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Mathematics Forum
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| illoyd Quote | Reply | | Prime fomulae posted on: 8/24/2007 4:20:51 AM Does anyone know some better formulae for describing primes than the ones which I have so far found? The two formulae x=6m+1 and m=6m+5 together account for all odd numbers except for those divisible by 3 which is therfore a reduced set of all odd numbers which still includes all primes. Does anyone know a better set? |
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cramwit
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Prime fomulae
replied on: 11/17/2007 1:46:44 AM I don't know how useful it is but you can simply take any primefactorial & pluck the available prime spots for those intervals. ie. for 2*3 = primefactorial p3! x mod 6 -> 1,5 for p5! x mod 30 -> 1,7,11,13,17,19,23,29 & i think it is just a list of primes in the interval with locality to the prime factorial having a wider sweep of non-primes on the low end. the high end is already intrinsically swept out. i think for [probably should scrutinize it just to be sure] p7! x mod 210 -> 1,11,13,17,19,23,29,31,37,41,43, 47,53,59,61,67,71,73,79,83,89,97,101,103,107, 109,113,127,131,137,139,149,151,157,163, 167,173,179,181,191,193,197,199 I don't know if that is much of a help, but if it is you can [should be able to] do that for pretty much any prime factorial. obviously as with p3! it doesn't work for the initial prime sequence ie. 2 & 3 but anything above that it holds true for as a prime filter I think what it misses is the actual prime factorial values themselves, so for p5! it would miss 2,3,5 for p7! it misses 2,3,5,7 since presumably one is considering reasonably large numbers it shouldn't be a problem. pls let me know if you think that makes sense. I really am getting sketchy here. |
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cramwit
Quote | Reply | This message was updated on 11/17/2007 2:33:22 AM by cramwit |
Prime fomulae
replied on: 11/17/2007 2:18:42 AM an easier way to do it is simply eliminate the prime factorial values for the interval's prime list for p3!=6 one just drops the 2 & 3 values from prime list for the interval of course one has to include 1 also which is not technically a prime to the modulo list for p5!=30 just drop 2,3,5 from prime list & include 1 on modulo list for p7!=210 drop 2,3,5,7 add 1 to list for p11!=2310 drop 2,3,5,7,11 add 1 to list for p13!=30030 drop 2,3,5,7,11,13 add 1 to list etc. I guess the real question is if having to compare any non-zero modulo with a long list of primes is useful. I suppose one could do a binary search lookup which would be fast. you could have a tree with the middle most prime value as the root, then the 1/4 & 3/4 most values as each branch from there which would be log2 time I think it might be useful. possibly one span of primes could leverage you rapidly up & into the next higher prime factorial interval. But the sheer amount of numbers to consider might still be overwhelming. |
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cramwit
Quote | Reply | This message was updated on 11/18/2007 12:12:42 AM by cramwit |
Prime fomulae
replied on: 11/17/2007 11:21:26 PM Prime factorials make natural folding points when dealing with primes. An apparent failure point of a primefactorial filter is at pure multiple powers or products of subsequent slightly larger primes. p3! breaks down at 25, 5^2 & at 35, 7*5 which are neatly symmetric about the p5!=30 but mislead me into a failed idea about a predictable failure at the p(n+1) iteration of the p3! folded interval. obviously for p5!=30 breaks down at 49, 7^2 it will fail at 7*11, 7*13, 11^2, 7etc. so maybe there is some kind of rule of the primes that must be considered as failure points for a given interval? If that was in any way constrained it might mean there was something approaching predictability, or at least mechanically useful for considering a given interval. If one knows all previous primes can the number of primes for an adjacent prime be predicted? I suppose we predict/determine them with crude brute force number crunching, but the question is, is there a more elegant spare articulation of known, accumulated principles that minimizes the effort required? Maybe a prime filter mask combined with some hopefully systematic use of subsequent primes beyond the primefactorial factors. p(n+k) to limited numbers of powers before it exceeds the interval under consideration. I suppose if you took the smallest next prime, p(n+1) & calculated the highest power of that that fell on the interval in question, call it the kth power, then any factor permutation of p(n+m) primes could not exceed k elements. |
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cramwit
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Prime fomulae
replied on: 11/18/2007 8:06:10 AM One can say with certainty that the iterative primefactorial pattern/mask is absolutely accurate until (p(N+1))^2 BUT (huge but) the trouble with that is that the quadratic, (p(N+1))^2 is vastly exceeded, even consumed in the iterative masks themselves, by pN!, which is steeper even than standard factorial. p7! pattern mask, includes 121,11^2,143,11*13,169,13*13,187,11*17 & while 221 & 289 (13*17 & 17^2) are in the first iteration. it reminds me of fractals where you get distorted repetitions of the initial pattern like mandelbrot echos that are to varying degrees distorted. since at the far ends of the interval it is determinedly [inversely] symmetric the trick becomes filling in the span between them. Sort of like building a bridge. Maybe with lesser degrees of certainty as you get further & further from the primefactorial pier foundations. Like you have to build out the middle by reaching from each end at the same time. Im not sure there is any backbuilding from the high part of the span though, it might be an absolutely one ended construction from the low end. although the fact that one can say with certainty what can not be prime immediately prior to the prime factorial point hints that something might be possible. |
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