(Replies to plug-discuss preferred)
<best viewed monospaced>
0th item: 0 places: 0 parity combinations,
2^0 possible
nil:nil: 1 :1
1th item: 2 places: 01
10
2 parity combinations,
sum of each row is 1,
2^2 possible
1:1: 2 :4
2th item: 4 places: 0011
0101
0110
1001
1010
1100
6 parity combinations,
sum of each row is 2
2^4 possible (16)
4: 6 :8:16
3th item: 6 places: 000111
001011
001101
001011
001110
010011
010101
010110
011001
011010
011100
100011
100101
100110
101001
101010
101100
110001
110010
110100
111000
20 parity combinations (if my spreadsheet is right)
sum of each row is 3
2^6 possible
16: 20 :32:64
4th item: 8 places: 00001111
00010111
.
.
11101000
11110000
70 parity combinations (if spreadsheet right)
sum of each row is 4
2^8 possible
64: 70 :128:256
========================
This is beyond my initially weak, and now rusted, skill at counting and
combinatorics.
NOTE each binary number (or representation of state) has an equal number of
'zeroes' and 'ones'.
Because of the parity of zeroes and ones we are only interested in binary
numbers with an even number of places.
The idea that the sum of symbols for each number equals the rank of the item
in the sequence is not proven, but is evident.
---------------
Question:
For a given item n in the sequence N, how many parity number combinations will
occur in the resulting set of all modulo 2^(2n) binary representations? (How
many parity combinations for a given item N?)
Is number of binary parities bounded underneath by 2^(2[n-1]) for all N?
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