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Re: RE To summarize the question

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  • Mark Underwood
    Hi Jose To clarify and answer your questions, yes we do need a solution for every possible mod configuration and set of n values. (as you suspect). Why do
    Message 1 of 2 , Apr 23, 2005
      Hi Jose

      To clarify and answer your questions, yes we do need a solution for
      every possible mod configuration and set of "n" values. (as you
      suspect).

      Why do two mods have to be blocked off for every prime? Actually it
      is a maximum of two mods blocked off. I don't know how this plays a
      part in the larger GB proof.

      So we are looking for a proof that there is always one R (gcd(p#,R)=1
      and R<p^2 ) and a k such that the integer portion of k*p#/R produces
      at least one "n" value from any one given set of n's.

      Mark



      --- In primenumbers@yahoogroups.com, Jose Ramón Brox <ambroxius@t...>
      wrote:
      > Let's see If I have understood the definition (I have some problems
      with "Chris chains"):
      >
      > 1) We select a prime p. We list in rows all the odd primes from 3
      to p.
      >
      > 2) We select two random mods for every prime and we put a 1 on
      their columns. For example,
      > if we take p = 5 and we "block off" 1,2 (mod 3) and 0,2 (mod 5)
      then we get:
      >
      > 3 011011011011
      > 5 101001010010
      >
      > Q1: If this is correct, then I have one question: do we need an
      affirmative answer for
      > Chris question for EVERY random selection of two mods, or it is
      enough to find ANY random
      > selection that satisfies his question? (I think we need it for
      everyone).
      >
      > Q2: Why two mods? Why random?
      >
      > 3) We search the columns where there are only zeros and we call
      them numbers n+1. So the
      > numbers n are those columns minus one.
      >
      > 4) We define R<p^2 to be a number with GCD(R,p#) = 1. We want to
      confirm that R exists so
      > that the integer part of a multiple of p# / R is in the set of
      numbers n.
      >
      > Q3: Do we really need one R or every R to fulfill this condition?
      (I think we need one).
      >
      > --------------------------------------------------------------------
      -----
      >
      > Is that it? I don't want to waste time in a wrong question (time is
      gold for me!). If we
      > could use any random selection of the patterns, we could find one
      that fits some useful
      > conditions and answer in the affirmative (for example, take always
      0,1 (mod p_i) ).
      >
      > Jose Brox
      >
      > ----- Original Message -----
      > From: "Mark Underwood" <mark.underwood@s...>
      > To: <primenumbers@yahoogroups.com>
      > Sent: Saturday, April 23, 2005 2:28 PM
      > Subject: [PrimeNumbers] To summarize the question
      >
      >
      >
      >
      >
      > Let's focus on the mathematical aspects of this problem.
      >
      > Let p# = 2*3*5*7*...*p
      >
      > We know that if we block off any two mods for each of the odd primes
      > there will be exactly (3-2)*(5-2)*(7-2)*(11-2)*...*(p-2) values
      > within
      > the p# length cycle which do not coincide with any blocks. These are
      > Chris' "zero columns". We take those values and subtract 1 from
      each
      > of them to get what Chris called our "n" numbers.
      >
      > Here's the question: Can we select a number t which is relatively
      > prime
      > to p# and less than p^2, such that the integer (truncated) portion
      of
      > the series of numbers
      >
      > k*p#/t for k=1,2,3....
      >
      > is coincident at least once with at least one of the "n" numbers?
      >
      > I'm sure the answer is in the affirmative but I'm still struggling
      > for a proof. (He's been patiently waiting on me to provide one in
      > private correspondence but I haven't produced!) If any one can prove
      > this, or if a respected theorist can say that the validity of this
      > proposition can be assumed, Chris has promised the list he will post
      > his attempted Goldbach proof.
      >
      > Any takers?
      >
      >
      >
      >
      >
      >
      >
      >
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      > The Prime Pages : http://www.primepages.org/
      >
      >
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