Break up all numbers into sets of 6, because every 3rd odd number is composite. Therefore, we can conduct twin prime tests on pairs of numbers that are mod 6 ±1. The odds of a number being prime is approximately equal to (1-1/P1)*(1-1/P2)*(1-1/P3)...(1-1/Pn) Where Pn is the largest prime less than or equal to √x. Since these numbers are written in the form 6x ± 1, they're not divisible by 2 or 3.
The probability that both these numbers (6x±1) are prime can be found using a variation of the earlier method. The odds that one of these numbers (picked before) is not divisible by 5 (it's not divisible by 2 or 3), is 4/5. The probability that a number not divisible by 5+2 is divisible by 5 is 3/4. Therefore, the probability that the numbers 6x±1 are both not divisible by 5 is 3/5. If one extends this for all primes, the probability of the pair 6x±1 not being divisible by a number P is (P-2)/P. Since the largest smallest prime factor of a number n can't be greater than √n, the probability of a pair 6x±1 being prime are approximately equal to (3/5)*(5/7)....*(Pn-2/Pn) Where Pn is the largest prime below √6x+1.
This sequence yields a greater result than the sequence (3/5)*(5/7)....*(On-2/On) Where On is the largest odd number below √6x+1. That sequence's solution is 3/On. Therefore, the lowest possible probability that numbers of the form 6x±1 are both prime is equal to 3/√6x+1. Since we break the numbers up into sets of 6, the probability of this being true for any set of 2 numbers is 1/2√x+1.
Therefore, the minimum number of paired primes between 1 and infinity is approximately equal to ∫1/(2[(x+1)] with limits 1 and infinity. Since the sum of probabilities ~ the average number of instances, as x tends to infinity. The integral of the above function=√(x+1), which becomes infinity as x reaches infinity. Therefore, there are an infinite number of twin primes.
No comments:
Post a Comment