Conjecture again but I said nothing about a partial proof. I attempt

to prove that any even number of the form n=2P0P1 Pk , Where Po, P1,

P2, ,Pk are the first k primes, can be written as the sum of two

primes. (Actually as all of you will soon see this can be generalized

quite easily to include even numbers of any form but if this proof is

incorrect then why bother to post the generalized form).

I begin by stating the definition of some sets that will be needed.

P_n = the set of all odd primes <= (n-3).

S_n = the set defined by taking n Pj where Pj runs through the set

P_n from the first to last element of P_n.

S'_n = the set of all odd numbers not in C_n but <= (n-3) excluding

the number 1.

C_n = the set of all odd composites less than n-3.

Notice that (C_n ^ S'_n) includes all of the equations of the form

n Ci = Cj. Where Ci and Cj are two not necessarily distinct

composite numbers. Where A ^ B is the intersection of the two sets A

and B. Notice also that

/S_n/ = /P_n/ = Pi(n-3) 1

because by the definition P_n includes the primes from 3 to n-3.

Where /A/ is the cardinality of the set A. Also notice

/S'_n/ = ((n-4)/2) - /S_n/

because any odd number not in S_n is in /S'_n/

Example:

N = 12

P_12 = {3,5,7}

C_12 = {9}

S_12 = {9,7,5}

S'_12 = {3}

/P_12 ^ S_12/ gives the number of primes such that n Pi = Pj where

Pi and Pj are primes. Notice that we can divide the lengths of these

sets by two to get the number of unique solutions. This fact can

also be written as

((/P_n/)/ 2) + ((/C_n/)/ 2) + ((/S_n/)/ 2)+ ((/S'_n/)/ 2) = ((n-

4)/2)/2 = ((n-4)/4)

We may now restate Goldbach's Conjecture in terms of these sets.

Goldbach's Conjecture can be stated as follows.

/P_n ^ S_n/ > 0 for n >= 4.

Consider the following Table.

Table 1.

N = 2*3*5 = P5#=30

Col.1 Col.2 Col.3 Col.4 Col.5

303=27 3015=15 307=23 3011=19 3013=17

30(3*3) = 21

30(3*5) = 15

Let us quickly analyze the table. Notice that this table includes

all of the unique ways n as the sum of two odd numbers between 3 and

n-3. Notice also that 30-15 = 15 appears once for every odd prime in

P#5. Also notice that for every odd prime Pi in P#5 N-(Pi*d), Where d

is an odd number, is divisible by Pi. This allows us to count the

elements in the first K columns; which allows us to give a lower

bound for /C_n ^ S'_n/. We state this in a lemma.

Lemma A

If n = P#k then we can give the following lower bound for /C_n ^

S'_n/.

(/C_n ^ S'_n/)/2 >= (N/4)[(1/P1)+(1/P2)+ +(1/Pk)] k^2 (k/2)

We need to count the elements in each column. For the 1st column we

have

(1) Col.1 = [N/(4*3)] 1.

(2)

We want the number of solutions N-(3*d) >= (n/2). We need to find

that unique number d1 such that N-(3*d1) = (n/2). Therefore d1 = n/

(2*3) but we only want the odd numbers less than this and we must

divide by two so we get (1). Finally We need to subtract 1 because N-

(3*1) which belongs to /P_n ^ S_n/, is included in this column. We

may repeat this argument for primes <= Pk and in fact for Pi we have:

(2)Col.i = [N/(4*Pi)] i - (1/2).

We subtract (i) because the ith prime has appeared as N-(pj*Pi) where

Pj is a prime less than Pi. We subtract (1/2) because for every

prime in P#k the trivial N (N/2) = (N/2) is included in that column.

To get the total number of equations in the k columns we must take

the sum

Col.1 + Col.2 + + Col.k

=

([N/(4*P1)] 1)+([N/(4*P2)] 2 (1/2))+ +([N/(4*Pk)] k -(1/2))

=

[N/(4*P1)] + [N/(4*P2)]+ + [N/(4*Pk)] (1+2+ +k) (k*(1/2))

=

(N/4)[(1/P1)+(1/P2)+ +(1/Pk)] k^2 - k (k/2)

=

(N/4)[(1/P1)+(1/P2)+ +(1/Pk)] k^2 (k/2)

but this only counts the composite numbers in the first k columns for

n sufficiently large there will exists Primes such that

N (Pm * dm) > (n/2) and is composite. This will not be in the

first k columns. An example is

n = 2*3*5*7*11 = P#11 = 2310

2310 13^2 = 2141 > (n/2) = 1155

The first k columns therefore hold the minimum number of equations

and therefore the lemma follows. #

We are now ready to prove the theorem.

Theorem 1

Any number n =P#k can be written as the sum of two primes.

Proof: Lets assume there exists a number g that contradicts theorem

1 then we know that

/C_g ^ S_g/ = /s/ = Pi(n-3)

where pi(n-3) is the prime counting function. Also notice that

/C_g ^ S_g/ + /C_g ^ S'_g/ = (n-4)/2

and

(/C_g ^ S_g/)/2 + (/C_g ^ S'_g/)/2 = (n-4)/4

(/C_g ^ S'_g/)/2 = (g-4)/4 - (/C_g ^ S_g/)/2

=(g-4)/4 (/s/)/2

=(g-4)/4 (Pi(g-3))/2

but by lemma A we know that

(/C_n ^ S'_n/)/2 >= (n/4)[(1/P1)+(1/P2)+ +(1/Pk)] k^2 (k/2)

Therefore

(g-4)/4 (Pi(g-3))/2 >= (g/4)[(1/P1)+(1/P2)+ +(1/Pk)] k^2 (k/2)

which implies that

(g/4) 1 + k^2 + (k/2) >=(g/4)[(1/P1)+(1/P2)+ +(1/Pk)] + (Pi(g-

3))/2

but this is impossible because

(g/4)[(1/P1)+(1/P2)+ +(1/Pk)] > (g/4)

and

(Pi(g-3))/2 > 1 + k^2 + (k/2)

because k is the index of the kth prime.

#

Any comments would be greatly appreciated. I know that this is a bit

long so I apologize for that. Thank you.