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12816my elementary method and Perry's /Goldbach's conjectures

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  • antonioveloz2
    Jul 3, 2003
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      The following is an elementary way to attack Goldbach's conjecture,
      the Twin Primes conjecture and many other problems in prime number
      theory.

      Let Sn={1,2,...,n}. We will be interested in studying the subsets of
      the set Sn, their cardinalities, and their characteristic functions.
      The following subsets of Sn will be useful.


      Pn={p| p is an odd prime and 3<=p<= n-3}
      |Pn| = Pi(n-3)-1 where Pi(n) is the prime counting function. Let p
      (x) = {1 if x is in Pn , 0 otherwise That is p(x) is the
      characteristic function of the set Pn.

      Cn = {c| c is composite and 3<=c<=n-3}
      |Cn|= n - |Pn|. Let c(x) ={1 if x is in Cn, 0 otherwise That is c(x)
      is the characteristic function of the set Cn.

      Tn={t| t is relatively prime to n and 3<=t<=n-3}
      |Tn| = Phi(n)-2 where Phi(n) is Euler's totient function. Let t(x) =
      {1 if x is in Tn, 0 otherwise

      Def. Sentence & equivalence
      A sentence is a finite products and sums of characterisic functions
      of subsets of Sn. Two sentences are equivalent iff the have the same
      truth tables.

      Example 1.

      The following is a sentence with its value. Let n be an integer then
      c(x)*c(n-x)*t(x) = {1 if x and n-x are both composite and (n,x)=1 and
      0 otherwise

      Example 2.
      Let n be an integer then
      c(x) is equivalent to 1-p(x) for all x | 2<= x <=n

      Def. Sum over a sentence
      Let Q be a sentence then we associate with Q the value Q+ which is
      defined as the sum over all x in Sn {Q(x)}.

      Lemma 1
      If Q and W are two equivalent sentences then Q+=W+.
      Proof: Since Q(x) = W(x) for all x in Sn the result follows trivially.

      From this theory we have the following applications

      Goldbach's Problem

      Let Gn = { p | n-p and p are in Pn and (n,p)=1 } that is the set of
      all primes p such that n-p is prime and p is not equal to n-p.
      Let g(x) = {1 if x is in Gn, 0 otherwise

      Lemma 2
      g(x) is equivalent to p(x)p(n-x)t(x) for all x such that 3<=x<=n-3

      Proof:
      We may prove the lemma by considering the truth tables associated
      with the sentences g(x) and p(x)p(n-x)t(x).

      Let Hn={c | c and n-c are composite and (n,c)=1}
      Let h(x) = {1 if x is in Gn, 0 otherwise
      This function is associated with Perry's conjecture which is
      available at http://www.primepuzzles.net/conjectures/conj_033.htm
      We will prove this conjecture below. This lemma will be useful.

      Lemma 3
      h(x) is equivalent to c(x)c(n-x)t(x) for all x in Sn

      Proof:
      Follows immediately by considering truth tables.

      Theorem 1
      |Hn| = |Gn| + Phi(n) - 2Pi(n) + 2Omega(n) - 2 where Omega(n) is the
      number of prime divisors of n.

      Proof:
      h(x) is equivalent to c(x)c(n-x)t(x) for all x in Sn by lemma 3.
      c(x) is equivalent to 1-p(x)
      so c(x)c(n-x)t(x) is equivalent to
      (1-p(x))(1-p(n-x))t(x)
      which is equivalent to
      Q = t(x) - p(x) - p(n-x) +p(x)p(n-x)t(x)
      now we find Q+ by taking the sum over Q and we find
      Q+ = Sum(t(x) - p(x)t(x) - p(n-x)t(x) + p(x)p(n-x)t(x) )
      splitting up this sum and noting that
      Sum(t(x)) = Phi(n) - 2
      Sum(p(x)t(x)) = Sum(p(n-x)t(x)) = Pi(n-3)- Omega(n) for n even (if
      you want a proof notify me)
      and
      Sum(p(x)p(n-x)t(x)) = |Gn|
      the result now follows by adding these identities together.
      #

      For many examples of Theorem 1 see my previous post.

      Theorem 2 (Perry's Conjecture)

      Every number number greater than 210 can be written as the sum of two
      relatively prime composites.

      Proof: (sketch)

      We know by Theorem 1 that
      |Hn| >= Phi(n) - 2Pi(n)

      The result follows immediately from inequalities for Phi(n) - 2Pi(n)
      for n sufficiently large then verifying the remaining cases.
      #

      I would give a complete proof to Theorem 2 but I am away from the
      university at this moment and do not have access to the online
      journals. I will post a complete proof at a later time.

      Other Theorems and Future Research

      Let R(n) be the number of representations of a number as the sum of
      two primes then

      Theorem 3

      R(n) =Sum(over primes x<= n-3) Pi(n-x+1)-Pi(n-x-1)
      Proof:
      we may note that Pi(n-x+1)-Pi(n-x-1) is equivalent to p(n-x) and
      since x is prime the result follows.
      #

      Theorem 4

      Sum(even x | 6<=x<=n)R(x) = Sum(primes p <=n-3)(Pi(n-p+1)

      The proof is a bit long and I will omit it here. There is a simple
      proof base on another method which I will post at a later time.

      I used this method mentioned here to study the sentence p(x)p(x+2) in
      its equivalent forms and the method has shown some promise. A proof
      of Goldbach's conjecture also seems possible by this method but my
      limited time does not allow me to work with on the conjecture.

      I would also like to study Dirichlet products of characteristic
      functions and formal power series of such functions. The methods
      outlined in Tom Apostol's Introduction to analytic number theory for
      partial summation also seem to be helpful to establish some importan
      inequalities. I am not ready to post any results on this front
      though.

      Please notify me with any questions or comments.
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