inverse is A338207
%C A322469   Level 3: A160016, inverse is A338206
%C A322469   Level >= 4: A000027, the positive integers
%C A322469 Conjectures (verified for k = 0..11):
%C A322469 a(A338186(k)) = 4^k.
%C A322469 If A338186(k) <= j < A338186(k+1) then a(A338186(k)) <= a(j).
%C A322469 (End)
%H A322469 Alois P. Heinz, <a href="/A322469/b322469.txt">Rows n = 1..10000, flattened</a>
%H A322469 <a href="/index/Per#IntegerPermutation">Index entries for sequences that are permutations of the positive integers</a>
%e A322469 Table T(i, j) begins:
%e A322469   i\j   1  2  3  4  5  6  7
%e A322469   -------------------------
%e A322469   1:    3  1  2
%e A322469   2:    7
%e A322469   3:   11
%e A322469   4:   15  5 10
%e A322469   5:   19
%e A322469   6:   23
%e A322469   7:   27  9 18  6 12  4  8
%p A322469 T:= proc(n) local m, l; m:= 4*n-1; l:= m;
%p A322469       while irem(m, 3, 'm')=0 do
%p A322469          l:= l, m; m:= m*2; l:=l, m;
%p A322469       od; l
%p A322469     end:
%p A322469 seq(T(n), n=1..40);  # _Alois P. Heinz_, Dec 10 2018
%t A322469 s={}; Do[a=4n-1; AppendTo[s,a]; While[Divisible[a, 3], a/=3; AppendTo[s, a]; a*=2; AppendTo[s, a]], {n, 1, 30}]; s (* _Amiram Eldar_, Dec 10 2018 *)
%o A322469 (PARI) apply( A322469_row(n,L=[n=4*n+3])={while(n%3==0,L=concat(L,[n\=3, n*=2]));L}, [0..99]) \\ Use concat(%) to flatten the table if desired. - _M. F. Hasler_, Dec 10 2018
%o A322469 (Perl) use integer; my $n = 1; my $i = 1;
%o A322469   while ($i <= 1000) { # next row
%o A322469     my $an = 4 * $i - 1; print "$n $an\n"; $n ++;
%o A322469     while ($an % 3 == 0) {
%o A322469       $an /= 3; print "$n $an\n"; $n ++;
%o A322469       $an *= 2; print "$n $an\n"; $n ++;
%o A322469     } # while divisible by 3
%o A322469     $i ++;
%o A322469 } # while next row - _Georg Fischer_, Dec 12 2018
%o A322469 (Sage)
%o A322469 def A322469_list(len):
%o A322469     L = []
%o A322469     for n in (1..len):
%o A322469         a = 4*n - 1
%o A322469         L.append(a)
%o A322469         while 3.divides(a):
%o A322469             a //= 3
%o A322469             L.append(a)
%o A322469             a <<= 1
%o A322469             L.append(a)
%o A322469     return L
%o A322469 A322469_list(28) # _Peter Luschny_, Dec 10 2018
%Y A322469 Cf. A066443, A084101, A160016 (level 3), A307048 (level 2), A307407 (level 0), A308709, A338186, A338206, A338207, A338208.
%K A322469 nonn,tabf,easy
%O A322469 1,1
%A A322469 _Georg Fischer_, Dec 09 2018