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Sum of reciprocals of 'superprimes'

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  • James Merickel
    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
    Message 1 of 4 , Jul 16 9:57 AM
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      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 using PARI, so if anybody wants to do this I think it's out there not computed yet. The first value over 1 (The sum is pretty close) occurs with term 148189304, the reciprocal of the 3081648379th prime (73898684653). A sharp enough estimate without actually summing is probably not hard, btw.
      JGM
    • 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 2 of 4 , Jul 16 3:25 PM
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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 3 of 4 , Jul 19 7:58 AM
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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 4 of 4 , Jul 19 8:08 AM
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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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