Goldbach Conjecture: Relationship to Even Numbers

[Paul Mackenzie]

I found the following relationship concerning goldbach's conjecture; viz that every even number is the sum of two primes.



If goldbach's conjecture is true then the following must hold for all 2N


\sum^{2N-1}_{l=0} ( \sum^{p < 2N-1}_{ p odd primes=3} cos (2πpl/2N) ])² > \sum^{2N-1}_{l=0} ( \sum^{p < 2N-1}_{ p odd primes=3} sin (2πpl/2N) ])²


Alternatively if goldbach's conjecture is false for some 2N then the following must hold true for that 2N


\sum^{2N-1}_{l=0} ( \sum^{p < 2N-1}_{ p odd primes=3} cos (2πpl/2N) ])² = \sum^{2N-1}_{l=0} ( \sum^{p < 2N-1}_{ p odd primes=3} sin (2πpl/2N) ])²


Unfortunately without some further knowledge concerning the distribution of primes I don't
think you can take this further.

Regards

 

[Paul Mackenzie]

I show below how I arrived at the relationships I mentioned in my previous post. My Apologies for not doing it before.

Firstly, I considered the properties in the Fourier domain of the following function.

f(x) = 1 if x is prime
f(x) = 0 if x is otherwise
for all real x.

In particular I perform a 2N-point DFT on the real values f(0), f(1), f(2), f(3),...,f(2N-1) to obtain the complex values F(0),F(1),F(2),...F(2N-1)


where F[l] = \sum^{p <= 2N-1}_{p odd prime = 3} e^{-j2πpl/2N} for l = 0, 1,2,3,...2N-1

Next I consider the sum

\sum^{l=2N-1}_{l=0}F^{2}[l]

That is F^{2}[l] = (e^{-j2π3l/2N} + e^{-j2π5l/2N} + e^{-j2π7l/2N} + e^{-j2π11l/2N} + ... + e^{-j2πp_{k}l/2N})(e^{-j2π3l/2N} + e^{-j2π5l/2N} + e^{-j2π7l/2N} + e^{-j2π11l/2N} + ... + e^{-j2πp_{k}l/2N})

where p_{k} is the last prime before 2N


Now multiplying out and re-arranging the last equation we get

F^{2}[l] = g(6)e^{-j2π6l/2N} + g(8)e^{-j2π8l/2N} + g(10)e^{-j2π10l/2N} + ... + g(2N)e^{-j2π2Nl/2N} + g(2N + 2)e^{-j2π(2N+2)l/2N} + k(2N+4)e^{-j2π(2N+4)l/2N} + ... + k(2p_{k})e^{-j2π(2p_{k})l/2N}

where the coefficient g(2i) is the number of goldbach partitions for the even number 2i. For instance when we multiply e^{-j2π3l/2N} and e^{-j2π7l/2N} together and vice versa and when we multiply e^{-j2π5l/2N} with itself we obtain g(10)e^{-j2π10l/2N} where g(10) is three. The coefiicient k(2i) may be typically less than the number of goldbach partitions for the equivalent even number 2i. This is because the goldbach partitions for an even number equal to or greater than 2N+4 may include prime numbers greater than 2N, whereas our sum is limited to primes less than 2N.

Returning to the sum \sum^{l=2N-1}_{l=0}F^{2}[l] we thus find that

\sum^{l=2N-1}_{l=0}F^{2}[l] = \sum^{l=2N-1}_{l=0}g(6)e^{-j2π6l/2N} + \sum^{l=2N-1}_{l=0}g(8)e^{-j2π8l/2N} + \sum^{l=2N-1}_{l=0}g(10)e^{-j2π10l/2N} + ... + \sum^{l=2N-1}_{l=0}g(2N)e^{-j2π2Nl/2N} + \sum^{l=2N-1}_{l=0}g(2N + 2)e^{-j2π(2N+2)l/2N} + \sum^{l=2N-1}_{l=0}k(2N+4)e^{-j2π(2N+4)l/2N} + ... + \sum^{l=2N-1}_{l=0}k(2p_{k})e^{-j2π(2p_{k})l/2N}

Each of these sums on the RHS is zero with the exception of \sum^{l=2N-1}_{l=0}g(2N)e^{-j2π2Nl/2N} which is equal to 2N.g(2N)

Thus we obtain g(2N) = \frac{1}{2N} \sum^{l=2N-1}_{l=0}F^{2}[l]

Thus we expressed the number of goldbach partitions in terms of the Fourier transform of the prime function f(x).

We now express the Fourier transform F[l] in terms of its real and imaginery parts.

where F[l] = \sum^{p <= 2N-1}_{p odd prime = 3} cos(2π pl/2N) + jsin(2π pl/2N)

Thus F²[l] = (\sum^{p <= 2N-1}_{p odd prime = 3} cos(2π pl/2N))^{2} - (\sum^{p <= 2N-1}_{p odd prime = 3} sin(2π pl/2N))^{2} + 2j\sum^{p <= 2N-1}_{p odd prime = 3} cos(2π pl/2N).\sum^{p <= 2N-1}_{p odd prime = 3} sin(2π pl/2N)


Thus we have g(2N) = \frac{1}{2N}\sum^{l=2N-1}_{l=0}F^{2}[l] = \frac{1}{2N} \sum^{l=2N-1}_{l=0}(\sum^{p <= 2N-1}_{p odd prime = 3} cos(2π pl/2N))^{2} - \sum^{l=2N-1}_{l=0}(\sum^{p <= 2N-1}_{p odd prime = 3} sin(2π pl/2N))^{2} +2j \sum^{l=2N-1}_{l=0}(\sum^{p <= 2N-1}_{p odd prime = 3} cos(2π pl/2N).\sum^{p <= 2N-1}_{p odd prime = 3} sin(2π pl/2N))

As the prime number sequence f[x] is in effect a real signal then it is well known that
F[-l]= F*[l], as the signal is a prior periodic then F[2N-l]= F*[l] or F[l] = F*[2N-l], then as a result the last mentioned sum of the imaginary components goes to zero

Thus if goldbach's conjecture is true then g(2N) > 0 for all 2N and thus according to the last equation the following must hold true for all 2N for goldbach's conjecture to be true

\sum^{l=2N-1}_{l=0}(\sum^{p <= 2N-1}_{p odd prime = 3} cos(2π pl/2N))^{2} > \sum^{l=2N-1}_{l=0}(\sum^{p <= 2N-1}_{p odd prime = 3} sin(2π pl/2N))^{2}


Alternatively for goldbach conjecture to be false for some 2N then the following must hold true for that 2N

\sum^{l=2N-1}_{l=0}(\sum^{p <= 2N-1}_{p odd prime = 3} cos(2π pl/2N))^{2} = \sum^{l=2N-1}_{l=0}(\sum^{p <= 2N-1}_{p odd prime = 3} sin(2π pl/2N))^{2}



Regards

 

[PAllen]

I haven't looked at your analysis in detail, but on the point of relationship between Goldbach and prime distribution, this is well known. There are a number of results relating Riemann Hypothesis or generalization thereof to Goldbach; and, of course, Riemann is closely related to prime distribution.

 

[Paul Mackenzie]

I have been able to take the aforementioned post one step forward.

As mentioned earlier I was able to show that the number of goldbach partitions g(2N) for some even number 2N was equal to

g(2N) = \frac{1}{2N} \sum^{l=2N-1}_{l=0}F²[l]

where F[l] is the Fourier transform of the prime function f(x)

where
f(x) = 1 if is x is prime
f(x) = 0 if x is otherwise for all real x

where F[l] = \sum^{p<=2N-1}_{p odd primes = 3}e^-j2πpl/2N for l = 0, 1, 2, ...2N-1

Now Multiplying and re-arranging the sum F²[l] we obtain [See previous post]

\sum^{l=2N-1}_{l=0}F²[l] = \sum^{l=2N-1}_{l=0}g(6)e^−j2π6l/2N + g(8)e^−j2π8l/2N + g(10)e^{−j2π10l/2N} + ... + g(2N)e^{−j2π2Nl/2N} + g(2N + 2)e^{−j2π(2N+2)l/2N} + k(2N+4)^{e−j2π(2N+4)l/2N} + ... + k(2pk)e−^{j2π(2p_k)l/2N}

where p_k is the last prime before 2N


The coefficient g(2i) is equivalent to the number of goldbach partitions for the even number 2i. As can be seen the coefficient k(2i) is also an integer. However, the coefiicient k(2i) may be typically less than the number of goldbach partitions for the equivalent even number 2i. It may even be zero. This is because the goldbach partitions for an even number equal to or greater than 2N+4 may include prime numbers greater than 2N, whereas our sum is limited to primes less than 2N. The coefficient k(2i) are in this sense equivalent to the number of incomplete goldbach partitions for the number 2i. As will be seen k(2i) is also dependent on the value of 2N.


The aforementioned sum can be re-written as

\sum^{l=2N-1}_{l=0}F²[l] = \sum^{l=2N-1}_{l=0} g(2)e^−j2π2l/2N + g(4)e^−j2π4l/2N + g(6)e^−j2π6l/2N + g(8)e^−j2π8l/2N + g(10)e^{−j2π10l/2N} + ... + g(2N)e^{−j2π2Nl/2N} + g(2N + 2)e^{−j2π(2N+2)l/2N} + k(2N+4)^{e−j2π(2N+4)l/2N} + ... + k(4N-4)e^{−j2π(4N-4)l/2N} + k(4N-2)e^{−j2π(4N-2)l/2N}

As will be appreciated the first two components are zero as g(2) and g(4) are zero. Also if any component of the components k(4N-2) etc are greater than 2pk then of course they will be zero. Thus the equation is equivalent to the aforementioned sum.

As mentioned in earlier post all of the components on the RHS of the sum sum to zero with the exception of g(2N)e^{−j2π2Nl/2N} which sums to 2N.g(2N) thus leading to the equation

g(2N) = \frac{1}{2N}\sum^{l=2N-1}_{l=0}F²[l]

It will be noted that the reason these sums sum to zero is that they are power series of the form 1 + z + z² + ...+z^2N-1 where z = e^-j2π.2k/2N where 2k ≠ 2N

Thus if we multiply the LHS and RHS of this sum by e^-j2π2/2N all these components will go to zero with the exception NOW of g(2N-2) and k(4N-2). Thus it follows

g(2N-2) + k(4N-2) = \frac{1}{2N}\sum^{l=2N-1}_{l=0}F²[l]e^-j2π2/2N

It should be noted k(4N-2) =1 when 2N-1 is a prime otherwise it is zero.

We can keep multiplying to obtain

g(2N-4) + k(4N-4) = \frac{1}{2N}\sum^{l=2N-1}_{l=0}F²[l]e^-j2π4/2N
g(2N-8) + k(4N-8) = \frac{1}{2N}\sum^{l=2N-1}_{l=0}F²[l]e^-j2π8/2N
.
.
.
.
g(8) + k(2N+8) = \frac{1}{2N}\sum^{l=2N-1}_{l=0}F²[l]e^{-j2π(2N-8)/2N}
g(6) + k(2N+6) = \frac{1}{2N}\sum^{l=2N-1}_{l=0}F²[l]e^{-j2π(2N-6)/2N}
g(4) + k(2N+4) = \frac{1}{2N}\sum^{l=2N-1}_{l=0}F²[l]e^{-j2π(2N-4)/2N} Of course in this case g(4)=0
g(2) + k(2N+2) = \frac{1}{2N}\sum^{l=2N-1}_{l=0}F²[l]e^{-j2π(2N-2)/2N} Of course in this case g(2)=0. And also it can be noted k(2N+2) = g(2N+2)


Lastly the sum of g(2) through to g(2N) and k(2N+2) through to k(4N-2) is a trivial result and is equal to π²[2N], where π[2N] is the number of primes less than 2N. This is consisent with the fact the number of all combinations of primes less than 2N is equal to π²[2N]

In particular we have shown for an even number 2N the goldbach partions g(2N-2),g(2N), and g(2N+2) are

g(2N) = \frac{1}{2N}\sum^{l=2N-1}_{l=0}F²[l]
g(2N+2) = \frac{1}{2N}\sum^{l=2N-1}_{l=0}F²[l]e^{-j2π(2N-2)/2N} = \frac{1}{2N}\sum^{l=2N-1}_{l=0}F²[l]e^j2π2/2N
g(2N-2) = \frac{1}{2N}\sum^{l=2N-1}_{l=0}F²[l]e^-j2π2/2N - k(4N-2) where k(4N-2) is 1 if 2N-1 is prime otherwise it is zero

Simplifying we get.

g(2N) = \frac{1}{2N}\sum^{l=2N-1}_{l=0}F²[l]
g(2N+2) = \frac{1}{2N}\sum^{l=2N-1}_{l=0}F²[l]e^j2π2/2N
g(2N-2) = \frac{1}{2N}\sum^{l=2N-1}_{l=0}F²[l]e^-j2π2/2N - k(4N-2) where k(4N-2) is 1 if 2N-1 is prime otherwise it is zero




Regards

 

[cosmik debris]

I found the following relationship concerning goldbach's conjecture; viz that every even number is the sum of two primes.
Regards

What about 2?

 

[Paul Mackenzie]

I just noticed I made a typo in my previous post of Jun3-12. The statement

"Thus if we multiply the LHS and RHS of this sum by e ^{-j2π2/2N} all these components will go to zero with the exception NOW of g(2N-2) and k(4N-2)"

contains a typo. It should read

"Thus if we multiply the LHS and RHS of this sum by e ^{-j2π2l/2N} all these components will go to zero with the exception NOW of g(2N-2) and k(4N-2)".

This frequency variable l is also missing from the shift components in the subsequent lines.

My Apologies

Regards

 

[marco39]

Relating to the strong Goldbach's Conjecture please see the following paper submitted to journal:

http://arxiv.org/abs/1208.2244

Is anybody there having comments about the approach followed in the proof?

 

[Paul Mackenzie]

I am able to take this matter a further step forward on this topic.
The whole aim of this exercise was to find relationships between the number of goldbach partitions g[2N-2], g[2N], and g[2N+2] for arbitrary consecutive even numbers 2N-2,2N, and 2N+2. and THEN somehow argue it is not possible that the number of goldbach partitions g(2N),g(2N-2), and g(2N+2) for any arbitrary even numbers 2N-2,2N,2N+2 can ALL be zero. I have been able to do the first part but the second part still very much eludes me. Perhaps someone else would like to try.

FIRST PART
As shown in my earlier posts I have found a relationship between the number of goldbach partitions and the Fourier transform of a prime number sequence. In particular I established that


g(2N) = \frac{1}{2N}\sum^{l=2N-1}_{l=0}F^{2}[l] Eqn(1)

g(2N+2) = \frac{1}{2N}\sum^{l=2N-1}_{l=0}F^{2}[l]e^{j2πl2/2N} Eqn(2)

g(2N-2) = - \nabla+\frac{1}{2N}\sum^{l=2N-1}_{l=0}F^{2}[l]e^{-j2πl2/2N} Eqn (3)

where F[l] is the Fourier transform of the prime function f(x)

where F[l] = \sum^{2N}_{p odd primes =3}f(x)e^{-j2pπl/2N} for l = 0, 1, 2, ...2N-1 Eqn(4)
and where
f(x) = 1 if x is an odd prime
f(x) = 0 if x is otherwise for all x = 0,1,2...2N-1 Eqn(5)

and where \nabla =1 if 2N-1 is prime

Now Eqn (1) can be expressed in the form

g[2N] = \frac{1}{2N}\sum^{l=2N-1}_{l=0}(Re(F[l]))^{2} -(Im(F[l]))^{2}+2jRe(F[l]).Im(F[l]) Eqn(6)

where Re(F[l]) is the real component of the lth harmonic of the Fourier transform F[l] of the prime function f(x) and Im(F[l] is the imaginary component of the lth

harmonic of the Fourier transform F[l] of the prime functiom f(x). For simplicities sake we let a_{l} = Re(F[l]) and b_{l} = Im(F[l]) so Eqn(6)

can be expressed as

g[2N]=\frac{1}{2N}\sum^{l=2N-1}_{l=0}a^{2}_{l} -b^{2}_{l} +2ja_{l}b_{l} Eqn(7)

As the prime number sequence f(x) is a real sequence it is well known that F[-l]=F^{*}[l] and since the prime number sequence f(x) is in effect a prior

periodic then F[2N-l]=F^{*}[l] and F^{*}[2N-l]=F[l] and consequently it follows that

\sum^{l=2N-1}_{l=0}2ja_{l}b_{l} = 0 Eqn(8)

Thus Eqn (7) becomes

g[2N] = \frac{1}{2N}\sum^{l=2N-1}_{l=0}a^{2}_{l}-b^{2}_{l} Eqn (9)

Now substituting well known trigometric identities in Eqns 2 and 3 and re-arranging we obtain [n.b. due to the symmetric nature of this Fourier transform the imaginary
components of eqns 2 and 3 go to zero] the following relationships connecting the number of goldbach partitions g(2N), g(2N-2), and g(2N+2)

g[2N] = \frac{1}{2N}\sum^{l=2N-1}_{l=0}(a^{2}_{l}-b^{2}_{l}) Eqn (10)

g[2N+2]= g[2N] - \frac{1}{2N}\sum^{l=2N-1}_{l=0}2(a^{2}_{l}-b^{2}_{l})sin^{2}(2πl/2N) - \frac{1}{2N}\sum^{l=2N-1}_{l=0}4a_{l}b_{l}sin(2πl/2N)cos(2πl/2N) Eqn (11)

g[2N-2]= g[2N] -\nabla - \frac{1}{2N}\sum^{l=2N-1}_{l=0}2(a^{2}_{l}-b^{2}_{l})sin^{2}<br />(2πl/2N) + \frac{1}{2N}\sum^{l=2N-1}_{l=0}4a_{l}b_{l}sin(2πl/2N)cos(2πl/2N) Eqn (12)

AN EXAMPLE

Firstly I have determined the number of goldbach partitions g(278), g(280), and g(282)in the normal manner by brute force, by counting all those possible partitions of

primes (p_{i},p_{j}) that satisfy the equation p_{i}+p_{j}=2N. Note those goldbach partitions (p_{i},p_{j}) where p_{i} ≠ p_{j} are counted twice whereas that goldbach partition (p_{i},p_{i}) where p_{i}+p_{i}=2N is counted only once.

I thus get the following values

g[278]=13
g[280]=28
g[282]=32


I now compute g[2N] via the Fourier transform route. In this particular example ∇=0 as 2N-1=279 is not prime.

Thus

g[278]=g[280]-5.5-9.5=28-15=13 (derived from Eqn 12)
g[280]=28 (derived from Eqn 10)
g[282]=g[280]-5.5+9.5=28+4=32 (derived from Eqn 11)

It should be appreciated that computing the number of goldbach partitions by the Fourier transform route is extremely computational intensive.

SECOND PART

It is straight forward that there are an infinite number of even numbers where goldbach's conjecture holds. For example 3+5,3+7,...3+p,...
The issue arises are there any other even numbers between these where goldbach's conjecture is false?. This issue arises in particular when
the gap between two consecutive primes is large.The idea is to show that if g(2N-2) and g(2N) are non-zero then from equations 10 to 12 g(2N+2)
is non-zero ad infinitum. Or show it is not possible that g(2N-2,g(2N), and g(2N+2) can all be zero. I am not sure but I doubt it can be done without using some other relationships concerning primes. Maybe some one else can take it from here?

SOME OTHER USEFUL RELATIONSHIPS
Using parseval'a theorem it can be shown that

π(2N) = \sum^{x=0}_{x=2N-1}f^{2}(x) = \frac{1}{2N}\sum^{l=0}_{l=2N-1}F[l]F^{*}[l] = \frac{1}{2N}\sum^{l=0}_{l=2N-1}a^{2}_{l} + b^{2}_{l} Eqn 13

and as from Eqn 9

g[2N] = \frac{1}{2N}\sum^{l=2N-1}_{l=0}a^{2}_{l}-b^{2}_{l}

combining Eqn 13 and Eqn 9 we get

π(2N) + g(2N) = \frac{2}{2N}\sum^{l=2N-1}_{l=0}a^{2}_{l}

and

π(2N) - g(2N) = \frac{2}{2N}\sum^{l=2N-1}_{l=0}b^{2}_{l}

Assuming that goldbach conjecture is false for some 2N then

π(2N)=\frac{2}{2N}\sum^{l=2N-1}_{l=0}a^{2}_{l}=\frac{2}{2N}\sum^{l=2N-1}_{l=0}b^{2}_{l}

It should also be noted that the components a_{l} and b_{l} are not trival [all zero] as we have a_{0} = π(2N) and a_{N}=-π(2N) etc.