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Re: [PrimeNumbers] 2 AP22 with the same difference

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  • Jens Kruse Andersen
    ... Congratulations! That is very impressive. http://hjem.get2net.dk/jka/math/aprecords.htm is updated. With the method in
    Message 1 of 7 , Dec 1, 2008
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      Jarek wrote:
      > For over a week I have been using 40-60 64-bit computers in an attempt
      > to find 2 AP22 with common difference. I have selected 43# as the
      > common difference. While the first AP22 was found early in the search,
      > the second one appeared today. It also improves the current AP22 record.
      >
      > First terms of progressions are given
      > 4 93490858594661729 AP22 !!!!!
      > 21 1351906725737537399 AP22 !!!!!

      Congratulations! That is very impressive.
      http://hjem.get2net.dk/jka/math/aprecords.htm is updated.

      With the method in
      http://tech.groups.yahoo.com/group/primenumbers/message/19719
      we hit the 44 primes for all 128 additive combinations of the
      8 distinct numbers: 860067786148634879, 629207933571437835,
      43#/2, 43#, 3/2*43#, 2*43#, 5/2*43#, 3*43#.

      --
      Jens Kruse Andersen
    • Mark Underwood
      ... Yes that is quite something, and with such a great application to the additive combination problem. Meanwhile, I m about 90 percent done my run with
      Message 2 of 7 , Dec 1, 2008
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        --- In primenumbers@yahoogroups.com, "Jens Kruse Andersen" <jens.k.a@...> wrote:
        >
        > Congratulations! That is very impressive.
        > http://hjem.get2net.dk/jka/math/aprecords.htm is updated.
        >
        > With the method in
        > http://tech.groups.yahoo.com/group/primenumbers/message/19719
        > we hit the 44 primes for all 128 additive combinations of the
        > 8 distinct numbers: 860067786148634879, 629207933571437835,
        > 43#/2, 43#, 3/2*43#, 2*43#, 5/2*43#, 3*43#.
        >

        Yes that is quite something, and with such a great application to the additive combination
        problem. Meanwhile, I'm about 90 percent done my run with finding seven relatively teenie
        weenie numbers up to 1365. :) No luck getting even another 63 primer, let alone a 64 primer.
        Oh well. But let me kick back and bask in the glow of Jens' 64 and 128 all-prime greatest
        hits.

        Mark
      • Mark Underwood
        ... primer. Patience would have been a virtue. If I had gone just a little further, to 1371, I would have found my first solution. A couple of days ago I
        Message 3 of 7 , Dec 16, 2008
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          --- In primenumbers@yahoogroups.com, "Mark Underwood" <mark.underwood@...>
          wrote:
          >(snip) Meanwhile, I'm about 90 percent done my run with finding seven relatively teenie
          > weenie numbers up to 1365. :) No luck getting even another 63 primer, let alone a 64
          primer.

          Patience would have been a virtue. If I had gone just a little further, to 1371,
          I would have found my first solution. A couple of days ago I started
          on the problem again from another angle, which resulted in a simpler
          program with what seems like twice the efficiency.
          Within a day it found the seven numbers

          1371, 1170, 825, 756, 700, 615, 414

          whose 64 possible additive combinations are all prime.
          The largest prime produced is 5851.
          My search was exhaustive in terms of the largest prime produced,
          so all other solutions must have the largest prime greater than 5851.

          Alas there are only forty eight distinct primes out of a possible 64,
          but beggars can't be choosers. :)

          Mark
        • Devaraj Kandadai
          Easy to test by programming inpari as follows: {p(k)=(%1)+k*(%3)} where (%1) represents the first term and (%3)=43#. Hope this is useful to all Devaraj On
          Message 4 of 7 , Dec 17, 2008
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            Easy to test by programming inpari as follows: {p(k)=(%1)+k*(%3)} where
            (%1) represents the first term and (%3)=43#. Hope this is useful to all
            Devaraj

            On Mon, Dec 1, 2008 at 5:26 PM, jarek372000
            <jaroslaw.wroblewski@...>wrote:

            > For over a week I have been using 40-60 64-bit computers in an attempt
            > to find 2 AP22 with common difference. I have selected 43# as the
            > common difference. While the first AP22 was found early in the search,
            > the second one appeared today. It also improves the current AP22 record.
            >
            > Known AP20+ with difference 43#
            > First terms of progressions are given
            >
            > 1 9372688136871853
            > 2 11735227242889999
            > 3 76240762416222539 AP21
            > 4 93490858594661729 AP22 !!!!!
            > 5 114146343711853721
            > 6 135651360264848653
            > 7 195625258610971297 AP21
            > 8 202860934798777373
            > 9 223789213311833843
            > 10 224957853888083671 AP21
            > 11 251672116721153519
            > 12 325435306756257757
            > 13 333012166298058323
            > 14 338275337330536643
            > 15 381336957506808803
            > 16 485191591159166291
            > 17 493052729074838717 AP21
            > 18 630202728690264047
            > 19 1006137974832655813
            > 20 1296572696606684143
            > 21 1351906725737537399 AP22 !!!!!
            > 22 1376274082220856487
            >
            > Jarek
            >
            >
            >


            [Non-text portions of this message have been removed]
          • Devaraj Kandadai
            A problem in programming: Let us take the first 9372688136871853; how do we test whether it is a Mangammal prime or not? To form the seq A123239 (Mangammal
            Message 5 of 7 , Dec 28, 2008
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              A problem in programming: Let us take the first 9372688136871853; how do
              we test whether it is a Mangammal prime or not? To form the seq A123239
              (Mangammal primes)(refer OEIS) I used the following programme in pari
              : {p(k)=(3^k-2)/p}; for (k=1,p-1,print(p(k))). Now this is ok when the
              primes are small; WE CHECK WHETHER WE GET ATLEAST ONE RATIONAL INTEGER- in
              which case the p being tested is NOT a Mangammal prime. If the prime to be
              tested is as big as that indicated in the begining of this post what should
              we do? I will be glad if any one can suggest a short cut in pari.
              A.K. Devaraj

              On Mon, Dec 1, 2008 at 5:26 PM, jarek372000
              <jaroslaw.wroblewski@...>wrote:

              > For over a week I have been using 40-60 64-bit computers in an attempt
              > to find 2 AP22 with common difference. I have selected 43# as the
              > common difference. While the first AP22 was found early in the search,
              > the second one appeared today. It also improves the current AP22 record.
              >
              > Known AP20+ with difference 43#
              > First terms of progressions are given
              >
              > 1 9372688136871853
              > 2 11735227242889999
              > 3 76240762416222539 AP21
              > 4 93490858594661729 AP22 !!!!!
              > 5 114146343711853721
              > 6 135651360264848653
              > 7 195625258610971297 AP21
              > 8 202860934798777373
              > 9 223789213311833843
              > 10 224957853888083671 AP21
              > 11 251672116721153519
              > 12 325435306756257757
              > 13 333012166298058323
              > 14 338275337330536643
              > 15 381336957506808803
              > 16 485191591159166291
              > 17 493052729074838717 AP21
              > 18 630202728690264047
              > 19 1006137974832655813
              > 20 1296572696606684143
              > 21 1351906725737537399 AP22 !!!!!
              > 22 1376274082220856487
              >
              > Jarek
              >
              >
              >


              [Non-text portions of this message have been removed]
            • Jaroslaw Wroblewski
              I do not know pari, using Mathematica instead. Take p = 9372688136871853 Then p-1 = 9372688136871852 If I can factor p-1 into primes: {{2, 2}, {3, 1}, {13, 1},
              Message 6 of 7 , Dec 29, 2008
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                I do not know pari, using Mathematica instead.

                Take
                p = 9372688136871853

                Then
                p-1 = 9372688136871852

                If I can factor p-1 into primes:
                {{2, 2}, {3, 1}, {13, 1}, {2321923, 1}, {25875679, 1}}

                I can get full list of divisors of p-1:

                {1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 78, 156, 2321923, 4643846, 6965769,

                > 9287692, 13931538, 25875679, 27863076, 30184999, 51751358, 60369998,

                > 77627037, 90554997, 103502716, 120739996, 155254074, 181109994,

                > 310508148, 336383827, 362219988, 672767654, 1009151481, 1345535308,

                > 2018302962, 4036605924, 60081334210717, 120162668421434, 180244002632151,

                > 240325336842868, 360488005264302, 720976010528604, 781057344739321,

                > 1562114689478642, 2343172034217963, 3124229378957284, 4686344068435926,

                > 9372688136871852}


                The rank(r) of a residue r mod p is the smallest of the above divisors
                of p-1 such that

                r^rank(r) = 1 (mod p)

                Using PowerMod routine I am arriving at:
                rank(2) = 3124229378957284 = (p-1)/3
                rank(3) = 781057344739321 = (p-1)/12

                Since rank(2) IS divisible by rank(3), some power of 2 (mod p) is equal to 3.

                Hence p is a divisor of a number of the form 2^n-3, but it is rather
                hard to find a particular n (although it should be possible in this
                case given that p-1 has no huge prime factors).

                Since rank(3) is NOT divisible by rank(2), NO power of 3 (mod p) is equal to 2.

                Hence p is NOT a divisor of any number of the form 3^n-2.

                Jarek



                2008/12/29 Devaraj Kandadai <dkandadai@...>:
                > A problem in programming: Let us take the first 9372688136871853; how do
                > we test whether it is a Mangammal prime or not? To form the seq A123239
                > (Mangammal primes)(refer OEIS) I used the following programme in pari
                > : {p(k)=(3^k-2)/p}; for (k=1,p-1,print(p(k))). Now this is ok when the
                > primes are small; WE CHECK WHETHER WE GET ATLEAST ONE RATIONAL INTEGER- in
                > which case the p being tested is NOT a Mangammal prime. If the prime to be
                > tested is as big as that indicated in the begining of this post what should
                > we do? I will be glad if any one can suggest a short cut in pari.
                > A.K. Devaraj
                >
                > On Mon, Dec 1, 2008 at 5:26 PM, jarek372000 <jaroslaw.wroblewski@...>
                > wrote:
                >>
                >> For over a week I have been using 40-60 64-bit computers in an attempt
                >> to find 2 AP22 with common difference. I have selected 43# as the
                >> common difference. While the first AP22 was found early in the search,
                >> the second one appeared today. It also improves the current AP22 record.
                >>
                >> Known AP20+ with difference 43#
                >> First terms of progressions are given
                >>
                >> 1 9372688136871853
                >> 2 11735227242889999
                >> 3 76240762416222539 AP21
                >> 4 93490858594661729 AP22 !!!!!
                >> 5 114146343711853721
                >> 6 135651360264848653
                >> 7 195625258610971297 AP21
                >> 8 202860934798777373
                >> 9 223789213311833843
                >> 10 224957853888083671 AP21
                >> 11 251672116721153519
                >> 12 325435306756257757
                >> 13 333012166298058323
                >> 14 338275337330536643
                >> 15 381336957506808803
                >> 16 485191591159166291
                >> 17 493052729074838717 AP21
                >> 18 630202728690264047
                >> 19 1006137974832655813
                >> 20 1296572696606684143
                >> 21 1351906725737537399 AP22 !!!!!
                >> 22 1376274082220856487
                >>
                >> Jarek
                >>
                >>
                >
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