Loading ...
Sorry, an error occurred while loading the content.
 

Re: form of "Mills' constant"? [Puzzle 1000]

Expand Messages
  • djbroadhurst
    ... https://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind1204&L=nmbrthry&P=R67 gives portmanteau probable primes that encode 1000 smaller probable primes in
    Message 1 of 49 , Apr 10, 2012
      --- In primenumbers@yahoogroups.com, "djbroadhurst"
      <d.broadhurst@...> wrote:

      > Puzzle 709, where a 5045-digit probable prime, ending in
      > ...2963978865589225213070090428982660368263, neatly stores
      > the 1,656,992 decimal digits of 709 smaller probable primes.

      https://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind1204&L=nmbrthry&P=R67
      gives portmanteau probable primes that encode 1000 smaller
      probable primes in sequences that grow more slowly than does
      Warren Smith's original choice of sequence, which used a
      divisor function log(x)^2.

      In that harder case, I am now able to provide a
      7464-digit probable prime, p[1000], such that
      p[n-1] = floor(p[n]/log(p[n])^2)
      is probably prime for 1001 > n > 0, with p[0] = 2.

      http://physics.open.ac.uk/~dbroadhu/cert/x1000.zip
      gives the unpacking and BPSW tests, with these timings:
      3480873 digits unpacked in 1569 milliseconds
      1001 tests in 3187 seconds

      David Broadhurst, 10 April 2012
    • 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
        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@...

        [Non-text portions of this message have been removed]
      Your message has been successfully submitted and would be delivered to recipients shortly.