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Mathematics Forum
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| Author | Message / Information |
| illoyd Quote | Reply | | Algorithm for primes posted on: 6/14/2006 3:29:40 PM Does anyone know if the formula for primes below has been disproved. Ip = product (p1,p2,p3,p4,p5...) +2 where "product" is the product of consecutive primes from 3 onwards. I have tried this up to 43 where my maths engine runs out of steam and it works. i.e P=(3*5) +2 is prime etc to: p=(3*5*7*11*13*17*19*23*29*31*37*43*47)+2 is prime |
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Euler
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Algorithm for primes
replied on: 6/14/2006 4:53:57 PM I am not familiar with this algorithm, nor its validity; however it is interesting, as a prime multiplied by n many primes would invariably be an odd number, and any odd number plus two is also odd, so it can at least be shown that only odd numbers can be produced, which in turn increases the probability of it being true. I will expand it a little further and see how it holds up..... |
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Euler
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Algorithm for primes
replied on: 6/14/2006 5:11:52 PM One thing I find particularly interesting with this though is that the number 2 is, by definition, a prime as well; however the algorithm does not include it, so should this be a naturally occuring phenomenon, so to speak, I find it odd that it does not occur with all the primes. Of course, that is assuming that there is some pattern governing the primes, so I guess the algorithm would make more sense if the prime sequence is chaotic. Nevertheless, it is interesting... oh! and I tried doing 3*5*...*53 + 2, and you get 16294579238595022367, which according to my sources is not a prime, but it is hard to tell how accurate that answer is because of how large the number is. Anyways, assuming my sources are correct, this statement does in fact fall apart after 47. |
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Euler
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Algorithm for primes
replied on: 6/14/2006 10:17:22 PM Also, I would like to mention that I may have developed an algorithm for the prime sequence itself; however I still need to refine it. I have found that it is approximately a constant raised to a natural log function, and ln functions level off as you approach infinity, so if I continue to expand it further until it levels off to a considerable flat level, then it should be adaquete for the rest of the primes as you approach infinity. Well see how it goes though.... and even if it doesn't predict the exact value of the nth prime, it can still be used to find the exact value of pi(N), so I will soon be able to write a paper on the distribution of primes below a certain number. |
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contrapositive
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Algorithm for primes
replied on: 8/25/2006 2:58:55 AM Does it work? - yes and no. It's very similar to Euclids proof that there are infinite prime numbers. Namely, assume their are a finite number of primes - multiply them together and add one. None of the primes on your list are a factor so either: a) this new number is prime, or b) there is a smaller prime not on your list which is a factor Ie, you can always find another prime. Where this algorithm falls down is it always assume a) occurs. If you include at least every prime up to and including the square root your new 'potential prime' it will be prime. If you miss a prime then it's not garrentied your 'potential prime' is prime - that missing number might be a factor. (The reason for the square root - assume n is composite and ab = n. a AND b can't both be bigger then sqrt(n) otherwise the product is bigger then n - ie, one of the pair of factors is smaller then sqrt(n)) However, this algorithm doesn't generate every prime number - for instance 3*5+2 = 17, you've missed 7, 11 and 13. Thus you can't be sure you've generated all primes up to the square root and removed them as potential factors. |
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