Loading ...
Sorry, an error occurred while loading the content.

Re: Brun's constants

Expand Messages
  • Adam
    (Shoot, I keep sending to the individual, not the group. This is the only list I am on that defaults to individual responses, rather than group responses.) A
    Message 1 of 5 , May 28 8:29 AM
    • 0 Attachment
      (Shoot, I keep sending to the individual, not the group. This is the
      only list I am on that defaults to individual responses, rather than
      group responses.)

      A different take on data analysis.

      If you plot sum of inverses versus product((p-1)/(p-2)) you get an
      almost perfect linear fit, with correlation coefficient r=1-(3*10^-
      5). For n=50 data points that's a pretty good fit.

      Adam

      --- In primenumbers@yahoogroups.com, "Mark Underwood" <marku606@y...>
      wrote:
      >
      > I just surveyed the gold mine of data which Adam presented below,
      the
      > quality of which went over and above what I was hoping for. And
      even
      > more that that I am elated that it so strongly corresponds with
      what
      > I was heuristically hanging my hat on. I was expecting somewhat
      > tighter results but I am still very happy with these results which
      > confirm my heuristics to a great degree of confidence. I have
      grouped
      > Adam's data to show how it falls into the pattern.
      >
      > To five significant digits (rounded),
      >
      > Primes separated by 2^n, that is 2, 4, 8, 16, 32, and 64 have
      inverse
      > sums of .017153, .017049, .017045, .016999, .016973 and .016981
      > respectively. That is about .0170. Call this B' for Brun
      > modified.
      >
      > Primes separated by (2^n) x (3^m), that is 6, 12, 18, 24, 36, 48,
      54,
      > 72 and 96 have inverse sums
      >
      of .033958, .034063, .034055, .034087, .034211, .034002, .034026, .03
      > 3998, and .034012 respectively. That is about .0340. The
      expected
      > result is B' x (3-1)/(3-2) = .0340
      >
      > Primes separated by (2^n) x (5^m), that is 10, 20, 40, 50, 80, and
      > 100 have inverse sums
      > of .022674, .022699, .022724, .022635, .022733, and .022683
      > respectively. That is about .0227. The expected result is B' x (5-
      1)/
      > (5-2) = .0227
      >
      > Primes separated by (2^n) x (7^m), that is 14, 28, 56, and 98 have
      > inverse sums of .020386, .020450, .020338 and 020352 respectively.
      > That is about .0204. The expected result is B' x ((7-1)/(7-2)
      > = .0204.
      >
      > Primes separated by (2^n) x (11^m), that is 22, 44 and 88 have
      > inverse sums of .018834, .018879 and .018920 respectively. That is
      > about .0189. The expected result is B' x (11-1)/(11-2) = .0189.
      >
      > Primes separated by (2^n) x (13^m), that is 26 and 52 have inverse
      > sums of .018643 and .018612 respectively. That is about .0186. The
      > expected result is B' x (13-1)/(13-2) = .0185.
      >
      > And so on.
      >
      > Primes separated by (2^n) x (3^m) x (5^r), that is 30, 60 and 90
      have
      > inverse sums of .045331, .045362 and .045378 respectively. That is
      > about .0453. The expected result is B' x (3-1)/(3-2) x (5-1)/(5-2)
      > = .0453.
      >
      > Primes separated by (2^n) x (3^m) x (7^r), that is 42 and 84 have
      > inverse sums of .040838 and .040786. That is about .0408. The
      > expected result is B' x (3-1)/(3-2) x (7-1)/(7-2) = .0408
      >
      > Primes separated by (2^n) x (5^m) x (7^r), that is 70 has an
      inverse
      > sum of about 0273. The expected result is B' x (5-1)/(5-2) x (7-1)/
      (7-
      > 2) = .272
      >
      > Good enough for me! Thanks Adam!
      >
      > Mark
      >
      >
      >
      >
      > --- In primenumbers@yahoogroups.com, "Adam" <a_math_guy@y...> wrote:
      > > Hmmm, I posted some data (somewhere) but it isn't showing up on
      the
      > > list. Did I send it to somebody's email, instead of the list?
      If
      > > so, please post for me. Anyway, it defintely looked like there
      > were
      > > patterns (wait, I might have saved the data)......here ya' go,
      sum
      > of
      > > (1/p) for p>10^6 and p<20 million when (p+k) is also prime,
      output
      > is
      > > (k,sum):
      > >
      > >
      > > 2, .01715295754
      > > 4, .01704852064
      > > 6, .03395755937
      > > 8, .01704518900
      > > 10, .02267391029
      > > 12, .03406331599
      > > 14, .02038594306
      > > 16, .01699946972
      > > 18, .03405521233
      > > 20, .02269903011
      > > 22, .01883421188
      > > 24, .03408741833
      > > 26, .01864293196
      > > 28, .02044996718
      > > 30, .04533148052
      > > 32, .01697346702
      > > 34, .01813592253
      > > 36, .03421070761
      > > 38, .01801860557
      > > 40, .02272393752
      > > 42, .04083825580
      > > 44, .01887947490
      > > 46, .01780359233
      > > 48, .03400242989
      > > 50, .02263451790
      > > 52, .01861219959
      > > 54, .03402561886
      > > 56, .02033752356
      > > 58, .01760087352
      > > 60, .04536210827
      > > 62, .01762533361
      > > 64, .01698090345
      > > 66, .03778604223
      > > 68, .01808951646
      > > 70, .02731960955
      > > 72, .03398823253
      > > 74, .01746662755
      > > 76, .01806215382
      > > 78, .03717973980
      > > 80, .02273260129
      > > 82, .01748673074
      > > 84, .04078555194
      > > 86, .01747334842
      > > 88, .01892019678
      > > 90, .04537844859
      > > 92, .01780730452
      > > 94, .01745072896
      > > 96, .03401206908
      > > 98, .02035243350
      > > 100, .02268286529
      > >
      > > Adam
      > >
    Your message has been successfully submitted and would be delivered to recipients shortly.