Derivation of a prime verification formula to prove the related open problems

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Mathematical Analysis and Applications
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Title: Derivation of a Prime verification formula to prove the related
open problems
Article Type: Regular Article
Section/Category: Miscellaneous
Keywords: Goldbach's conjecture; prime; twin prime conjecture; Polignac;
Carl Pomerance; Mersenne prime; Euler; Euclid; divisor function; aliquot;
perfect number; helix; odd and even harmonic divisor; ore number; complex
plane
Corresponding Author: Mr. Chris De Corte,
Corresponding Author's Institution: KAIZY BVBA
First Author: Chris De Corte
Order of Authors: Chris De Corte
Abstract: In this document, we will develop a new formula to calculate
prime numbers and use it to discuss open problems like Goldbach, Polignac
and Twin prime conjectures, perfect numbers, the existence of odd
harmonic divisors, ...

Derivation of a Prime verification formula
to prove the related open problems
By: Chris De Corte
E-mail: chrisdecorte@yahoo.com
Cover Letter

Abstract
In this document, we will develop a new formula to calculate prime numbers and
use it to discuss open problems like Goldbach, Polignac and Twin prime
conjectures, perfect numbers, the existence of odd harmonic divisors, ...
Key-words
Goldbach’s conjecture; prime; twin prime conjecture; Polignac; Carl Pomerance;
Mersenne prime; Euler; Euclid; divisor function; aliquot; perfect number; helix;
odd and even harmonic divisor; ore number; complex plane.
Introduction
The following document originated during our study of primes.
We tried to create a schematic representation of the mechanism that lead to the
creation of primes. Following, we tried to mathematically describe this graphical
representation. It resulted in a new formula to calculate if a number is prime or
not.
We guessed that our new formula could be useful in understanding the
mechanism of primes and possibly in proving some open questions.
After primes, we read about divisibility and open problems related to that subject
and found that we were on the right track to discuss these problems as well.
After deriving our formula, we will discuss about:
- Goldbach’s conjecture: “Every even integer greater than 2 can be
expressed as the sum of two primes.”
- Polignac’s conjecture: “For any positive even number n, there are infinitely
many prime gaps of size n. In other words: There are infinitely many cases
of two consecutive prime numbers with difference n.”
- Twin Prime conjecture: “for every natural number k, there are infinitely
many prime pairs p and p′ such that p′ − p = 2k. The case k = 1 is the twin
prime conjecture.”
- Perfect numbers: “a perfect number is a positive integer that is equal to
the sum of its proper positive divisors excluding the number itself (also
known as its aliquot sum).”
- Harmonic divisor numbers: “Ore conjectured that no odd harmonic divisor
numbers exist other than 1.”
Methods & Techniques
Step 1: development of a new formula to check if a number is prime or not
*Manuscript

We will refer to the following picture as a graphical aid to derive our formula:
Let’s draw an X&Y axis with on:
- X-axis: the natural numbers n of which we want to calculate if they are prime or
not
- Y-axis: the divisors i of the corresponding natural number n on X
Suppose, we want to draw the line that shows us all the numbers that can be
divided by 1. In that case, we draw a line that starts at (0,0) and that goes
through (1,1). We will mark all the valid divisors on the line by a full dot. Since
every natural number can be divided by 1, we can mark the following dots on line
1: (1,1), (2,2), …(n,n), ... So, all possible dots are marked on that line.
divided by 2. In that case, we draw a line that starts at (0,0) and that goes
through (2,1). Only the even numbers will be divisible by 2. Hence we can mark
the following dots on line 2 in full: (2,1), (4,2), …, (n,n/2), …
divided by i (ϵ N). In that case, we draw a line that starts at (0,0) and that goes
through (i,1).
The full dots that represent the divisors by i will again be on equal distant and
can be marked on line i as: (i,1), (2.i,2), … (n.i,n), …
This line will have the following formula for line i:

After we draw a few lines and highlight the divisors with full dots, we can easily
see that the prime numbers will be these x=n that have only a full dot on line 1
and on line n and not on the lines i from 2 to n-1. We could mark these couples
with a different sign (triangle).
We can now “sew” the X,Y plane starting from the origin through the full dots,
that represent the divisors of the natural numbers on the X-as. We will sew the
X,Y plane on each line along the Z axis using a sinus function with following
characteristics:
Period T of sinus function covering 3 dots:
Hence we get for our sinus function:
With d denoted as the distance we have traveled on the line i while sewing the
dots.
This can be expressed as:
Note: our sinus will become zero at each cross through the X&Y-plane on a full
dot.
Working out the above, we get:
Natural numbers n that are not prime will have at least one line i with a full dot
that will represent a divisor different than 1 or itself and hence an i between 1 and
n where the above sinus will be zero. This will not be the case for prime
numbers.
Hence we can state the formula for determining if a natural number is prime or
not as:

(formula 1.)
In literature, it is said that there is no known useful formula that sets apart all of
the prime numbers from composites. This proves then the contrary.
If P(n)=0, this means that at least one divisor was found between 1 and n and
hence that n is not prime.
The meaning of this formula can be graphically represented by drawing vectors
from (0,0) in the direction of the different angles for i=2 to n-1.
We will show the main different cases with an examples:
Example 1: Prime number (example n=7):

There is no vector along the x-axis (if we consider only n=2 to n-1), so n is a
prime number.
Example 2: even non Prime number with only even divisors (example n=8):
There is at least one vector along the x-axis and it is pointed to the right.
Example 3: even non Prime number with odd and even divisors (example n=6):

There is more than one vector along the x-axis and they are pointed to the left
and to the right.
Attention: 6 is also a perfect number. This means that the sum of its aliquot
divisors equals itself (1+2+3=6). Unfortunately, this can not immediately be seen
in the graphic.
Example 4: odd non Prime number (example n=9):

There is at least one vector along the x-axis and it is pointed to the left.
Out of the above graph follows that as soon as i>n/2, there is no more chance of
a vector hitting the 180° line as it’s angle will always be between >180° and
<360°.
It is clear from the above graphical representations of our formula 1 that every
vector set out starting from 2, 3, … will or will not hit the 180° line. The first vector
δ that will hit the 180° line (for i= δ) will determine n as a non prime. δ will then be
the first divisor of n.
It is clear that if δ is the first divisor of n then n- δ will be a multiple of δ as well.
Since it is a multiple it cannot be a prime. If δ is the first divisor then it follows that
δ-1 will not be a divisor. Hence δ-1 is a relative prime. Following the logic, it also
follows that n- δ+1 will be a relative prime. The opposite is also true: if δ is not a
divisor of n then n- δ will also not be a multiple of δ as well in n.
In the case when there is no vector that hits the 180° line, n will be a prime
number. When we choose the variable n to be 2n, then it is obvious that there will
always be a vector that hits the 180° line before it reaches 2n-1. Would this not
be the case then 2n would be a prime which is a contradiction since it is even.

It is clear that we already proved Golbach’s conjecture out of the above.
Remark 1:
The above graphical representations can be interpreted as a spring that is
twisted in counter clockwise direction for a maximum number of turns for i=2 and
then slowly released in clockwise direction in ever smaller steps until i=n-1.
Actually for better understanding we can say that the spring has been twisted n
turns (nx360°) and released till 180° (in ever smaller steps) but we won’t take into
account the first and the last position (i=1 and i=n) for calculating primes.
The angle for an arbitrary i is:
The step taken between i-1 and i equals:
The smallest step will be the last one for i=n-1:
At that moment, the vector will be the closest to 180° but not reaching it:
So this last vector will not contribute to n being not prime. In fact none of the
vectors where i>n/2 will contribute to n being not prime.
In graphical representations, it sometimes looks like the vectors are randomly
distributed but this is because they take bigger jumps for smaller i and that way
they look like they have mixed up. But if we would represent it as being on a
helix, things would become more obvious. So the previous examples can be
seen as the projections on the xy-plane of what happens in reality on a helix.

Remark 2:
For i=1, the vector representing will be on positive x-axis for even n and on
negative x-axis for odd n. This is mentioned here because this vector is not
drawn in the above representations but it might have a use to check if n is a
perfect number.
Carl Pomerance has presented a heuristic argument which suggests that no odd
perfect numbers exist.
So for perfect numbers, the vectors will start on positive x-axis and end on
negative x-axis.
Each time that a vector hits the x-axis, it seems that i is a divisor of n.
Remark 3:
Using Euler’s formula, we could also write:
This means as well that if we set out the vectors for n in the direction for i=2 to
n-1 and one of these vectors is entirely on the x-axis then n is not a prime.
Step 2: splitting an even number into the sum of 2 primes

Goldbach’s conjecture states that every even number can be written as the sum
of two primes.
We can agree on following convention:
2n every even number
q prime number greater than 2 (and q≤p)
p prime number smaller than 2n-2
With: 2n=p+q
So for 2n, p and q, we will be able to write:
This is obvious since for i=2, the sinus term will equal 0.
The above 2 equations are also obvious since a prime number is not divisible,
hence the argument will never be a multiple of π.

Moreover, if we find one of the 2 primes (p or q) then it follows automatically that
the other one (2n-p or 2n-q) is also prime:
The above is true because the sinus in the first term will become 0 for i=2 and the
second term will never become 0.
The above is true because of the same reasons.
This means that we have to find only one prime below each even number as the
other prime will follow from the equation.
Since 2 is the first prime and 2 is even, it follows that we will have to start looking
for the first prime >= 3.
Summary, if we can prove that there always exists a prime under following
condition:
3<=q<=n(=2n/2)
Then we have proven Goldbach’s conjecture.
From the above, it is obvious that we have to consider only even numbers where
2n>=6.
As 3 is the first prime >2 that fulfills the requirement, it follows that we have
proved Goldbach’s conjecture.
Step 3: relationship between the sum of primes
From the above formula’s, we can deduct an interesting relationship:

Since and as well as , we have a linear
equation that will need to have solutions for as well as
.
This was to be expected since our formula for calculating if a number is prime is
nothing more as a consecutive multiplication of vectors until one multiplication
equals zero as a result whereby we know that n is not a prime.
In formula’s:
The imaginary part is 0 if
This is illustrated in the example below:

This means that out of 2 prime numbers (or vectors), a non prime number (or
vector) can come if the surface made from the imaginary part of the first vector
with the real part of the second vector is the same but opposite as the surface
made from the imaginary part of the second vector with the real part of the first
vector.
The above can be shown in a small example demonstrating the sum of 2 primes
7 and 11. One can see in the picture that the 2 factors are equal but opposite for
i=3, 6 and 9 resulting in a non prime sum:
Step 4: graphical representation of divisibility and prime
We will refer to the following picture as a graphical aid to talk about divisibility:
In the above picture we will draw a line from (0,0) to (2n,2n) as a starting point.
We notice that this line goes through all the couples (i,i) from i=0 to 2n.
This means that 2n is divisible by 2n under all circumstances.
We will mark all integer couples with a dot.

If we want to know if 2n is divisible by a certain number q (ex. q=4), we draw a
line from (0,0) to (2n,q) as well. If the line hits one or multiple times couples (i,j)
with i,j between 0 to 2n, we know that 2n is divisible by q.
In fact, any line between (0,0) and (2n,q) that hits integer couples (i,j) shows that
there is a linear relationship between q and 2n using only integers. The
relationship is: q=2n*j/i.
One of the interesting matters is that the 2n-1 line (as well as the 2n+1 line) will
never hit any integer (i,j) except (0,0) and (2n,2n-1). This is because 0≤Δy(x)<1
for all x<2n. In fact Δy(x)=1*x/2n. So, it doesn’t hit an integer value anymore
before it reaches x=0. Unfortunately, 1 is not considered a prime number
otherwise Goldbach’s conjecture would also be proved by this.
But what it does show is that for every 2n (which is obviously not a prime), 2n+1
and 2n-1 will not have a linear relationship in the form of q=2n*j/i with ni.
Therefore 2n-1 will either be a prime or a composite of primes.
Sometimes, n=2n/2 happens to be a prime number. In that case, it is obvious that
2n can be split into n and n as prime numbers.
For the other case (where n=2n/2 ≠ prime), we claim that every line that goes
through (0,0) and (2n,n+δ) with δ an integer ≥0 and ≤ n-3, this line together with
the line that goes through (0,0) and (2n,n-δ) with δ an integer ≥0 and ≤ n-3 will
equal to the line that goes through (0,0) and (2n,2n). This means that if we find a
δ where n- δ=q represents a prime line, then the other line n+ δ=p will also
represent a prime line. Both lines will represent the proof of Goldbach’s
conjecture.
We will now use formula 1 from above to verify our findings:
for 2n:
This is because for i=n, sin(2π) will equal to 0 (as well as for i=2).
for n:

The above =0 if one of i (between 2 and n-1) is a divisor of n. This is the case if
on the line from (0,0) to (2n,n) on the above drawing there is one or more dots
marked. If the above formula doesn’t equal 0 then n is the prime number also
mentioned above.
In the case where n is not the prime we are looking for, then we will look for a
prime, a little higher or lower than n.
for n :
using the following trigonometric identity:
We get:
The first term will be 0 since n was not a prime (and hence divisible by i) but if we
choose such that it is not divisible by i, then will seem to be one of the
primes that we were looking for. The other one can be found by subtracting the
first one from 2n as shown before.
Case for n- :
The above equation is ≠ 0 if:

Or if:
n-δ i
Because δ can go from 0 to n-2 and i can go from 2 to n-1, it follows from the
below picture that n-δ ≠ i.
The above as well already illustrates the sum of the twin primes.
Step 5: Polignac’s conjecture
Polignac’s conjecture states: “For any positive even number n, there are infinitely
many prime gaps of size n. In other words: There are infinitely many cases of two
consecutive prime numbers with difference n.”
To highlight that n is even, we will write n as 2n so that 2n=p-q.
To illustrate our proof, we will show an example below:
Using the previous formula’s, this would mean that we are looking for:
; ;
Let’s take any prime number (out of the infinite amount of existing primes) and
call it p if it is >2n or call it q if it is <2n.
Suppose, we call it p then:

However the first term in the equation is 0 because .
This means that the second term cannot be 0.
This means that or that q is also prime.
This means that for every prime, we can find a corresponding prime such that
2n=p-q. Since there are an infinite amount of primes to choose from at the start, it
means that we have proven Polignac’s conjecture.
Step 6: Twin primes
The Greeks speculated that there are an infinitude of twin primes. But they
weren’t able to prove that.
Using the previous formula’s, this would mean:
; ;
The first factor in this equation will be 0 because P(2n)=0.
The second factor will never be 0 because the cosines argument will never reach
π/2 and the sinus argument will never reach π.
Hence P(2n- ) will never be 0.

The similar can be proved for P(2n+ ).
The above again proves that there are an infinitude of twin primes.
Step 7: General twin prime conjecture
The general twin prime conjecture states: “for every natural number k, there are
infinitely many prime pairs p and p′ such that p′ − p = 2k. The case k = 1 is the
twin prime conjecture.”
Based on the above, we can write: p′ = p + 2k.
We again work out our formula:
The left term will be 0 because for i=p, the sinus will become 0.
The right term will never become 0 since p/p’ will never become ½ otherwise p’
wouldn’t be prime. Neither will 2k/p or 2k/p’ ever become a natural number for
the same reason.
We conclude that the equation will always be different from 0 and hence that we
can choose any k to find a p’.
So, an infinite amount of possibilities are available.
Step 8: 2p
−1 is prime if p is prime

We will try to prove this here using our formula:
For i=2, the first term will be 0 due to the cosines.
In the second term, the sinus will never be 0 as the argument will never reach kπ.
So, we can write:
For all i different from multiples of 2, it is clear that the cosines term will never
reach π/2 or 3π/2. So the cosines will never be 0.
For all i that are a multiple of 2, the value in the denominator can be eliminated
against the value in the teller. What remains in the teller will be a multiple of 2 as
well. So, this cosines value will never be 0 either. Furthermore, i will never reach
2p+1
.
So, it is clear that the result will be <>0 and that 2p
-1 will be prime if p is prime.
Step 9: Perfect numbers
Another open question is whether there are infinitely many perfect numbers.
In number theory, a perfect number is a positive integer that is equal to the sum
of its proper positive divisors, that is, the sum of its positive divisors excluding the
number itself (also known as its aliquot sum). Equivalently, a perfect number is a
number that is half the sum of all of its positive divisors (including itself) i.e. σ1(n)
= 2n.
It is also unknown whether there are any odd perfect numbers.
First of all lets write down the formula for the divisor function:

with r the number of distinct prime factors of n, is the ith prime factor, and is
the maximum power of by which n is divisible.
Suppose n is a non prime and hence it has r divisors that satisfy the above
formula.
Then we can write our formula 1 as follows:
or:
We have separated all the divisors of n in the first term after the equation and
all these sinus will be 0. This is the reason why P(n)=0.
The second term is ≠0.
Step 10: even Perfect numbers
In this part, we will only talk about even perfect numbers (2n). We will say
something about odd perfect numbers in the next part.
Since in this part, we are talking only about even numbers, it means that 2 is at
least a divisor (2n/2=n), together with 1 and itself.
We will show how to calculate the divisors starting from the number of divisors
that we want to calculate.
We will use following general presentation as a guide:
number of divisors: 3

Solving this gives:
Which is a known solution.
This gives:
This gives
So no solutions in natural numbers.
In general, we expect no solutions for even number of divisors (excluding itself!).

It seems like can be freely chosen to come to the other unknowns and n
should be even.
Solving for n, we get the following formula:
It looks like the only natural solutions for the unknowns are then:
1,2,3,12,18,36 and 1,2,4,7,14,28.
This follows from calculations in Excel:

number of divisors: r
we select only r odd so that there is a middle column (after separating 1 and n).
We can rewrite n as:

since:
We come to the general formula for perfect numbers:
So, we will have to look for solutions of this equation where n becomes a natural
number for the chosen divisors.
Remark: other perfect numbers are: 6,28,7056, …
Step 11: Odd Perfect numbers
Another open question is whether there are any odd perfect numbers.
We will first of all present two examples with the perfect numbers 6 and 28:
We can generalize the above using the following diagram:
Out of the examples we can generalize that:

- the divisors of n will be listed from left to right in an ascending order
starting with 1, …, p1, …, p2 until we reach the last (improper) divisor n
itself.
- the , which represent the number of times the divisor will go into n, will
be listed from left to right in a descending order starting with n, …, p2, …,
p1 until we reach 1.
- if there are r divisors then we can say that
- So, the row of will be the same as the row of but in opposite order.
-
Since the sum of the proper divisors needs to be n in order to be a perfect
number and since we are looking for an n which is odd, it means that except for 1
(which is odd) and n itself, there can only be an even number of odd divisors of n.
The easiest way to represent this is by taking 2 odd divisors p1 and p2 in the
representation above.
Since n needs to be odd, it can either be a prime (case 1) or a composite (case
2).
Case 1: n is an odd prime, say n=p3
We replace n in the excel everywhere with p3:
The first row says that 1+p1+p2=p3. This could be possible.
The first column says that 1.p3=p3. This is ok.
Out of the second column, we then conclude that p1.p2=p3.
But since p3 is a prime, this is not possible.
So, this is a contradiction.
Hence case 1 is not possible.
Case 2: n is a composite, say n=k.p2
We replace n in the excel everywhere with k.p2:

The first column says 1.k.p2= k.p2. This is ok.
The second column says: p1.p2=k.p2. This is possible. It means k=p1.
The first row says: 1+p1+p2=k.p2. Using k=p1, we then get: 1+k=p2(k-1)
I then follows that p2= (k+1)/(k-1).
We split this up in 2 sub cases: k=2 or k>2.
Sub case a: k=2
This gives: p2=3. It follows then that p1=2 and n=6.
Although this is a perfect number, n is in this case even what is not the solution
we are looking for.
Sub case b: k>2 say p1=3
In this case p2 will never reach an integer value any more.
So there are no odd solutions for a perfect number.
An alternative method to prove that there are no odd perfect numbers is as
follows:
Once you multiply an odd number with an even number, you will get an even
number. This means that an odd number can only have odd divisors.
In order to get a perfect number, the sum of all proper divisors of our odd number
must equal our odd number. We can only get an odd number again with odd
divisors if we add up an odd number of divisors (1 inclusive). So, 1 exclusive, we
have an even number of odd divisors.
Suppose, we have 2 divisors apart from 1:
Then we can write the following 2 equations:

In order to get a perfect odd number, we must equal both equations. This will
give:
Solving this equation for gives:
We concluded above that the divisors needed to be odd. So we start with
assuming that is odd. This however means that is even and that
is also even.
A first possible solution is for . This results in . However, odd is
not acceptable.
For , and for .
So no real solutions can expected for 3 real divisors.
Suppose, we have r divisors:
Then we can write the following 2 equations:
In order to get a perfect odd number, we must equal both equations. This will
give:
Solving this equation for gives:

So no real solutions can expected for r (odd) divisors.
Step 12: Mersenne numbers
Another open question is: No proof is known whether there are infinitely many
Mersenne primes.
A Mersenne prime is any prime that is of the form:
where n is a prime itself.
Using our formula:
For i=2, the left term will be 0 and the right term will be <> 0.
So, in general, hence, will be prime.
This means that n=3 will be a Mersenne prime.
For i>2 the equation will be prime if:
or if:
This will be the case if i is a divisor of n.
Example:

The equation:
becomes (for i=21):
(-0.1490).(0.9888)-(0.9888).(-0.1490)=0
Which means that is not a prime (because 12 is not a prime).
Generally speaking, will be a Mersenne prime if it has no divisors between 2
and .
A theorem about Mersenne numbers states that if is prime then the exponent
p must also be prime.
Since there are infinitely many prime numbers above n, it means that there will
be infinitely many Mersenne primes as well.
Step 13: odd harmonic divisor numbers
Using a small example, we will derive the formula for a harmonic divisor:
Divisors of 28 are: 1,2,4,7,14,28. So, we have 5 proper divisors (=r) and including
28 itself, we will have 6 (=r+1).
Using a small example, we recapitulate the conventions for the variables:
The harmonic mean (HM) for 28 is:
So, we can write the general formula for the harmonic mean using our
conventions:

This last expression is true because:
In the case of perfect numbers, the sum of all the divisors of n equals 2n.
Then the harmonic mean for perfect numbers becomes:
Oystein Ore, who defined the harmonic divisor number (also called ore number),
conjectured that no odd harmonic divisor numbers exist other than 1. If the
conjecture is true, this would imply the nonexistence of odd perfect numbers.
As a matter of fact this is true. If we use our previously derived formula:
We try to list all the possible scenario’s in terms of odd/even:
We see that the only chance to have an odd harmonic mean is in the case where
all the 3 components of the formula are odd.

We know that an odd number of divisors (r+1) is only possible for squares.
Since every perfect number is harmonic and an harmonic number can only be
odd in the case of n being a square but a perfect number cannot be a square it
follows that no odd perfect numbers exist.
In the following, we explore possible odd harmonic numbers a little further.
We look at the following example for n=81:
Another example for n=9:
Lets generalize our form a bit for purpose of finding possible odd solutions.
This gives for 2+1 divisors:
So, we get:

This gives:
The above equations has only real solutions for HM≤4.
For HM=4,
For HM=3,
…
We can try the next generalized form for 5 divisors:
From here, we get:
The harmonic mean then becomes:
Since , the value for HM will never reach a value of 5. No possible integer
solutions lie within this range.
Since solutions for more than 5 divisors can be written using similar formula’s as
above, we can summarize by saying that no odd harmonic means will ever exist.
Results

It is clear from the above that our formula for differentiating primes from non
primes is sufficiently strong to prove the existing open conjectures. This is done
using basic trigonometric formula’s and some simple interpretations.
Discussions:
It is our strong belief that in the representation of complex numbers an extra
factor would be useful to represent the spin. Although in most calculations φ and
φ+360 or φ+720 represent the same vector, this might not necessarily always be
the case.
The working out of this will however be for a next paper as it seems to be a little
complicated due to the rules of group theory involved.
The working out of the relation between divisibility and standing waves could be
further developed as well.
Conclusion:
1. Discussing about primes and prime gaps is not so difficult if one uses the
right formula to demonstrate prime.
2. The natural numbers are just a special case in the complex plane which
seems to be a projection from the complex helix.
3. Divisibility is not about the smaller number fitting into the larger one but
about the larger wave from the smaller numbers coming together with the
smaller wave from the larger numbers in a standing wave for the different
i.
Acknowledgements
I would like to thank this publisher, his professional staff and his volunteers for all
the effort they take in reading all the papers coming to them and especially I
would like to thank this reader for reading my paper till the end.
I would like to thank my wife for having faith in my work.
References
1. God created the integers; Stephen Hawking; Penguin Books;p1-5
2. De zeven grootste raadsels van de wiskunde;Alex van den Brandhof,
Roland van der Veen, Jan van de Craats, Barry Koren; Uitgeverij Bert
Bakker
3. http://en.wikipedia.org/wiki/Goldbach's_conjecture

4. http://mathworld.wolfram.com/GoldbachConjecture.html
5. http://en.wikipedia.org/wiki/Twin_prime
6. http://en.wikipedia.org/wiki/De_Polignac%27s_conjecture
7. http://en.wikipedia.org/wiki/Complex_number
8. http://en.wikipedia.org/wiki/Perfect_number
9. http://en.wikipedia.org/wiki/Aliquot_sum#Definition
10.http://en.wikipedia.org/wiki/Aliquot_sequence
11.http://en.wikipedia.org/wiki/Composite_number
12.http://en.wikipedia.org/wiki/Harmonic_divisor_number
13.http://en.wikipedia.org/wiki/Helix
14.http://www.intechopen.com/books/an-update-on-power-quality/harmonics-
generation-propagation-and-purging-techniques-in-non-linear-loads; figure 1

Derivation of a prime verification formula to prove the related open problems

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