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Re: [PrimeNumbers] 2 AP22 with the same difference

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  • 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 1 of 7 , Dec 29, 2008
      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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