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twin primes and 2modp

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  • Robert
    I have been having a mathematical long weekend, first the prime race 1mod3 and 2 mod3 which went nowhere new, and now, 2modp, which will probably also go
    Message 1 of 3 , May 1, 2006
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      I have been having a mathematical long weekend, first the prime race
      1mod3 and 2 mod3 which went nowhere new, and now, 2modp, which will
      probably also go nowhere. I beg your patience and indulgence a second
      time and ask for help.

      Can the twin primes conjecture be stated another way?

      It seems to me that possible that the following conjecture is equivalent.

      "Every prime p, greater than 3, is 2mod(at least one prime less than p)"

      It seems to hold for all primes up to prime=250031. I haven't checked
      further.

      Other than twin primes discussed below, in the range 5-250031 the
      prime 239119 gives the highest first 2modp as 2mod437.
      The first instance of non twin prime 2modp with highest first 2modp
      for p= 3,5,7..97 are
      11,37,79,211,223,631,439,853,1249,1459,1741,2503,2281,3433,5779,4663,
      5431,4759,7741,16573,8929,8053,12373,13291. This series is not listed
      at Sloane's.

      I did some fiddling about with the conjecture:

      If the conjecture holds then:

      1. Take consecutive primes with p0 (>2), p1, and the odd number x = p1+2

      2. Then x==2mod(p1)

      3. For x prime, p1 is not 0mod(any prime smaller than p1) by
      definition of prime numbers

      4. Then x is not 2mod(any other prime except p1)

      5. An odd number x (not necessarily prime) can be 0mod3, 1mod3 or 2mod3,

      6. then the chances of x not being 2mod3 are 2/3 or (3-1)/3

      7. The chances of x not being 2modp are (p-1)/p

      8. Cumulatively, it appears that the chance, Ca, of x not being
      2mod(any prime less than p1) would be given by:

      Ca = the product (for primes from p=3 to p=p0) of (p-1)/p

      9. If x is prime, then the odds are different, as x cannot be 0mod3,
      0modp etc

      10. Therefore the chance that x is not 2mod3 is 1/2 is (3-2)/(3-1)

      11. chance that x is not 2modp are (p-2)/(p-1)

      12. And cumulatively, the chance Cb of a prime x is given by

      Cb= the product (for primes from p=3 to p=p(0) of (p-2)/(p-1)

      Note: These calculations seem to be heading towards the
      Hardy-Littlewood twin prime constant

      c2 = product (from p=3 upward) of (p-2)/(p-1)*p/(p-1)

      I want to look into this further, but I want to make sure that this
      conjecture as stated is not well known before I do so. After
      yesterday's debacle concerning prime races, I have gotten shy about
      going down well trodden paths.

      Regards

      Robert Smith
    • Phil Carmody
      ... That s equivalent to: for every prime p 3, p-2 has at least one prime factor and thus is always satisfied. Phil () ASCII ribbon campaign ()
      Message 2 of 3 , May 1, 2006
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        --- Robert <rw.smith@...> wrote:
        > "Every prime p, greater than 3, is 2mod(at least one prime less than p)"

        That's equivalent to:
        for every prime p>3, p-2 has at least one prime factor
        and thus is always satisfied.

        Phil


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      • Robert
        ... than p) ... Phil, as always, to the rescue. Good man! Think I ll stick to prime searching rather than mathematics. I am not yet at class 101 stage, and it
        Message 3 of 3 , May 1, 2006
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          --- In primenumbers@yahoogroups.com, Phil Carmody <thefatphil@...> wrote:
          >
          > --- Robert <rw.smith@...> wrote:
          > > "Every prime p, greater than 3, is 2mod(at least one prime less
          than p)"
          >
          > That's equivalent to:
          > for every prime p>3, p-2 has at least one prime factor
          > and thus is always satisfied.
          >
          > Phil

          Phil, as always, to the rescue. Good man!

          Think I'll stick to prime searching rather than mathematics. I am not
          yet at class 101 stage, and it shows. Still, managed to waste 3 hours
          even if non-productively.

          For what its worth, the jumping champions (refer to top message) were
          as below.

          Regards

          Robert

          11 3
          37 5
          79 7
          211 11
          223 13
          439 19
          853 23
          1249 29
          1459 31
          1741 37
          2281 43
          3433 47
          4663 59
          4759 67
          7741 71
          8053 83
          12373 89
          12829 101
          13591 107
          14281 109
          17401 127
          18211 131
          22501 149
          25219 151
          28201 163
          32233 167
          32401 179
          38011 191
          43933 197
          47563 199
          52963 211
          53299 223
          53359 229
          57601 239
          60493 241
          68023 251
          71191 257
          72901 269
          77839 277
          86269 281
          91711 293
          95479 307
          99223 313
          114859 331
          129553 353
          146833 359
          158803 379
          159199 397
          164011 401
          176401 419
          194083 421
          194479 439
          205111 443
          227431 467
          239023 479
          239119 487
          253993 499
          263071 503
          275371 509
          284989 521
          297589 523
          307831 541
          321091 547
          324901 569
          352309 571
          356311 587
          367069 593
          367189 599
          378223 613
          396733 617
          407923 619
          416023 643
          439471 653
          463363 661
          497011 701
          531343 719
          553249 727
          571369 743
          585163 757
          622123 769
          650851 787
          679843 797
          680431 811
          680611 821
          680623 823
          720703 839
          756853 863
          777901 877
          804511 887
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