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Permutations with two variables:

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Permutation table A for x=2 binary variables has 2^x=2^2=4 permutations, ie 00,01,10,11.

Table A: binary table for two variables

n	variable1	variable2
1	0	0
2	0	1
3	1	0
4	1	1

Proportions:

For the range 1 to 4 for n:
00 occurs 1/4 of the time
01 occurs 1/4 of the time
10 occurs 1/4 of the time
11 occurs 1/4 of the time

These proportions sum to 1, corresponding to all natural numbers.

1/4+1/4+1/4+1/4=1. 

Sequences:

The n values where these permutations occur give the four sequences: (only first 7 values shown)

00: n: {1, 5, 9, 13, 17, 21, 25}
01: n: {2, 6, 10, 14, 18, 22, 26}
10: n: {3, 7, 11, 15, 19, 23, 27}
11: n: {4, 8, 12, 16, 20, 24, 28}

The difference sequences of the above:
00: Differences[n]: {4, 4, 4, 4, 4, 4}
01: Differences[n]: {4, 4, 4, 4, 4, 4}
10: Differences[n]: {4, 4, 4, 4, 4, 4}
11: Differences[n]: {4, 4, 4, 4, 4, 4}

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Permutation table B for x=2 prime divisors has maximum 2^x=2^2=4 distinct permutations, ie
for two prime divisors (2 and 3) the divisibility pattern 00,10,01,10,00,11 gives the 4 distinct permutations in order of first appearance in Table B: 00,10,01,11.
Ie. in Table B, for n=3, 3 is divisible by 3 and not 2, so the value of the two columns are 0,1 for row n=3.

Table B: Prime Factor table for 2 prime factors

n	prime2	prime3
1	0	0
2	1	0
3	0	1
4	1	0
5	0	0
6	1	1

Proportions:

For the range 1 to 6 for n:
00 occurs 1/3 of the time, so 1/3 of numbers don't have either prime factor 2 or 3.
10 occurs 1/3 of the time, so 1/3 of numbers have prime factor 2 and not 3.
01 occurs 1/6 of the time, so 1/6 of numbers have prime factor 3 and not 2.
11 occurs 1/6 of the time.  So 1/6 of numbers have both prime factors 2 and 3.

These proportions sum to 1, corresponding to all natural numbers.

1/3+1/3+1/6+1/6=1. 

Sequences:

The n values where these permutations occur give the four sequences: (only first 20 values shown)

00: n: {1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 55, 59}
10: n: {2, 4, 8, 10, 14, 16, 20, 22, 26, 28, 32, 34, 38, 40, 44, 46, 50, 52, 56, 58}
01: n: {3, 9, 15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99, 105, 111, 117}
11: n: {6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96, 102, 108, 114, 120}

First three values in each sequence: {{1, 5, 7}, {2, 4, 8}, {3, 9, 15}, {6, 12, 18}}

Difference Sequences:

The difference sequences of the above:

00: Differences1[n]: {4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4}
10: Differences1[n]: {2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2}
01: Differences1[n]: {6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6}
11: Differences1[n]: {6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6}

The count of how many difference sequences are required to give a sequence of all the same absolute value numbers:
Ie two difference sequences are required for 00:
00: n: {1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 55, 59}
00: Differences1[n]: {4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4}
00: Abs[Differences2[n]]: {2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2}
 
00: 2
10: 2
01: 1
11: 1

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