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Special Set of Eleven Consecutive Primes

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  • w_sindelar@juno.com
    Here is a set of 11 consecutive primes consisting of 8 contiguous subsets of 4 primes. Each subset has 3 distinct differences (gaps) between adjacent primes
    Message 1 of 5 , Jul 8, 2011
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      Here is a set of 11 consecutive primes consisting of 8 contiguous subsets
      of 4 primes. Each subset has 3 distinct differences (gaps) between
      adjacent primes and 4 distinct rightmost digits.
      (2213, 2221, 2237, 2239, 2243, 2251, 2267, 2269, 2273, 2281, 2287).
      For example, the first subset of 4 primes is (2213, 2221, 2237, 2239).
      The 3 distinct gaps are (8, 6, 2). The 4 distinct rightmost digits are
      (3, 1, 7, 9). The next contiguous subset is (2221, 2237, 2239, 2243) with
      distinct gaps (6, 2, 4) and distinct rightmost digits (1, 7, 9, 3). And
      so on for 8 subsets.
      Can theory hint where on the number line one is most likely to find sets
      like this? Anyone care to try to find a bigger one? Thanks folks.
      Bill Sindelar
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    • djbroadhurst
      ... Why stop at 11? Here is the same sort of thing, but with 12 consecutive primes: [2123971, 2123983, 2123999, 2124007, 2124011, 2124013, 2124019, 2124037,
      Message 2 of 5 , Jul 8, 2011
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        --- In primenumbers@yahoogroups.com,
        w_sindelar@... wrote:

        > Here is a set of 11 consecutive primes consisting
        > of 8 contiguous subsets of 4 primes. Each subset
        > has 3 distinct differences (gaps) between adjacent primes
        > and 4 distinct rightmost digits.
        > (2213, 2221, 2237, 2239,
        > 2243, 2251, 2267, 2269,
        > 2273, 2281, 2287).

        Why stop at 11? Here is the same sort of thing,
        but with 12 consecutive primes:

        [2123971, 2123983, 2123999, 2124007,
        2124011, 2124013, 2124019, 2124037,
        2124041, 2124043, 2124049, 2124127]

        David
      • Jens Kruse Andersen
        ... The ending digit must go through a cycle of length 4, for example the above 3179 3179 ... If the gap requirement is satisfied modulo 10 for the cycle then
        Message 3 of 5 , Jul 8, 2011
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          Bill Sindelar wrote:
          > Here is a set of 11 consecutive primes consisting of 8 contiguous subsets
          > of 4 primes. Each subset has 3 distinct differences (gaps) between
          > adjacent primes and 4 distinct rightmost digits.
          > (2213, 2221, 2237, 2239, 2243, 2251, 2267, 2269, 2273, 2281, 2287).

          > Can theory hint where on the number line one is most likely to find sets
          > like this? Anyone care to try to find a bigger one? Thanks folks.

          The ending digit must go through a cycle of length 4, for example the above
          3179 3179 ...

          If the gap requirement is satisfied modulo 10 for the cycle then
          it will automatically be satisfied for the whole sequence.
          The record setters up to 10^12 except length 19 are all of this type.

          12:
          2123971
          2123983
          2123999
          2124007
          2124011
          2124013
          2124019
          2124037
          2124041
          2124043
          2124049
          2124127

          13:
          66945301
          66945313
          66945349
          66945407
          66945421
          66945443
          66945449
          66945467
          66945491
          66945503
          66945509
          66945517
          66945521

          14:
          66945301
          66945313
          66945349
          66945407
          66945421
          66945443
          66945449
          66945467
          66945491
          66945503
          66945509
          66945517
          66945521
          66945553

          15:
          184171621
          184171643
          184171679
          184171697
          184171711
          184171733
          184171739
          184171747
          184171751
          184171753
          184171759
          184171777
          184171781
          184171843
          184171849

          16:
          2493412841
          2493412847
          2493412849
          2493412853
          2493412891
          2493412937
          2493412939
          2493413023
          2493413051
          2493413057
          2493413059
          2493413063
          2493413071
          2493413077
          2493413099
          2493413113

          17:
          2732087209
          2732087227
          2732087231
          2732087263
          2732087279
          2732087317
          2732087321
          2732087333
          2732087359
          2732087447
          2732087461
          2732087483
          2732087489
          2732087537
          2732087551
          2732087563
          2732087569

          18:
          10206934519
          10206934523
          10206934541
          10206934567
          10206934579
          10206934583
          10206934601
          10206934607
          10206934609
          10206934613
          10206934631
          10206934697
          10206934709
          10206934723
          10206934741
          10206934757
          10206934769
          10206934783

          19:
          554804585227
          554804585233
          554804585299
          554804585311
          554804585347
          554804585393
          554804585449
          554804585471
          554804585507
          554804585513
          554804585579
          554804585651
          554804585737
          554804585743
          554804585779
          554804585831
          554804585837
          554804585873
          554804585899

          20:
          692245399661
          692245399687
          692245399709
          692245399723
          692245399751
          692245399787
          692245399789
          692245399793
          692245399801
          692245399837
          692245399849
          692245399873
          692245399891
          692245399907
          692245399919
          692245399943
          692245399991
          692245399997
          692245400009
          692245400023

          --
          Jens Kruse Andersen
        • w_sindelar@juno.com
          Thank you David, and thank you Jens.You never cease to astound me. In the 9 examples you so quickly calculated, the sets 12, 13, 14 and 15 have the pattern (1,
          Message 4 of 5 , Jul 9, 2011
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            Thank you David, and thank you Jens.You never cease to astound me. In the 9 examples you so quickly calculated, the sets 12, 13, 14 and 15 have the pattern (1, 3, 9, 7) for the beginning set of 4 consecutive primes. For sets 16 and 20 the pattern is (1, 7, 9, 3). For set 11 the pattern is (3, 1, 7, 9). For set 19 the pattern is ((7, 3, 9, 1). For set 18 the pattern is (9, 3, 1, 7). For set 17 the pattern is (9, 7, 1, 3).


            There are 24 possible patterns into which the 4 digits 1, 3, 7 and 9 can be arranged, so that each pattern contains distinct digits. The sets 11 to 20 account for six.


            The obvious question comes up. Is it possible to find additional sets 21, 22, 23, 24...N, such that for the sets 11 to N all 24 possible patterns are used in the begining sets of 4 consecutive primes?


            Bill Sindelar



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          • djbroadhurst
            ... That was indeed my methodology, adopted in order quickly to attain length 12, with gaps in Z/Z10 given by [2, 6, 8, 4, 2, 6, 8, 4, 2, 6, 8] It s truly
            Message 5 of 5 , Jul 9, 2011
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              --- In primenumbers@yahoogroups.com,
              "Jens Kruse Andersen" <jens.k.a@...> wrote:

              > The ending digit must go through a cycle of length 4 ...
              > If the gap requirement is satisfied modulo 10 for the cycle then
              > it will automatically be satisfied for the whole sequence.

              That was indeed my methodology, adopted
              in order quickly to attain length 12,
              with gaps in Z/Z10 given by

              [2, 6, 8, 4, 2, 6, 8, 4, 2, 6, 8]

              It's truly impressive that Jens attained length 20.
              Moreover, he had the foresight to study gaps in Z.
              Videlicit:

              > 19:
              > 554804585227
              > 554804585233
              > 554804585299
              > 554804585311
              > 554804585347
              > 554804585393
              > 554804585449
              > 554804585471
              > 554804585507
              > 554804585513
              > 554804585579
              > 554804585651
              > 554804585737
              > 554804585743
              > 554804585779
              > 554804585831
              > 554804585837
              > 554804585873
              > 554804585899

              where the gaps in Z/Z10 are

              [6, 6, 2, 6, 6, 6, 2, 6, 6, 6, 2, 6, 6, 6, 2, 6, 6, 6]

              but are still OK in Z:

              [6, 66, 12, 36, 46, 56, 22, 36, 6, 66, 72, 86, 6, 36, 52, 6, 36, 26]

              David (well and truly outdone by Jens, as so often before)
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