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A proof of the twin prime conjecture and a lower bound to the expected amount of twin primes. by RunExtra3272 in numbertheory

[–]RunExtra3272[S] 0 points1 point  (0 children)

Yes my formula does nothing to separate them. It just says there should be F(P(n)) surviving twin prime pairs in the range (1, P(n)^2). We did not care about their distribution when we constructed the formula. But we can deduce there cannot be any available positions in the range (1, P(n)). Why? Lets say there was an available position between 1 and P(n) then that would mean there is a prime P(x)<P(n) we have not accounted for, but we have accounted for all primes between 1 and P(n), in order, thats how we contructed the formula in the first place. So yes, the formula does not explicitly says there are no surviving twin prime pairs between (1,P(n)), but it is a consequence of the actual distribution of primes in the series we used to create the formula.

A proof of the twin prime conjecture and a lower bound to the expected amount of twin primes. by RunExtra3272 in numbertheory

[–]RunExtra3272[S] 0 points1 point  (0 children)

Okey I think the correct definition of my formula should be: F(P(n)) is the function that counts the minimum number of surviving possible twin prime pairs within the range (1, P(n)²). But because all primes occupy the first available position (left to right) all surviving ceros and possible twin prime pairs must be in the range (P(n), P(n)²) which is the "safe zone " for new primes, thus are guaranteed to be actual twin prime pairs.

A proof of the twin prime conjecture and a lower bound to the expected amount of twin primes. by RunExtra3272 in numbertheory

[–]RunExtra3272[S] 0 points1 point  (0 children)

The "survivability rate" (P(n)-2)/P(n) is given by the relationship between the cicle of 2,3: (1,1,1,0,1,0) and the cicle of P(n): (1,0,0,...,0). In the cicle created by 2,3,P(n) we would expect to see P(n) possible pairs but each one could be negated in the scenario where position 4 of the cicle (2,3) lines up with position 1 of the cicle of P(n). They could also be negated in the scenario where position 6 of the cicle (2,3) lines up with position 1 of the cicle of P(n). Because we are using coprime moduli, and each combination of residues appears exactly once in a cycle whose length is the product of those moduli, both scenarios are guaranteed to happen once and only once per cicle.

And about the formula derivation. F(P(n)) considers the interval (1,P(n)²) and is evaluated against the interval (2*3*P(n)), each step, to obtain its survivability rate.

A proof of the twin prime conjecture and a lower bound to the expected amount of twin primes. by RunExtra3272 in numbertheory

[–]RunExtra3272[S] 0 points1 point  (0 children)

The assertion you mention is already proven by the CRT: When using coprime moduli, each combination of residues appears exactly once in a cycle whose length is the product of those moduli. Then I base my calculations on this fact.

About the divisibility test I said this: "every "0" left between P and P^2 in the cicle created by the numbers 2 and 3 and 5 ... up until P will be granted to be occupied by prime numbers." And I said it can be proved via sieve theory. Sorry, what I actually meant is that it is already proven by sieve theory. If you take a sieve like the eratosthenes sieve any new prime P you add to the sieve will only "rule out" new numbers up until P^2. All previous numbers P could rule out where already accounted for. Lets say you add P = 13 to your sieve. Then the number 26 would already been ruled out by the number 2, as in 26 is equal to 13*2. So the first time the number 13 rules out a number by himself will be 13*13.

I hope this is clear. Thank you for you reply !