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Re: form of "Mills' constant"? [Puzzle 604]

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  • djbroadhurst
    ... Infinity is too far away for me, as a mere physicist, to bother much about it. Suppose that there is infinitude of prime gaps [p,p+d] with 2*exp(-Euler)
    Message 1 of 49 , Mar 19, 2012
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      --- In primenumbers@yahoogroups.com,
      "WarrenS" <warren.wds@...> wrote:

      > --and then the constant C is g(g(g(...p[n]))) with n-fold g
      > in the limit n-->infinity

      Infinity is too far away for me, as a mere physicist,
      to bother much about it.

      Suppose that there is infinitude of prime gaps [p,p+d] with
      2*exp(-Euler) > d/log(p)^2 > 1. (We have no idea whether this
      might be the case.) What then might be the probability of hitting
      one of these hypothetical Warren-defying gaps in an hypothetical
      infinitude of Warren's iterations? And who might ever care?

      For myself, I am content to investigate finite sequences of
      primes and happily report that Puzzle 604 is already solved.

      When I stop, I'll post a modest conjecture-free Prime Curio, with
      due credit to Warren for what we can all agree was a neat idea.

      David (finitely)
    • kadhirvel
      Hi, I had a methodology(may be not efficient had a intuition it will be useful) to generate all odd composite. Factors of all odd composites are predetermined
      Message 49 of 49 , May 7, 2012
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        Hi,
        I had a methodology(may be not efficient had a intuition it will be useful) to generate all odd composite. Factors of all odd composites are predetermined through multiples of 4. Let A(n) = 4n where n = 1,2,3... Then A(n) = P1Q1 = P2Q2 = ..... = PkQk be the 'K' possible set of distinct even factors(means both Pk,Qk  should even) such that Pk > Pk-1 > Pk-2 > ....> P1 then  (Pk  + 1) (Qk  + 1),   (Pk-1 + 1)(Qk-1  + 1), ......   (P1 + 1)(Q 1  + 1)    will eventually generate all odd composite numbers. For e.g. A(1) = 4 = 2x2 here P1 = 2, Q1  = 2. Then (P1 + 1)(Q 1  + 1) = (2 + 1)(2 +1) = 9 First odd composite. A(2) = 8 = 2x4 here P1 = 2, Q1  = 4. Then (P1 + 1)(Q 1  + 1) = (2 + 1)(4 +1) = 15 second odd composite A(3) = 12 = 2x6 => 3x7 = 21  third odd composite A(4) = 16 = 2x8 = 4x4 => 5x5 =25 fourth odd composite and 3x9 = 27 fifth odd composite like that it continues. Based on this i found a methodology to
        sieve Primes in given range 'a' to 'b' a,b both odd numbers. Write all the odd composite between 'a' to 'b' Find 'm' even such that 3(m + 1) <= 'a' < 3(u + 3) Now v= u/2 Calculate 4v = P1Q1 = P2Q2 = ..... = PkQk Then omit the following numbers  (Pk  + 1) (Q k  + 1),   (Pk-1 + 1)(Q k-1  + 1), ......    (P1 + 1)(Q 1  + 1)  which falls in the range 'a' to 'b' Now increment 'k' and repeat the procedure till (Pk  + 1) (Q k  + 1) > b. Now the remaining numbers are prime. 

        With regards
        yourskadhir@...

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