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

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  • ratwain
    ... 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. ...
    Message 1 of 22 , Oct 1, 2003
      --- In primenumbers@yahoogroups.com, "Zak Seidov" <seidovzf@y...> wrote:
      > 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,
      > 43,46,49,52,55,58,61,64,67,70,73,76,
      > 79,82,83,85,88,89,91,94,97,100,101,
      > 103,106,107,109,112,113,115,118,121,
      > 124,127,130,131,133,136,137,139,142,
      > 145,148,149,151,154,157,158,160,163,
      > 166,167,169,172,173,175,178,179,181,
      > 184,187,190,191,193,194,196,197,199
      >
      > 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
    • Zak Seidov
      ... All of that is lost (unless, Zak, you want to cut and paste and post ... Also I did not receive this! ... message, ... a ... post ... is ... the
      Message 2 of 22 , Oct 1, 2003
        >
        All of that is lost (unless, Zak, you want to cut and paste and post
        > for me).

        Also I did not receive this!

        --- In primenumbers@yahoogroups.com, "Adam" <a_math_guy@y...> wrote:
        > Shoot, last Friday I sent this fairly long response to this
        message,
        > which isn't posted here, because this is the only yahoo group I am
        a
        > part of for which "reply" defaults to the poster and not the whole
        > group.
        >
        > Sigh.
        >
        > All of that is lost (unless, Zak, you want to cut and paste and
        post
        > for me).
        >
        > Anyway, in that intended post I reasoned that the +1 mod 3 residue
        is
        > impossible, and that the other residues should happen, if searched
        > deep enough. As a proof of concept I list the following p values
        > that have [p(n+2)-p(n+1)]*[p(n+1)-p(n)]/4=194:
        >
        > p(n) from {1374538987513, 1374682325293, 1374446799889,
        > 1374488450563, 1374402345229, 1374600127909, 1374415063363,
        > 1374735847123, 1374546494143, 1374429446293, 1374472611199,
        > 1374810758323, 1374520532653, 1374801030553, 1374811265683,
        > 1374679474459, 1374724625689, 1374637137469, 1374748849219,
        > 1374445952569, 1374505434703, 1374691452103, 1374702725089,
        > 1374770337163, 1374598127599, 1374757434889, 1374584932489,
        > 1374696888559, 1374406298959, 1374724738573, 1374429655633,
        > 1374721538119, 1374775427743, 1374475007509, 1374593340163,
        > 1374523688173, 1374670818133, 1374665432893, 1374734529673,
        > 1374786320833, 1374515455399}
        >
        > I believe that many, if not all, of the other -1 mod 3 numbers on
        the
        > list would also 'fall' to a search of sufficient depth.
        >
        > Adam
        >
        > --- In primenumbers@yahoogroups.com, "Zak Seidov" <seidovzf@y...>
        > wrote:
        > > Just sent to OEIS:
        > >
        > > 4,7,10,13,16,19,22,25,28,31,34,37,40,
        > > 43,46,49,52,55,58,61,64,67,70,73,76,
        > > 79,82,83,85,88,89,91,94,97,100,101,
        > > 103,106,107,109,112,113,115,118,121,
        > > 124,127,130,131,133,136,137,139,142,
        > > 145,148,149,151,154,157,158,160,163,
        > > 166,167,169,172,173,175,178,179,181,
        > > 184,187,190,191,193,194,196,197,199
        > >
        > > Numbers which can not be of form
        > > (p[n+2]-p[n+1])(p[n+1]-p[n])/4.
        > >
        > > Some cases are proved easily,
        > > some are only guesses checked
        > > for n<=2,000,000.
        > >
        > > SHANA TOVA!!
        > >
        > > Zak
      • 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 3 of 22 , Oct 1, 2003
          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
          7)Zak

          %%%%%%%%%%%%%%%%%%%%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,
          <skip>
          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),
          and
          neither of those can be the fingerprint for an admissible 3-tuple
          apart
          from {3,5,7} -> (2)*(2)/4 = 1.

          (2) is probably true because any {0,2,6n+2} is an admissible 3-tuple,
          and
          using the same heuristics as are behind the k-tuple conjecture, but
          based
          on the principles behind Dirichlet's Theorem. An arithmetic
          progression
          a+i.b can be created with known composites at a+i.p+1, and a+i.p+
          {3..6n+1}
          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
          because

          (00:34) gp > p=257987875972449177033341526073139
          257987875972449177033341526073139
          (00:35) gp > isprime(p)
          1
          (00:35) gp > q=nextprime(p+1)
          257987875972449177033341526073141
          (00:35) gp > r=nextprime(q+1)
          257987875972449177033341526073307
          (00:35) gp > (r-q)*(q-p)/4
          83

          (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...>
          wrote:
          > > 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,
          <skip>
          > >
          > > 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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