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Prime Numbers Equation

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  • Sebastian Martin
    I have obtained a short equation for the prime numbers. For all n   (Prime[Prime[n]]-1)*Product[Prime[Floor[(Prime[n]-1)/i]]-1,{i,1,Prime[n]-1}]=
    Message 1 of 7 , Jul 19, 2008
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      I have obtained a short equation for the prime numbers. For all n
       
      (Prime[Prime[n]]-1)*Product[Prime[Floor[(Prime[n]-1)/i]]-1,{i,1,Prime[n]-1}]=
      (Prime[Prime[n]-1]-1)*Product[Prime[Floor[Prime[n]/i]]-1,{i,1,Prime[n]-1}]
       
       
      attached  MSWORD documment
       
      Sincerely:
       
      Sebastián Martín Ruiz


      ______________________________________________
      Enviado desde Correo Yahoo! La bandeja de entrada más inteligente.

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    • Sebastian Martin
      Let x a integer number 1. x is a prime number if and only if it is a solution of this equation:
      Message 2 of 7 , Nov 8, 2008
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        Let x a integer number>1.

        x is a prime number if and only if it is a solution of this equation:

        Sum[Floor[x/i],{i,1,Sqrt[x]}]=1+Sum[Floor[(x-1)/i],{i,1,Sqrt[x]}]


        Sincerely

        Sebastian Martin Ruiz




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      • Phil Carmody
        ... That s, in its shorter form, is the oldest one in the book! You ve attempted to obfuscate it by separating it into two sums, but we can see through that.
        Message 3 of 7 , Nov 8, 2008
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          --- On Sat, 11/8/08, Sebastian Martin <sebi_sebi@...> wrote:
          > Let x a integer number>1.
          >
          > x is a prime number if and only if it is a solution of this
          > equation:
          >
          > Sum[Floor[x/i],{i,1,Sqrt[x]}]=1+Sum[Floor[(x-1)/i],{i,1,Sqrt[x]}]

          That's, in its shorter form, is the oldest one in the book!

          You've attempted to obfuscate it by separating it into two sums, but we can see through that.

          Sum[Floor[x/i]-Floor[(x-1)/i],{i,1,Sqrt[x]}] = 1

          The difference of floors is either 0 or 1, and only 1 if i divides x.
          So the left hand side is just a count of the divisors between 1 and sqrt(x) inclusive. Only for primes can that be 1.

          I'm sure you've posted this to the list in the past, but am too lazy to check.

          Phil
        • Sebastian Martin Ruiz
          Hello all: n positive integer n=/=1 and 4   n is prime  if and only if 8*Product[(k*n-1)/(Floor[n/k]+1), {k,1,n}]=Floor[8*Product[(k*n-1)/(Floor[n/k]+1),
          Message 4 of 7 , Mar 30, 2009
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            Hello all:
            n positive integer n=/=1 and 4  
            n is prime 
            if and only if
            8*Product[(k*n-1)/(Floor[n/k]+1), {k,1,n}]=Floor[8*Product[(k*n-1)/(Floor[n/k]+1), {k,1,n}]]

            Sincerely

            Sebastian Martin Ruiz






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          • Sebastian Martin Ruiz
            Hello all:   I send you a equation whit primes numbers solutions.   Let x an integer x =5 and pi=i-th prime number.   If x is solution of the equation:   
            Message 5 of 7 , May 12, 2009
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              Hello all:
               
              I send you a equation whit primes numbers solutions.
               
              Let x an integer x>=5 and pi=i-th prime number.
               
              If x is solution of the equation: 
               
              x = Product[Numerator[(x+pi)/(x-pi)]+Denominator[(x+pi)/(x-pi)],
              {i=2 to Pi[x]-1}]^(1/(Pi[x]-2))
                                                                  Pi[x] is the prime counting function.
              Then x IS PRIME; x=p(Pi[x]) 
               
               
              Note that i in the primes of the product is to Pi[x]-1
              and x is the next prime with i=Pi[x], then we can obtain a recurrent algorithm
              to obtain the prime numbers.
               
              Mathematica Software:
               
              F[x_]:=Product[
                    Numerator[(x+Prime[i])/(x-Prime[i])]+
                      Denominator[(x+Prime[i])/(x-Prime[i])],
              {i,2,PrimePi[x]-1}]^(1/(PrimePi[x]-2))
               
              Do[If[IntegerQ[F[x]],Print[F[x]," ",x]],{x,5,40}]
              5   5
              4   6
              7   7
              16   8
              11   11
              13   13
              32   16
              17   17
              19   19
              23   23
              29   29
              31   31
              64   32
              37   37
               
              Sincerely
               
              Sebastián Martín Ruiz
               
              PD:  2x=F[x] has solutions powers of 2.




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            • Sebastian Martin Ruiz
              Hello:   If Goldbach Conjecture is True then:   For all p prime number p =7 exists q and r also primes such that:   p**2+4(q+r)**2=4p(q+r)+1   (It is easy
              Message 6 of 7 , Jun 12, 2009
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                Hello:
                 
                If Goldbach Conjecture is True then:
                 
                For all p prime number p>=7 exists q and r also primes such that:
                 
                p**2+4(q+r)**2=4p(q+r)+1
                 
                (It is easy to prove)
                 
                Sincerely
                 
                Sebastian Martin Ruiz
                 
                 




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              • Ali Adams
                Dear Sebastain, Did you mean by  p**2+4(q+r)* *2=4p(q+r) +1 this                    p^2 + 4(q+r)^2 = 4p(q+r) + 1    ??? if so, maybe
                Message 7 of 7 , Jun 12, 2009
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                  Dear Sebastain,

                  Did you mean by  p**2+4(q+r)* *2=4p(q+r) +1
                  this                    p^2 + 4(q+r)^2 = 4p(q+r) + 1    ???

                  if so, maybe you need to clarify that:

                  r or q can equal p
                  or
                  1 is prime

                  see below:

                  11^2 + 4(q+r)^2 - 44(q+r) + 1 = 0

                  let w = q+r

                  121 + 4w^2 - 44w - 1 = 0

                  120 + 4w^2 - 44w = 0

                  divide both sides by 4

                  30 + w^2 - 11w = 0

                  as a quadratic equation:
                  w^2 - 11w + 30 = 0
                  a = 1
                  b = -11
                  c = 30

                  w = [-b +- sqrt(b^2 - 4ac)] / 2a
                  w = [11 +- sqrt(-11^2 - 4*1*30)] / 2*1
                  w = [11 +- sqrt(121 - 120)] / 2
                  w = [11 +- sqrt(1)] / 2
                  w = [11 +- 1] / 2

                  w1 = 12 / 2 = 5
                  w2 = 10 / 2 = 10

                  so

                  either:
                  w1 = q1 + r1
                  12 = 1 + 11    or    5 + 7
                  but but q != p and r != p
                  and 1 is not 1 since 1899
                  so this option is out

                  or:
                  w2 = q2 + r2
                  10 = 3 + 7   or   5 + 5
                  but q != p and r != p
                  and r != q
                  so this fails too unless you calrify that they can :)

                  Sorry if misunderstood ** and * * oparators.

                  Ali
                  <prime numbers are God's signature>
                   




                  ________________________________
                  From: Sebastian Martin Ruiz <s_m_ruiz@...>
                  To: Claudi Alsina <claudio.alsina@...>; Azmy Ariff <azmyarif@...>; Chris Caldwell <caldwell@...>; Pierre Deligne <deligne@...>; Gaussianos <gaussianos@...>; Andrew Granville <andrew@...>; Lista NMBRTHRY <nmbrthry@...>; Antonio Pérez Sanz <aperez4@...>; primenumbers@yahoogroups.com; Carlos Rivera <crivera@...>; CarlosB Rivera <cbrfgm@...>
                  Sent: Saturday, June 13, 2009 12:05:30 AM
                  Subject: [PrimeNumbers] Prime Numbers Equation





                  Hello:
                   
                  If Goldbach Conjecture is True then:
                   
                  For all p prime number p>=7 exists q and r also primes such that:
                   
                  p**2+4(q+r)* *2=4p(q+r) +1
                   
                  (It is easy to prove)
                   
                  Sincerely
                   
                  Sebastian Martin Ruiz
                   
                   

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