This should not take that long: I will drop this quickly so I can go back to being bored:
If one has been following my previous posts, then one will know the terms by now; if not, then I suggest that one reads the previous posts for definition of terms.
Come on, this is NOT that difficult to prove!!!
Now, give me a REAL challenge!!!
If one has been following my previous posts, then one will know the terms by now; if not, then I suggest that one reads the previous posts for definition of terms.
About even numbers; let Σ be the set of all even numbers; it follows then that, if 'e' the element of Σ is sufficiently large that there are primes less than e such that the greatest prime, pn, is the one just less than the square root of 'e' then the following factor sets FULLY define the even number 'e' ( in the sense that every number from 1 to e is accounted for... a bit tricky but you had better get this before you proceed, because if you are lost here you are lost for good)
e = [Fp1] + [Fp2] +[Fp3] +...+[Fpn] +[Rpn]
The sizes of the factor sets and the remainder set are from '1' to 'e' as I have mentioned.
Now, it is a known fact that every even number is 'double' some number, or that there are numbers less than the evn number that when added will give the sum of the even number. The greater the even number, the more the possible pairs, both even and odd, yes?
The number of possible pairs for the even number are exactly half the value of the evn number, and, if one removes the even number pairs...the pairs due tyo even numbers pairing with other even numbers less than the even number in question, then the remaining half is also halved... work that out.
Suppose the even number has other factors that are greater than two, and up to pn, for example. These factors will have their multiples pairing with other multiples which have THOSE factors as common, when THOSE multiples add up to give the sum of 'e'.
The point? Only NON- factors of the particular even number will have their factors pair with the elements of Rpn, which will contain only primes, in case you haven't noticed.
So, to find the NUMBER of prime pairs that, per even number will give a sum of the even number, factor out the factor sets of the evn number, remove them from the above equation, then use what is left to find the "proportion" which are claimed by the factor sets which have their primes as non-factors of 'e'. The remainder in the remainder will be the number of prime pairs which give the sum of the even. You can then identify the primes using the laborious method that I used in proving that the primes are infinite.
If you want an equation, then it will be something like
p(pairs) = (1/2)(e/4)(1 - { [Fp2] +[ Fp3] +[Fp4]+...+[ Fpn]} [Rpn]))
(where p(pairs) are the number of prime pairs that give the sum of e for every even number)
provided that none of the primes, p2;p3;p4;...;pn are factors of e, in which case they will not be included in the equation. Note the factor sets have an "upper boundary" of 'e'.
e = [Fp1] + [Fp2] +[Fp3] +...+[Fpn] +[Rpn]
The sizes of the factor sets and the remainder set are from '1' to 'e' as I have mentioned.
Now, it is a known fact that every even number is 'double' some number, or that there are numbers less than the evn number that when added will give the sum of the even number. The greater the even number, the more the possible pairs, both even and odd, yes?
The number of possible pairs for the even number are exactly half the value of the evn number, and, if one removes the even number pairs...the pairs due tyo even numbers pairing with other even numbers less than the even number in question, then the remaining half is also halved... work that out.
Suppose the even number has other factors that are greater than two, and up to pn, for example. These factors will have their multiples pairing with other multiples which have THOSE factors as common, when THOSE multiples add up to give the sum of 'e'.
The point? Only NON- factors of the particular even number will have their factors pair with the elements of Rpn, which will contain only primes, in case you haven't noticed.
So, to find the NUMBER of prime pairs that, per even number will give a sum of the even number, factor out the factor sets of the evn number, remove them from the above equation, then use what is left to find the "proportion" which are claimed by the factor sets which have their primes as non-factors of 'e'. The remainder in the remainder will be the number of prime pairs which give the sum of the even. You can then identify the primes using the laborious method that I used in proving that the primes are infinite.
If you want an equation, then it will be something like
p(pairs) = (1/2)(e/4)(1 - { [Fp2] +[ Fp3] +[Fp4]+...+[ Fpn]} [Rpn]))
(where p(pairs) are the number of prime pairs that give the sum of e for every even number)
provided that none of the primes, p2;p3;p4;...;pn are factors of e, in which case they will not be included in the equation. Note the factor sets have an "upper boundary" of 'e'.
Come on, this is NOT that difficult to prove!!!
Now, give me a REAL challenge!!!
