(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? ==AzTechList============================================================ AzTechList Discussion List http://www.aztechlist.org Subscribe by sending an email to: [EMAIL PROTECTED] Report list related problems/concerns: [EMAIL PROTECTED] Become a Member and/or Sponsor: http://paypal.azipa.org * Co-sourcing with Inflow is the future - http://www.inflow.com * PublicOpinion.com. Share with the world - http://www.publicopinion.com * Contactlink Keeps You Connected! - http://www.contactlink.com * $7.85 Domain Name Registrations/Transfers - http://www.domaindo.com ============================================================AzTechList== Yahoo! Groups Links <*> To visit your group on the web, go to: http://groups.yahoo.com/group/aztechlist/ <*> To unsubscribe from this group, send an email to: [EMAIL PROTECTED] <*> Your use of Yahoo! Groups is subject to: http://docs.yahoo.com/info/terms/ _______________________________________________ http://www.mccmedia.com/mailman/listinfo/brin-l