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Re: Sum of reciprocals of 'superprimes'

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  • djbroadhurst
    ... http://math.stackexchange.com/questions/430369/what-is-the-value-of-sum-limits-i-1-infty-frac1p-p-i-where-p-i-i claims a value between 1.0444 and 1.0478.
    Message 1 of 4 , Jul 16, 2013
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      --- In primenumbers@yahoogroups.com, James Merickel <moralforce120@...> wrote:

      > I don't have this value (Superprimes--by wikipedia
      > definition--are primes of prime index, but I don't
      > think much of the term) and it will take a while
      > to get close on my machine

      http://math.stackexchange.com/questions/430369/what-is-the-value-of-sum-limits-i-1-infty-frac1p-p-i-where-p-i-i

      claims a value between 1.0444 and 1.0478.

      David
    • djbroadhurst
      ... James did not give a refrence. As far as I can determine this result was first obtained by Richard Fischer, late in the evening of 17 October 2004.
      Message 2 of 4 , Jul 19, 2013
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        --- In primenumbers@yahoogroups.com,
        James Merickel <moralforce120@...> wrote:

        > The first value over 1 (The sum is pretty close) occurs with term
        > 148189304, the reciprocal of the 3081648379th prime (73898684653).

        James did not give a refrence. As far as I can determine
        this result was first obtained by Richard Fischer,
        late in the evening of 17 October 2004.

        http://229540.forumromanum.com/member/forum/forum.php?action=std_show&entryid=1089500177&USER=user_229540&threadid=2

        >> Seit gestern spät abends kenne ich nun die Stelle,
        wo die Summe 1 überschritten wird.
        Dies ist bei der 148189304. prim-ten Primzahl.
        n = 148189304
        p(n) = 3081648359
        p(p(n)) = 73898684653
        Die Summe nach der Addition von 1/73898684653 lautet: 1.0000000000089
        Die Extrapolationrechnung an dieser Stelle ergibt 1.043204331...
        Wahrscheinlich dürfte die wahre Summe (bis n = unendlich) bis zur zweiten 4 korrekt sein, also 1.043204...
        Gruss Richard <<

        David
      • djbroadhurst
        PS: Jens confirmed and extended this result in 2005: http://tech.groups.yahoo.com/group/primenumbers/message/17241
        Message 3 of 4 , Jul 19, 2013
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          PS: Jens confirmed and extended this result in 2005:
          http://tech.groups.yahoo.com/group/primenumbers/message/17241

          --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:

          > --- In primenumbers@yahoogroups.com,
          > James Merickel <moralforce120@> wrote:
          >
          > > The first value over 1 (The sum is pretty close) occurs with term
          > > 148189304, the reciprocal of the 3081648379th prime (73898684653).
          >
          > James did not give a refrence. As far as I can determine
          > this result was first obtained by Richard Fischer,
          > late in the evening of 17 October 2004.
          >
          > http://229540.forumromanum.com/member/forum/forum.php?action=std_show&entryid=1089500177&USER=user_229540&threadid=2
          >
          > >> Seit gestern spät abends kenne ich nun die Stelle,
          > wo die Summe 1 überschritten wird.
          > Dies ist bei der 148189304. prim-ten Primzahl.
          > n = 148189304
          > p(n) = 3081648359
          > p(p(n)) = 73898684653
          > Die Summe nach der Addition von 1/73898684653 lautet: 1.0000000000089
          > Die Extrapolationrechnung an dieser Stelle ergibt 1.043204331...
          > Wahrscheinlich dürfte die wahre Summe (bis n = unendlich) bis zur zweiten 4 korrekt sein, also 1.043204...
          > Gruss Richard <<
          >
          > David
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