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[PrimeNumbers] Re: primeth primes

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  • Jens Kruse Andersen
    ... My program agrees with Andrey s table. An extension shows his estimate for sum 1 is good (it s n=148189304): n p(n) p(p(n))
    Message 1 of 23 , Nov 25, 2005
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      Andrey Kulsha wrote:

      > reaches 1 for p(p(n)) near exp(25) (so n is about 145e6),
      > and converges to the value of 1.04...
      >
      > n p(n) p(p(n)) sum(1/p(p(n)))
      > 125791 1666589 26717767 0.978334479698888415
      ...
      > 6037968 105097561 2147483563 0.992403845704372124

      My program agrees with Andrey's table.
      An extension shows his estimate for sum>1 is good (it's n=148189304):

      n p(n) p(p(n)) sum(1/p(p(n)))
      10000000 179424673 3767321791 0.993782079779048
      20000000 373587883 8132396761 0.995553008182270
      30000000 573259391 12736743289 0.996528886890023
      40000000 776531401 17500350067 0.997196394520154
      50000000 982451653 22383430879 0.997700609931158
      60000000 1190494759 27363009697 0.998104122267683
      70000000 1400305337 32423573141 0.998439516689021
      80000000 1611623773 37553793989 0.998725881592459
      90000000 1824261409 42745397623 0.998975327892169
      100000000 2038074743 47991893393 0.999196010613537
      110000000 2252945251 53287844179 0.999393678625278
      120000000 2468776129 58629206579 0.999572530777542
      130000000 2685457421 64011376427 0.999735723538573
      140000000 2902958801 69432251137 0.999885688879310
      148189304 3081648379 73898684653 1.000000000008883
      150000000 3121238909 74889755137 1.000024339564021
      160000000 3340200037 80380156427 1.000153205905392
      170000000 3559788179 85901366807 1.000273531253509
      180000000 3780008329 91452739153 1.000386339616163
      190000000 4000846301 97033105289 1.000492482024935
      200000000 4222234741 102639953363 1.000592674128579

      --
      Jens Kruse Andersen
    • Adam
      Retraction: Whoops, somebody pointed out my integration mistake. My wife suggests I should say I was too busy talking to my wife or I just put that there to
      Message 2 of 23 , Nov 26, 2005
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        Retraction:

        Whoops, somebody pointed out my integration mistake.

        My wife suggests I should say "I was too busy talking to my wife"
        or "I just put that there to see if anyone was paying attention."
        All I can say in my own defense is don't integrate in your head
        while typing email and whoops classic intro calc mistake.

        Of course int(1/[x*log(x)^2]) should be -1/log(x) and not 1/(3*log(x)
        ^3). That would change the error term dramatically.

        Sorry for the mistake.

        Adam

        --- In primenumbers@yahoogroups.com, "Adam" <a_math_guy@y...> wrote:
        >
        > Message 17230 of 17232 indicates that pn>=n(log(n)+log(log(n))+1)
        so
        > we conclude that pn>n*log(n) and then "compose this with itself"
        to
        > get p(pn)>pn*log(pn)>n*log(n)*log[n*log(n)]>n*log(n)^2 so 1/p(pn)
        <1/
        > [n*log(n)^2].
        >
        > The tail end of the sum sum(1/p(pn),n=k+1..infinity) is bounded
        above
        > by sum(1/[n*log(n)^2],n=k+1..infinity) which is bounded above by
        > integral(1/[x*log(x)^2],x=k..infinity) which is 1/[3*log(k)^3].
        >
        > You add up the first four terms 1/p(p1)+1/p(p2)+1/p(p3)+1/p(p4)=1/p
        (2)
        > +1/p(3)+1/p(5)+1/p(7)=1/3+1/5+1/11+1/17<.68306596 and then you
        have
        > the error term of int(1/[x*log(x)^2],x=3..infinity)=1/[3*log(3)^3]
        > <.25138849 so the whole sum is <.68306596+.25138849=.93445445.
        >
        > For n=7 I get an upper bound of .814611570570018210618938079323
        >
        > Adam
        >
        > --- In primenumbers@yahoogroups.com, "Werner D. Sand"
        > <Theo.3.1415@w...> wrote:
        > >
        > > The sum of the reciprocal primeth primes p(p(n))
        > >
        > > s = 1/3 + 1/5 + 1/11 + 1/17 + 1/31 + … seems to
        > >
        > > converge to 1.
        > >
        >
      • Dick
        ... Well, obviously Milton has proven the twin prime conjecture to be false! -Dick
        Message 3 of 23 , Nov 26, 2005
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          --- In primenumbers@yahoogroups.com, "crgreathouse"
          <crgreathouse@g...> wrote:
          >
          > > Any series like this, where you select from an infinite number of
          > > primes, will diverge.

          > What about the reciprocals of the twin primes
          >
          > B=(1/3+1/5)+(1/5+1/7)+(1/(11)+1/(13))+(1/(17)+1/(19))+....
          >
          > which converges to about 1.902160583?

          Well, obviously Milton has proven the twin prime conjecture to be false!

          -Dick
        • Milton Brown
          Thanks for your comments. However, you are missing part of my message. Please include the entire message in the future. And try to think positively. Thanks.
          Message 4 of 23 , Nov 27, 2005
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            Thanks for your comments.

            However, you are missing part of my message.

            Please include the entire message in the future.
            And try to think positively.

            Thanks.

            Here it is:

            "As I indicated, this sum does not converge.

            The comparable sum to compare it to is the reciprocal of the squares.

            Consider to the denominators compared to the squares.

            3 5 11 17 31 41 59 67 83 10- 127 157 179 191 211 241 277 283 331 ...

            1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 289 384 361 400 ...

            After 191<196 the squares remain larger so 1/196 < 1/191

            Any series like this, where you select from an infinite number of primes,
            will diverge."

            Milton L. Brown

            > [Original Message]
            > From: Dick <richard042@...>
            > To: <primenumbers@yahoogroups.com>
            > Date: 11/26/2005 9:54:24 PM
            > Subject: [PrimeNumbers] Re: primeth primes
            >
            > --- In primenumbers@yahoogroups.com, "crgreathouse"
            > <crgreathouse@g...> wrote:
            > >
            > > > Any series like this, where you select from an infinite number of
            > > > primes, will diverge.
            >
            > > What about the reciprocals of the twin primes
            > >
            > > B=(1/3+1/5)+(1/5+1/7)+(1/(11)+1/(13))+(1/(17)+1/(19))+....
            > >
            > > which converges to about 1.902160583?
            >
            > Well, obviously Milton has proven the twin prime conjecture to be false!
            >
            > -Dick
            >
            >
            >
            >
            >
            >
            > Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
            > The Prime Pages : http://www.primepages.org/
            >
            >
            > Yahoo! Groups Links
            >
            >
            >
            >
            >
            >
          • Jens Kruse Andersen
            ... I used my own C program with the sieve of Eratosthenes. It took below a GHz hour with 3 MB ram. I have not analyzed whether the exact n is reliable after
            Message 5 of 23 , Nov 27, 2005
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              > > My program agrees with Andrey's table.
              > > An extension shows his estimate for sum>1 is good (it's n=148189304):

              Jose Ramón Brox brox wrote:

              > I was trying to compute "Werner's constant" up to 10^10.
              > It was running in PARI-GP for near three days and then
              > Windows crashed (blame it on Billy :P). How did you
              > do the computation so quickly? A fast processor,
              > a lot of memory, a nice computation trick?

              I used my own C program with the sieve of Eratosthenes.
              It took below a GHz hour with 3 MB ram.
              I have not analyzed whether the exact n is reliable after rounding errors.

              PARI/GP does not have functions suited to compute more small primes
              than fit in memory after increasing primelimit. nextprime(x) is too slow.

              --
              Jens Kruse Andersen
            • Dick Boland
              ... Hi Milton, I responded to crgreathouse s message which was a response to yours--your blather had already been excised. Are you suggesting I am responsible
              Message 6 of 23 , Nov 27, 2005
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                --- Milton Brown <miltbrown@...> wrote:
                > Thanks for your comments.
                > However, you are missing part of my message.
                > Please include the entire message in the future.
                > And try to think positively.

                Hi Milton,

                I responded to crgreathouse's message which was a response to
                yours--your blather had already been excised. Are you suggesting
                I am responsible to go back and reassemble a thread into a single
                historical document everytime I post inside that thread?

                > "As I indicated, this sum does not converge.
                > The comparable sum to compare it to is the reciprocal of the
                > squares.
                > Consider to the denominators compared to the squares.
                > 3 5 11 17 31 41 59 67 83 10- 127 157 179 191 211 241 277 283
                > 331 ...
                > 1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 289 384 361
                > 400 ...
                > After 191<196 the squares remain larger so 1/196 < 1/191
                > Any series like this, where you select from an infinite number
                > of primes, will diverge."

                Three things:

                1) Other posters prior to you provided sufficient proof to
                demonstrate that the sequence does indeed converge, and in fact
                the limit was calculated and posted.

                2) Zeta[1] diverges and Zeta[1+c] converges for any positive real
                value of c. As c ranges over 0+-->1, there are an infinite
                number of convergent series that can have denominators
                consistently larger than the squares on a 1 to 1 basis. You seem
                to be thinking that the pivot point between convergence and
                divergence is Zeta[2], corresponding to c=1, but it is not, the
                pivot point is Zeta[1], corresponding to c=0. There are infinite
                possibilities for convergent series between c=0 and c=1.

                3) What on God's green Earth makes you think that primeth primes
                and the squares progress more or less with the same magnitude
                when followed on a one to one basis?

                Everyone knows your village idiot routine is just an act
                Milton--why do you bother to keep it up?

                Regards

                Dick Boland

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              • Kermit Rose
                From: Milton Brown Date: 11/25/05 17:57:22 To: crgreathouse; primenumbers@yahoogroups.com Subject: RE: [PrimeNumbers] Re: primeth primes As I indicated, this
                Message 7 of 23 , Nov 27, 2005
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                  From: Milton Brown
                  Date: 11/25/05 17:57:22
                  To: crgreathouse; primenumbers@yahoogroups.com
                  Subject: RE: [PrimeNumbers] Re: primeth primes


                  As I indicated, this sum does not converge.

                  The comparable sum to compare it to is the reciprocal of the squares.

                  Consider to the denominators compared to the squares.

                  3 5 11 17 31 41 59 67 83 10- 127 157 179 191 211 241 277 283 331 ...

                  1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 289 384 361 400 ...

                  After 191<196 the squares remain larger so 1/196 < 1/191

                  Any series like this, where you select from an infinite number of primes,
                  will diverge.

                  Milton L. Brown


                  Milton, I'm not sure what you are saying here.

                  Suppose I select primes by the following rule.

                  Pick not more than 10 primes between consecutive squares to get the
                  sequence p1, p2, p3,...

                  Then would you expect the series

                  1 + 1/P1 + 1/P2 + 1/P3 + .... to converge or diverge?

                  It's easy to prove that it will converge.

                  [Non-text portions of this message have been removed]
                • Jens Kruse Andersen
                  ... Based on the above (the entire message quoted to please Milton :-), I guess he assumes the sum of the reciprocals of squares diverges, and then he thinks
                  Message 8 of 23 , Nov 27, 2005
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                    Milton wrote:

                    > As I indicated, this sum does not converge.
                    >
                    > The comparable sum to compare it to is the reciprocal of the squares.
                    >
                    > Consider to the denominators compared to the squares.
                    >
                    > 3 5 11 17 31 41 59 67 83 10- 127 157 179 191 211 241 277 283 331 ...
                    >
                    > 1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 289 384 361 400 ...
                    >
                    > After 191<196 the squares remain larger so 1/196 < 1/191
                    >
                    > Any series like this, where you select from an infinite number of primes,
                    > will diverge.
                    >
                    > Milton L. Brown

                    Based on the above (the entire message quoted to please Milton :-),
                    I guess he assumes the sum of the reciprocals of squares diverges,
                    and then he thinks the discussed primeth sum must also diverge
                    because the terms soon become larger.
                    One minor problem with that argument:
                    The sum of reciprocal of squares converges to pi^2/6.

                    I guess Milton was thinking of the harmonic series
                    1 + 1/2 + 1/3 + 1/4 + ... which is famous for diverging.
                    (The pi^2/6 sum is also a little famous.)

                    I wonder whether I'm better at mind reading than Milton is at maths ;-)

                    --
                    Jens Kruse Andersen
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