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Re: (p[n+2]-p[n+1])(p[n+1]-p[n])/4

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  • Zak Seidov
    Shalom R. A. Twain, 1) OEIS assumes scientific level or smth like, according which you may reject most of seqs in it... 2) guesses in any sci sense are
    Message 1 of 22 , Oct 1, 2003
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      Shalom R. A. Twain,

      1) OEIS assumes "scientific level" or smth like,
      according which you may reject most of seqs in it...
      2) "guesses" in any sci sense are more important
      than "established facts"
      3) my this particular seq was labeled as "bad sequence"
      (see copy of Neil's message -
      sorry if someone reads it twise or trice)
      and I guess that it will be removed from OEIS
      4)in my (weakest) exuse -
      i am only amateur and fan
      5)"knowledge causes only sadness"
      may i only add: "extra" knowledge
      6)My deepest respects to all NT gurus

      %%%%%%%%%%%%%%%%%%%%end of copy of Neil's message%%%%%%%%%%%%
      From: N. J. A. Sloane [njas@...]
      Sent: 1:34 02/10/03
      Subject: bad sequence
      this just came in from a correspondent:
      ID Number: A087656
      URL: http://www.research.att.com/projects/OEIS?Anum=A087656
      Sequence: 4,7,10,13,16,19,22,25,28,31,34,37,40,43,46,49,52,55,58,61,
      Name: Numbers that can not be of form (p[n+2]-p[n+1])(p[n+1]-p
      [n])/4, where p[i] is i-th prime.
      Comments: Some cases are proved easily, some are only guesses checked
      only up n<=2000000.
      See also: Adjacent sequences: A087653 A087654 A087655 this_sequence
      A087657 A087658 A087659
      Keywords: nonn,new
      Offset: 4
      Author(s): Zak Seidov (seidovzf(AT)yahoo.com), Sep 26 2003

      The sequence is bogus because:

      1) All numbers of the form 3n+1 (>1) are in it
      2) Subject to heuristically safe conjecture, no number not of the form
      3n+1 can be in it.

      In more detail -

      (1) is true as 12n+4 factors into either(3a+1)(3b+1) or (3a+2)(3b+2),
      neither of those can be the fingerprint for an admissible 3-tuple
      from {3,5,7} -> (2)*(2)/4 = 1.

      (2) is probably true because any {0,2,6n+2} is an admissible 3-tuple,
      using the same heuristics as are behind the k-tuple conjecture, but
      on the principles behind Dirichlet's Theorem. An arithmetic
      a+i.b can be created with known composites at a+i.p+1, and a+i.p+
      using the chinese remainder theorem. This AP should have a density of
      prime k-tuples in proportion (by a fixed constant, realted to the
      multiplier b) with the number of tuplets that arbitrary integers would
      yield - which by Hardy & Littlewood's (2nd) conjecture is infinite.

      e.g. the first number not of the form 3n+1 in the list, 83 fails

      (00:34) gp > p=257987875972449177033341526073139
      (00:35) gp > isprime(p)
      (00:35) gp > q=nextprime(p+1)
      (00:35) gp > r=nextprime(q+1)
      (00:35) gp > (r-q)*(q-p)/4

      (not the smallest, I made a typo in my script, but a number jumped out
      within seconds anyway!)
      %%%%%%%%%%%%%%%%%%%%end of copy of Neil's message%%%%%%%%%%%%

      --- In primenumbers@yahoogroups.com, "ratwain" <ratwain@y...> wrote:
      > --- In primenumbers@yahoogroups.com, "Zak Seidov" <seidovzf@y...>
      > > Just sent to OEIS:
      > I hope you haven't. I don't really think the Online Encyclopedia
      of Integer Sequences was created for "guesses" -- your sequence
      doesn't belong there.
      > > 4,7,10,13,16,19,22,25,28,31,34,37,40,
      > >
      > > Numbers which can not be of form
      > > (p[n+2]-p[n+1])(p[n+1]-p[n])/4.
      > All numbers of the form 3n+1 are part of the set, and the rest
      aren't. I'm sure there are people on the list who can give you some
      more hints if you need them, so I'll just say "k-tuple conjecture."
      > R. A. Twain
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