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Re: [PrimeNumbers] 2 AP22 with the same difference

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  • 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 1 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 2 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 3 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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