A338206 -id:A338206 - cp's OEIS frontend

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.

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, 34, 55, 59, 63, 21, 42, 14, 28, 67, 71, 75, 25, 50, 79, 83, 87, 29, 58, 91, 95, 99, 33, 66, 22, 44, 103, 107, 111, 37, 74

Offset: 1

Georg Fischer, Dec 09 2018

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:
    j := 1; T(i, j) := 4 * i - 1;
    while T(i, j) is divisible by 3 do
        T(i, j + 1) := T(i, j) / 3;
        T(i, j + 2) := T(i, j + 1) * 2;
        j := j + 2;
    end while
The algorithm always stops.
The first rows which are longer than any previous row are 1, 7, 61, 547, 4921 ... (A066443).
Property: The sequence is a permutation of the natural numbers > 0.
Proof: (Start)
The values in the columns j of T for row indexes i of the form i = e * k + f,
k >= 0, if such columns are present, have the following residues modulo some power of 2:
j | Op.  | Form of i    |  T(i, j)     |  Residues  | Residues not yet covered
--+------+ -------------+--------------+------------+-------------------------
1 |      |  1 * k +  1  |   4 * k +  3 |   3 mod  4 |   0,  1,  2     mod  4
2 | / 3  |  3 * k +  1  |   4 * k +  1 |   1 mod  4 |   0,  2,  4,  6 mod  8
3 | * 2  |  3 * k +  1  |   8 * k +  2 |   2 mod  8 |   0,  4,  6     mod  8
4 | / 3  |  9 * k +  7  |   8 * k +  6 |   6 mod  8 |   0,  4,  8, 12 mod 16
5 | * 2  |  9 * k +  7  |  16 * k + 12 |  12 mod 16 |   0,  4,  8     mod 16
6 | / 3  | 27 * k +  7  |  16 * k +  4 |   4 mod 16 |   0,  8, 16, 24 mod 32
7 | * 2  | 27 * k +  7  |  32 * k +  8 |   8 mod 32 |   0, 16, 24     mod 32
8 | / 3  | 81 * k + 61  |  32 * k + 24 |  24 mod 32 |   0, 16, 32, 48 mod 64
9 | * 2  | 81 * k + 61  |  64 * k + 48 |  48 mod 64 |   0, 16, 32     mod 64
..| ...  |  e * k +  f  |   g * k +  m |   m mod  g |   0, ...
The variables in the last, general line can be computed from the operations in the algorithm. They are the following:
  e = 3^floor(j / 2)
  f = A066443(floor(j / 4)) with A066443(n) = (3^(2*n+1)+1)/4
  g = 2^floor((j + 3) / 2)
  m = 2^floor((j - 1) / 4) * A084101(j + 1 mod 4) with A084101(0..3) = (1, 3, 3, 1)
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.
(End)
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).
Length of row n = 1, 2, ... is 1+2*A007949(A004767(n-1)). - M. F. Hasler, Dec 10 2018
From Georg Fischer, Oct 16 2020: (Start)
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.
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:
  Level 0: A307407, nodes in the graph of the "3x+1" or Collatz problem
  Level 1: A322469 (this sequence), inverse is A338208
  Level 2: A307048, inverse is A338207
  Level 3: A160016, inverse is A338206
  Level >= 4: A000027, the positive integers
Conjectures (verified for k = 0..11):
a(A338186(k)) = 4^k.
If A338186(k) <= j < A338186(k+1) then a(A338186(k)) <= a(j).
(End)

			Table T(i, j) begins:
  i\j   1  2  3  4  5  6  7
  -------------------------
  1:    3  1  2
  2:    7
  3:   11
  4:   15  5 10
  5:   19
  6:   23
  7:   27  9 18  6 12  4  8

Alois P. Heinz, Rows n = 1..10000, flattened
Index entries for sequences that are permutations of the positive integers

Cf. A066443, A084101, A160016 (level 3), A307048 (level 2), A307407 (level 0), A308709, A338186, A338206, A338207, A338208.

T:= proc(n) local m, l; m:= 4*n-1; l:= m;
      while irem(m, 3, 'm')=0 do
         l:= l, m; m:= m*2; l:=l, m;
      od; l
    end:
seq(T(n), n=1..40);  # Alois P. Heinz, Dec 10 2018

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 *)

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

use integer; my $n = 1; my $i = 1;
  while ($i <= 1000) { # next row
    my $an = 4 * $i - 1; print "$n $an\n"; $n ++;
    while ($an % 3 == 0) {
      $an /= 3; print "$n $an\n"; $n ++;
      $an *= 2; print "$n $an\n"; $n ++;
    } # while divisible by 3
    $i ++;
} # while next row - Georg Fischer, Dec 12 2018

def A322469_list(len):
    L = []
    for n in (1..len):
        a = 4*n - 1
        L.append(a)
        while 3.divides(a):
            a //= 3
            L.append(a)
            a <<= 1
            L.append(a)
    return L
A322469_list(28) # Peter Luschny, Dec 10 2018

cp's OEIS Frontend

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.

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A338208 Inverse permutation to A322469.

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