A160016 -id:A160016 - cp's OEIS frontend

A338206 Inverse of permutation in A160016.

0, 2, 1, 6, 3, 10, 4, 14, 5, 18, 7, 22, 8, 26, 9, 30, 11, 34, 12, 38, 13, 42, 15, 46, 16, 50, 17, 54, 19, 58, 20, 62, 21, 66, 23, 70, 24, 74, 25, 78, 27, 82, 28, 86, 29, 90, 31, 94, 32, 98, 33, 102, 35, 106, 36, 110, 37, 114, 39, 118, 40, 122, 41, 126, 43, 130, 44, 134, 45, 138, 47, 142, 48, 146, 49, 150, 51, 154, 52, 158

Offset: 0

Georg Fischer, Oct 16 2020

Permutation of the nonnegative integers.

Index entries for sequences that are permutations of the natural numbers
Index entries for linear recurrences with constant coefficients, signature (0,1,0,0,0,1,0,-1).

Cf. A160016.

gf := (x*(1 + x^2)*(2 + x + 2*x^2 + x^3 + 2*x^4))/((-1 + x)^2*(1 + x)^2*(1 - x + x^2)*(1 + x + x^2)): ser := series(gf, x, 82):
seq(coeff(ser, x, n), n=0..79); # Peter Luschny, Oct 16 2020

LinearRecurrence[{0,1,0,0,0,1,0,-1}, {0,2,1,6,3,10,4,14},80]

my(x='x+O('x^80)); Vec((x*(2+x+4*x^2+2*x^3+4*x^4+x^5+2*x^6))/((1-x^2)^2*(1+x^2+x^4)))

Blocks of 6 numbers: a(6*k+0 .. 6*k+5) = (4*k+0, 12*k+2, 4*k+1, 12*k+6, 4*k+3, 12*k+10) for k >= 0.
O.g.f.: x*(1 + x^2)*(2 + x + 2*x^2 + x^3 + 2*x^4)/((1 - x^2)^2*(1 + x^2 + x^4)).
If n is odd, then a(n) = 2*n; otherwise, a(n) = nearest integer to 2*n/3. - Philippe Deléham, Nov 09 2023

A307048 Permutation of the positive integers derived from the terms of A322469 having the form 6*k - 2.

Original entry on oeis.org

2, 1, 6, 5, 10, 4, 14, 7, 18, 13, 22, 8, 26, 3, 30, 21, 34, 12, 38, 9, 42, 29, 46, 16, 50, 23, 54, 37, 58, 20, 62, 19, 66, 45, 70, 24, 74, 17, 78, 53, 82, 28, 86, 39, 90, 61, 94, 32, 98, 15, 102, 69, 106, 36, 110, 25, 114, 77, 118, 40, 122, 55

Offset: 1

Georg Fischer, Mar 21 2019

The sequence is the flattened form of an irregular table U(i, j) similar to table T(i, j) in A322469. U(i, j) = k is defined only for the elements T(i, j) which have the form 6*k - 2, so the table is sparsely filled.
Like in A322469, the columns in table U contain arithmetic progressions.
a(n) is a permutation of the positive integers, since A322469 is one, and since there is a one-to-one mapping between any a(n) = k and some A322469(m) = 6*k - 2.
There is a hierarchy of such permutations of the positive integers derived by mapping the terms of the form 6*k - 2 to k:
  Level 1: A322469
  Level 2: A307048 (this sequence)
  Level 3: A160016 = 2, 1, 4, 6, 8, 3, ... period of (3 even, 1 odd number)
  Level 4: A000027 = 1, 2, 3, 4 ... (the positive integers)
  Level 5: A000027

			Table U(i, j) begins:
   i\j   1  2  3  4  5  6  7
   -------------------------
   1:
   4:          2
   7:                   1
  10:
  13:          6
  16:                5
  19:
  22:         10
  25:             4
  28:
  31:         14
-----
T(4, 3) = 10 = 6*2 - 2, therefore U(4, 3) = 2.
T(7, 6) =  4 = 6*1 - 2, therefore U(7, 6) = 1.

Cf. A000027, A160016, A322469.

# Derived from A322469
use integer; my $n = 1; my $i = 1; my $an;
while ($i <= 1000) { # next row
  $an = 4 * $i - 1; &term();
  while ($an % 3 == 0) {
    $an /= 3; &term();
    $an *= 2; &term();
  } # while divisible by 3
  $i ++;
} # while next row
sub term {
  if (($an + 2) % 6 == 0) {
    my $bn = ($an + 2) / 6;
    print "$n $bn\n"; $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.

Original entry on oeis.org

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

A307407 Irregular table read by rows: rows list terms that map to the nodes in the graph of the "3x+1" (or Collatz) problem.

Original entry on oeis.org

16, 4, 5, 1, 10, 2, 3, 40, 12, 13, 64, 20, 21, 88, 28, 29, 9, 58, 112, 36, 37, 136, 44, 45, 160, 52, 53, 17, 106, 34, 35, 11, 70, 22, 23, 7, 46, 14, 15, 184, 60, 61, 208, 68, 69, 232, 76, 77, 25, 154, 50, 51, 256, 84, 85, 280, 92, 93

Offset: 1

Georg Fischer, Apr 14 2019

The construction is similar to that in A322469. The sequence is the flattened form of an irregular table S(i, j) (see the example below) which has rows i >= 1 consisting of subsequences of varying length.
Like Truemper (cf. link), we denote the mapping x -> 2*x by "m" ("multiply"), the mapping x -> (x - 1)/3 by "d" ("divide"), and the combined mapping "dm" x -> (x - 1)/3 * 2 by "s" ("squeeze").  The d mapping is defined only if it is possible, that is, if x - 1 is divisible by 3. We write m, d and s as infix operation words, for example "4 mms 10", and we use exponents for repeated operations, for example "mms^2 = mmss".
Row i in table S is constructed by the following algorithm: Start with 6 * i - 2 in column j = 1. The following columns j are defined in groups of four by the operations:
  k   j=4*k+2    j=4*k+3    j=4*k+4    j=4*k+5
  --------------------------------------------------
  0   mm         dmm        mmd        dmmd
  1   mms        dmms       mmsd       dmmsd
  2   mms^2      dmms^2     mms^2d     dmms^2d
  ...
  k   mms^k      dmms^k     mm(s^k)d   dmm(s^k)d
The construction for the row terminates at the first column where a d operation is no longer possible. This point is always reached. This can be proved by the observation that, for any row i in S, there is a unique mapping x -> (x + 2)/6 of the terms in column 1, 2, 5, 9, 13, ... 4*m+1 to the terms in row i of table T in A322469. The row construction process in A322469 stops, therefore it stops also in the sequence defined here.
Conjecture: The sequence is a permutation of the positive numbers.

			Table S(i, j) begins:
  i\j    1   2   3   4   5   6   7   8   9  10  11  12  13  14  15
  ----------------------------------------------------------------
  1:    16   4   5   1  10   2   3
  2:    40  12  13
  3:    64  20  21
  4:    88  28  29   9  58
  5:   112  36  37
  6:   136  44  45
  7:   160  52  53  17 106  34  35  11  70  22  23   7  46  14  15
  8:   184  60  61

Georg Fischer, Perl program for the generation of related sequences.
Manfred Trümper, The Collatz Problem in the Light of an Infinite Free Semigroup, Chinese Journal of Mathematics, Vol. 2014, Article ID 756917, 21 pages.

Cf. A160016 (level 3), A307048 (level 2), A322469 (level 1).

Perl
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cf. link.
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A338208 Inverse permutation to A322469.

Original entry on oeis.org

2, 3, 1, 16, 7, 14, 4, 17, 12, 8, 5, 15, 21, 33, 6, 126, 26, 13, 9, 69, 31, 50, 10, 124, 38, 22, 11, 34, 43, 67, 18, 127, 48, 27, 19, 122, 55, 86, 20, 70, 60, 32, 23, 51, 65, 103, 24, 125, 74, 39, 25, 179, 79, 120, 28, 287, 84, 44, 29, 68, 91, 143, 30, 1100, 96, 49, 35, 232, 101, 160, 36, 123, 108, 56, 37, 87, 113, 177, 40, 611

Offset: 1

Georg Fischer, Oct 16 2020

nonn

Permutation of the positive integers.
There is a hierarchy of such permutations derived by selecting and mapping the terms of the form 6*k - 2 to k:
  Level 0: A307407
  Level 1: A322469, inverse is A338208 (this sequence)
  Level 2: A307048             A338207
  Level 3: A160016             A338206
  Level 4: A000027  (the positive integers)

Cf. A160016, A307048, A307407, A322469, A338206, A338207.

A160079 Lodumo_3 of Fibonacci numbers.

Original entry on oeis.org

0, 1, 4, 2, 3, 5, 8, 7, 6, 10, 13, 11, 9, 14, 17, 16, 12, 19, 22, 20, 15, 23, 26, 25, 18, 28, 31, 29, 21, 32, 35, 34, 24, 37, 40, 38, 27, 41, 44, 43, 30, 46, 49, 47, 33, 50, 53, 52, 36, 55, 58, 56, 39, 59, 62, 61, 42, 64, 67, 65, 45, 68, 71, 70, 48, 73, 76, 74, 51, 77, 80, 79, 54

Offset: 0

Philippe Deléham, May 01 2009

Permutation of nonnegative integers.

OEIS wiki, Lodumo transform
Index entries for sequences that are permutations of the nonnegative integers
Index entries for linear recurrences with constant coefficients, signature (0,0,0,1,0,0,0,1,0,0,0,-1).

Cf. A000045, A082115, A160016, A160051, A367651 (inverse permutation).

a(n) = lod_3(A000045(n)).
a(n) = 2*a(n-8) - a(n-16) for n >= 16. - Philippe Deléham, Mar 09 2023
a(8*n) = 6*n, a(8*n+1) = 9*n+1, a(8*n+2) = 9*n+4, a(8*n+3) = 9*n+2, a(8*n+4) = 6*n+3, a(8*n+5) = 9*n+5, a(8*n+6) = 9*n+8, a(8*n+7) = 9*n+7. - Philippe Deléham, Nov 24 2023
From Philippe Deléham, Nov 29 2023 : (Start)
a(n) = a(n-4) + a(n-8) - a(n-12) for n >= 12.
G.f. : (x + 4*x^2 + 2*x^3 + 3*x^4 + 4*x^5 + 4*x^6 + 5*x^7 + 3*x^8 + 4*x^9 + x^10 + 2*x^11) / (1 - x^4 - x^8 + x^12). (End)

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A338206 Inverse of permutation in A160016.

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A307048 Permutation of the positive integers derived from the terms of A322469 having the form 6*k - 2.

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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.

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A307407 Irregular table read by rows: rows list terms that map to the nodes in the graph of the "3x+1" (or Collatz) problem.

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A338208 Inverse permutation to A322469.

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A160079 Lodumo_3 of Fibonacci numbers.

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