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Re: Prime Hunt

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  • Dick
    Hello, ... for some ... So then k = p_1^(x1-1)*p_2^(x2-1)*.p_m^(xm-1) ... Implicitly understood ... is ... I realized after posting yesterday that I was using
    Message 1 of 5 , Nov 29, 2002
      Hello,

      > Consider the generalization of your idea which I have considered
      for some
      > time.
      > Let me change the notation somewhat.
      > Let p_3^(x1) be the third prime raised to the power x1,
      > P# = p_1*p_2*.p_m be the primorial function,
      > k*P# = p_1^(x1)*p_2^(x2)*.p_m^(xm) where all xi>=1,

      So then k = p_1^(x1-1)*p_2^(x2-1)*.p_m^(xm-1)

      > abs( y) represent the absolute value of y.
      > Following is based on the "theorem" that for integers a, b, & c
      > where a +/- b = c then it is not possible that exactly two of the
      > integers
      > are divisible by p_i
      > since this would imply that the third is also divisible by p_i..

      Implicitly understood

      > If one factors k*P# into any two relatively prime factors a & b
      >
      > (including b=1) such that abs(a +/- b) < p_(m+1)^2 then abs(a - b)
      is
      > either 1 or it is prime

      I realized after posting yesterday that I was using < p_m^2 whereas,
      as you state above, < p_(m+1)^2 is better and quite valid.

      > since it is not divisible by any p_i <= p_m. The plus is only
      applicable
      > for rather small m.
      >
      > I believe all primes less than p_m^2 , and so all primes as m
      increases, can
      > be generated
      >
      > this way although I have verified this only for very small primes.

      I should think you are probably correct.

      > Note that it is possible that a prime p_i can be generated in more
      than one
      > way so that
      >
      > abs(k*P#/a - a) is not a unique representation; see the two
      > examples below which generate 59.

      > Some examples of such primes are 2^3*3-1=23 , 2^2*3*7-5= 79,
      > and 2^2*3*7-5^2= 59. Multiple patterns can be observed such as
      the last
      > two primes.

      O.K., so we've rewritten what I wrote and extended it further to
      include powers x_i, which is good (using @ for notation originally
      suggested by Jon has the drawback that the system interprets it as an
      email address, turns it blue and appends... - annoying).

      My question on uniqueness was framed within the context of all x_i=1
      and as such k=1. By the way, I only see one example that gives 59
      above. I'd be interested in a counterexample to the k=1 case.

      > Also 2^(y-1)*P#/2^y - 2^y is a pattern for primes. Examples are
      > 2*3*5-2=103,
      > 2*3*5-2^2=101, 2*3*5-2^3=97, 2*3*5-2^4=89, 2*3*5-2^5=73, etc .

      Any formula that can take on prime values is a pattern for primes.
      Why not state "2^(y-1)*P#/2^y - 2^y" as "P#/2 - 2^y"?
      If y=1, we are back to the k=1 case, otherwise, the above is
      equivalent to all x_i=1 except x_1=y.

      Yes, the difference of co-prime integers is co-prime to the original
      integers, I get that.

      > It "seems" that it should be possible to prove that there is always
      such a
      > prime
      > n^2< abs(k*P#/a-a) < (n+1)^2 for any integer n, but I haven't been
      able to
      > prove this.

      Nor anyone else, guaranteeing a prime between every pair of
      consecutive squares should also prove the Twin Prime Conjecture,
      Goldbach's Conjecture and a slew of others - so don't take it too
      hard.

      I think your conjecture of being able to represent every prime by the
      difference of two co-prime products when the x_i can take on values >
      1 is a good one - probably true.

      As for me, I am interested in the all x_i=1 case and the questions of
      unique representation and the behaviour of the minimum difference
      amongst all possibilities of the two morialish products for any given
      n. If the minimums can be modeled such that we know where to look or
      such that the computational path needed to get there can be followed,
      and it can be proven that such minimum is always < P(Pi(n)+1)^2, the
      value is more theoretical than anything else. Unless we can find an
      efficient method to compute the difference of 2 massive products
      without actually having to compute the whole products - something
      akin to the modular reduction that allows us to prove massive primes
      by tackling the even more massive and otherwise inaccessible number a^
      (p-1). With these pieces in place, one could generate vs. discover
      large prime numbers.

      Overall, I like your notation better - Thank you.


      -Dick

      > ----- Original Message -----
      > From: "Dick" <richard042@y...>
      > To: <primenumbers@y...>
      > Sent: Friday, November 29, 2002 5:10 AM
      > Subject: [PrimeNumbers] Re: Prime Hunt
      >
      >
      > > Hello,
      > >
      > > I wrote a post in response to Jon's:
      > >
      > > --- In primenumbers@y..., "Jon Perry" <perry@g...> wrote:
      > > > Anyone want to find large primes of the form:
      > > > p@+/-1
      > > > where @ represents the alternating primorial function, i.e.
      > > > p_1.p_3.p_5.p_7....
      > > > or p_2.p_4.p_6.p_8...
      > >
      > > And inadvertantly sent it only to Jon. I sent an email to Jon to
      ask
      > > him to pass it along so I wouldn't have to retype it, but - my
      > > apologies to you Jon - I've been thinking about it more and have
      > > answered some of my own questions, realized a couple of mistakes,
      > > plus there's some compelling things going on and I can't sleep,
      so to
      > > summarize my first post:
      > >
      > > I have played with this idea and some generalizations of it in the
      > > past. Some suggestions on additional notation were to call n@e
      the
      > > product of primes P(k) <= n with even indices k and n@o similarly
      for
      > > odd indices. Also, n@f for the product of primes <= n such that
      > > p+1=0Mod4 and n@t the product of primes <= n such that p+1=2Mod4.
      > > (in retrospect, I should have noted that p(1)=2 would have to be
      > > handled separately for the n@f and n@t notations)
      > >
      > > These notations would seem to be manageable if programmed into
      PFGW
      > > and used to express prime values.
      > > If they were there, I would definitely play with them. I am
      > > interested in the general case of two products n@' and n@'' such
      that
      > > every prime <= n is used as a factor only once in one of the two
      > > products. Since there are a great number of ways of splitting Pi
      (n)
      > > primes into two primorial like products, most of the possible
      splits
      > > would be too complex for simple notation.
      > >
      > > Numbers of the form n@'-n@'', must be such that
      > > GCD(n@'-n@'',n@')=GCD(n@'-n@'',n@'')=1, thus it is obvious that
      > > n@'-n@'' must always be > n.
      > > What I didn't say, but also must be true, is that n@'-n@'' cannot
      > > possess a prime factor < n. Thus if ABS(n@'-n@'') < n^2,
      > > then ABS(n@'-n@'') must be prime.
      > >
      > > I had made the following list of PRP's to post on the top ten PRP
      > > pages one of these days when I got around to it (could someone
      send
      > > me the link please?).
      > >
      > > differences
      > > (P(12*282)#/P(282)#)-P(282)# = 12771 digits
      > > (P(8*492)#/P(492)#)-P(492)# = 14556 digits
      > > (P(11*531)#/P(531)#)-P(531)# = 23318 digits
      > > (P(16*607)#/P(607)#)-P(607)# = 41988 digits
      > >
      > > sums
      > > (P(6*540)#/P(540)#)+P(540)# = 11236 digits
      > > (P(9*345)#/P(345)#)+P(345)# = 11313 digits
      > > (P(10*383)#/P(383)#)+P(383)# = 14453 digits
      > > 2*(P(9*445)#/P(445)#)+P(445)#/2 = 15038 digits
      > > 2*(P(16*243)#/P(243)#)+P(243)#/2 = 15181 digits
      > > 2*(P(11*371)#/P(371)#)+P(371)#/2 = 15641 digits
      > > (P(11*408)#/P(408)#)+P(408)# = 17391 digits
      > > 2*(P(13*406)#/P(406)#)+P(406)#/2 = 21091 digits
      > >
      > > The differences are the fascinating ones for me. Amongst all the
      > > possible combinations for a given n, what is the minimum n@'-n@''?
      > > Shouldn't it be possible, even for arbitrarily large n, to find
      some
      > > n@'-n@'' such that ABS(n@'-n@'') < n^2. If so, one has just
      found a
      > > prime number - no testing in the formal sense would be necessary.
      > > If a difference can be found that is slightly higher than n^2,
      > > proving it prime might be feasible with Newton's method.
      > >
      > > I had also suggested the question
      > > "does it ever happen that n@'-n@''= P(Pi(n)+1)?".
      > > And of course, looking at small numbers, it does:
      > >
      > > 2*5-3=7
      > > 3*7-2*5=11
      > >
      > > Taking this a step further, we can imagine an algorithm, which I
      will
      > > try to implement in Pari sometime soon, unless one of our real
      > > programmers can beat me to it (hint, hint),
      > > that goes simply like this:
      > >
      > > Given initial n@'< n@'',
      > > IF ABS(n@'*P(Pi(n)+1)-n@'')< P(Pi(n)+1)^2
      > > Then print ABS(n@'*P(Pi(n)+1)-n@'')< P(Pi(n)+1)^2 is prime!
      > > set n@'=Min(n@'*P(Pi(n)+1),n@'')
      > > set n@''=Max(n@'*P(Pi(n)+1),n@'')
      > > set next prime = P(Pi(n)+2)
      > > iterate loop
      > >
      > > Continuing the example above:
      > > 2*5-3=7
      > > 3*7-2*5=11
      > > 2*5*11-3*7=89 < 11^2 thus 89 is prime
      > > 3*7*13-2*5*11=163 < 13^2 thus 163 is prime
      > > 2*5*11*17-3*7*13=1597 > 17^2 (way over, but 1597 is a prime)
      > > 3*7*13*19-2*5*11*17=3317 > 19^2 (way,way over and 3317 is
      composite)
      > >
      > > Seems to grow very large quickly, but are there times when the two
      > > products come close to each other again?
      > > If you wanted to get fancy, the program could try to bring n@' and
      > > n@'' closer to each other by swapping factors. For example, if
      n@' <
      > > n@'' and n@'' has the higher of a twin as a factor and n@' has the
      > > lower of the twin as a factor, swapping these factors might do the
      > > trick. If you wanted to get really fancy, the program could
      guess,
      > > based on the magnitude of the difference, how big and how far
      apart
      > > the swapped factors should be to bring the products close to each
      > > other.
      > >
      > > Swapping 2 and 3 in the last line above yields:
      > > 2*7*13*19-3*5*11*17=653>19^2 (much closer and 653 is a prime))
      > > now swapping 11 and 13
      > > 2*7*11*19-3*5*13*17=389>19^2 (almost made it and 389 is a prime)
      > >
      > > Now we are getting into a program to search for the minimum n@'-
      n@'',
      > > which I think is worth writing since it gets messy real fast by
      hand.
      > > Maybe there's a theoretical approach that would prove that the
      > > minimum difference can never be less than n^2 for sufficiently
      > > large n, but a heuristic/computational approach to get our feet
      wet
      > > with some small n should be revealing in any event.
      > >
      > > O.K., well I've had my say - good night.
      > >
      > > In the back of my mind, I'm getting a glimpse of this idea as
      being
      > > able to shed light on the prime gap problem-I'll sleep on that.
      > >
      > >
      > > -Dick Boland
      > >
      > >
      > >
      > > Unsubscribe by an email to: primenumbers-unsubscribe@y...
      > > The Prime Pages : http://www.primepages.org/
      > >
      > >
      > >
      > > Your use of Yahoo! Groups is subject to
      http://docs.yahoo.com/info/terms/
      > >
      > >
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