Counting

(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?
---------------------------------------------------
PLUG-discuss mailing list - PLUG-discuss@lists.plug.phoenix.az.us
To subscribe, unsubscribe, or to change you mail settings:
http://lists.PLUG.phoenix.az.us/mailman/listinfo/plug-discuss