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

http://www.primepuzzles.net/puzzles/puzz_278.htm

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

primes.

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

affirmative.

Result1:

Let n be an even number greater than 6. Then

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

Example.

If n = 34=2*17. So phi(34)=16 and w(34)=2 and Pi(34-2)=11.

Then 34=9+25 =25+9

J(34)=2;

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

variables.

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

Example.

N=2^9

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)

Example.

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

J[N]=6

S1 + 2*S2=34

4(9-1)-2(2)=28

and the formula above holds.

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

results.

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)).

p.s.

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!