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An interesting formula

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  • antonioveloz2
    Hello all, I wish to share with you a couple of formulas that I have been working with for some time. Let J(n) = the number of ways of writing n as the sum
    Message 1 of 1 , Mar 3, 2008
      Hello all,

      I wish to share with you a couple of formulas that I have
      been "working" with for some time.

      Let J(n) = the number of ways of writing n as the sum of two
      relatively prime composites. In all that follows we count a + b and
      b + a as two different representations.

      Puzzle 278 at the prime puzzles website
      asks if J(n) > 0 for n sufficiently large.

      Let Pi(n) be the number of primes less than n, w(n) be the number of
      prime factors of n and phi(n) be Euler's Totient function. Let R(n)
      be the number of ways of writing n as the sum of two distinct

      We have the following simple result that answers puzzle 278 in the

      Let n be an even number greater than 6. Then

      J(n) = phi(n) – 2Pi(n-2) +2 w(n) + R(n) - 2


      If n = 34=2*17. So phi(34)=16 and w(34)=2 and Pi(34-2)=11.
      Then 34=9+25 =25+9
      34= 3+31=5+29=11+23=31+3=29+5=23+11
      So R(34)=6;

      Now J(34)= 2 = 16 – 2*11 +2 *2 + 6 – 2 = phi(34) – 2Pi(34-2) +2 w(34)
      + R(34)-2 as the formula predicts.

      If n = 328 you may check that phi(318)=104 and w(318)=3 and Pi(318-2)
      =65 and also J(318)=8 and R(318)=30.

      And J(318) = 8
      = 104 – 2*65 +2*3 + 30 - 2
      = phi(318) – 2Pi(318-2) +2 w(318) + R(318) – 2

      as we expected.

      Result 1 would lead us another way to attack Goldbach's Conjecture
      since we would only need to show that J(n) > phi(n) – 2Pi(n-2) +2 w
      (n) – 2 since then R(n) > 0. This led me to the second result, which
      is an alternate formula for J(n).

      Again some quick notation…

      Let Mu(n) be the mobius function and {x} be the fractional Part
      function. Let a* be the inverse of a modulo b (In the statement of
      the result this will cause no confusion.). Let r[a] be the function
      that is 1 when a is prime and 0 otherwise.

      Then we have the following result.

      Result 2:
      Let n=2^r where r >= 3. Define the following function of two
      n[a,b] = (n – 2*a*r[a]– 2*b*r[b] – a – b)/(2)

      Let SUM[..] be the double sum over a,b such that n[a,b] is
      nonnegative and both a,b are relatively prime to n.

      Let S1 = SUM[(Mu[a]*Mu[b]*((n[a,b]/ab) + 1))]

      Let S2 = SUM[(Mu[a]*Mu[b]*({n[a,b]a*/b}]

      Then J(n) = S1+2*S2



      S1= - 688116791584/19841545815
      S2= 443266124867/19841545815
      J[N]= 10 = (- 688116791584/19841545815) + 2(443266124867/19841545815)

      Result 3

      Let N=(2^r)(p^s) Where p is a prime then J[N] = S1+2*S2 - 2^(r-1)(p^
      (s-1)-1) - 2*(s-1)


      Let N=(2^3)(3^3)

      S1 + 2*S2=34
      and the formula above holds.

      There are more general formulas for J[N] than in the previous

      The last results allow a method of attacking Goldbach's Problem.
      While it appears that S1 and S2 have lots of terms most of the terms
      are 0. There are also ways of simplifying the sums by using symmetry
      to reduce the number of terms. The sum S1 can be estimated fairly
      well but S2 is more dificult to estimate. What we would like to do is
      estimate S1 and S2 with an error term less than O(N/log^2(N)).


      I hope someone might be able to use these ideas, since I will not be
      working on Goldbach's Conjecture anymore. Goldbach's problem is so
      difficult that it is addictive and has taken up too much of time. I
      wish someone had warned me!
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