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

Pi product?

Expand Messages
  • Shane
    I was just playing around with factors of F(p), and noticed a curious onvergence towards pi. 2*5*13*37*73*89*... ___________________ = pi 1*4*12*36*72*88*...
    Message 1 of 16 , Oct 19, 2002
    • 0 Attachment
      I was just playing around with factors of F(p), and noticed a curious
      onvergence towards pi.



      2*5*13*37*73*89*...
      ___________________ = pi

      1*4*12*36*72*88*...



      Product of F(p)
      ___________________________________________ = pi

      Product of, one less than the factors of F(p)



      I have checked this to F(1000) ofcourse, to get pi~ 3.103...
      and it's slowing down fast. Could this be?
    • Shane
      Factors of F(p) In linear order
      Message 2 of 16 , Oct 19, 2002
      • 0 Attachment
        Factors of F(p)


        In linear order<2000:
        2,5,13,37,73,89,113,149,157,193,233,269,277,313,353,389,397,457,557,61
        3,673,677,733,757,877,953,977,997,1069,1093,1153,1213,1237,1453,1597,1
        657,1753,1873,1877,1933,1949,1993,...

        Notice that there are only 5 ending in a 9:
        89,269,389,1069,1949.
        There are none ending in a 1, yet!

        Why is there a discrimination,in linear order?
      • Jack Brennen
        ... No. If you keep computing it, it passes pi when you multiply by 11617/11616. 11617 is the 153rd prime which is a factor of a Fibonacci number with prime
        Message 3 of 16 , Oct 19, 2002
        • 0 Attachment
          Shane wrote:
          >
          > I was just playing around with factors of F(p), and noticed a curious
          > onvergence towards pi.
          >
          > 2*5*13*37*73*89*...
          > ___________________ = pi
          >
          > 1*4*12*36*72*88*...
          >
          > Product of F(p)
          > ___________________________________________ = pi
          >
          > Product of, one less than the factors of F(p)
          >
          > I have checked this to F(1000) ofcourse, to get pi~ 3.103...
          > and it's slowing down fast. Could this be?

          No. If you keep computing it, it passes pi when you multiply by
          11617/11616. 11617 is the 153rd prime which is a factor of a
          Fibonacci number with prime index.

          But you do raise an interesting question... Does this product
          converge? Computing it for all primes < 10^6, of which 4878 meet
          the criterion, it only reaches a product of approx. 3.228752.

          The answer depends on how sparse such primes are, and how fast
          their density drops off (assuming that it does).
        • Jack Brennen
          ... Just past your horizon: 2221. And more: 5381, 8861, 10061, 15761, 21481, 23021, 26141, 27941, ... ... Another good question. Of the first 1200 such
          Message 4 of 16 , Oct 19, 2002
          • 0 Attachment
            Shane wrote:
            >
            > Factors of F(p)
            >
            > In linear order<2000:
            > 2,5,13,37,73,89,113,149,157,193,233,269,277,313,353,389,397,457,557,61
            > 3,673,677,733,757,877,953,977,997,1069,1093,1153,1213,1237,1453,1597,1
            > 657,1753,1873,1877,1933,1949,1993,...
            >
            > Notice that there are only 5 ending in a 9:
            > 89,269,389,1069,1949.
            > There are none ending in a 1, yet!

            Just past your horizon: 2221.

            And more: 5381, 8861, 10061, 15761, 21481, 23021, 26141, 27941, ...

            > Why is there a discrimination,in linear order?

            Another good question. Of the first 1200 such primes:

            Ending in 1: 40
            Ending in 2: 1
            Ending in 3: 506
            Ending in 5: 1
            Ending in 7: 488
            Ending in 9: 164

            Why such a heavy prevalance of numbers ending in 3 or 7? Note that
            the ratios are also close to integers: Ratio of xxx1:xxx9 is about
            1:4, ratio of xxx9:xxx7 is about 1:3, and ratio of xxx7:xxx3 is about
            1:1... Interesting...
          • Shane
            I will just list a few of L(p), because it is just to tedious. 3,7,11,29,59,139,179,199,359,461,509,521,619,709,809,1031,1049,1511,.. . With an abundance,
            Message 5 of 16 , Oct 19, 2002
            • 0 Attachment
              I will just list a few of L(p), because it is just to tedious.

              3,7,11,29,59,139,179,199,359,461,509,521,619,709,809,1031,1049,1511,..
              .


              With an abundance, approx 1/2 ratio between 9, and 1.
            • Shane
              This could be of use. We would need to know if this trend continues or reverses itself, where the primes ending in 1,3,7,9 have equal probability, over time.
              Message 6 of 16 , Oct 19, 2002
              • 0 Attachment
                This could be of use.

                We would need to know if this trend continues or reverses itself,
                where the primes ending in 1,3,7,9 have equal probability, over time.

                Cycles maybe?
              • mikeoakes2@aol.com
                In a message dated 20/10/02 05:27:34 GMT Daylight Time, TTcreation@aol.com ... Relevant to this, and to your Fib observations, is p. 67 of Ribenboim s NBPNR,
                Message 7 of 16 , Oct 20, 2002
                • 0 Attachment
                  In a message dated 20/10/02 05:27:34 GMT Daylight Time, TTcreation@...
                  writes:


                  > I will just list a few of L(p), because it is just to tedious.
                  >
                  > 3,7,11,29,59,139,179,199,359,461,509,521,619,709,809,1031,1049,1511,..
                  > .
                  >
                  > With an abundance, approx 1/2 ratio between 9, and 1.
                  >

                  Relevant to this, and to your Fib observations, is p. 67 of Ribenboim's
                  NBPNR, where he records that Jarden (1958), Ward (1961) and others have
                  characterised the set of primes which divide Lucas numbers.
                  In particular, primes = 3, 7, 11, 19 mod 20 divide some L(n), primes = 13 or
                  17 mod 20 don't divide any L(n), and for p = 1, 9 mod 20 "there is no finite
                  set of divisibility conditions to describe completely" [whether they do or
                  don't]. Isn't that remarkable?

                  I know you are restricting your attention to L(p) for p _prime_, but with
                  these previous results in mind, it would seem interesting to do your analyses
                  w.r.t. mod 20, not mod 10.

                  Mike



                  [Non-text portions of this message have been removed]
                • Shane
                  Subject: Re: Linear factors of F(p), L(p) ADVERTISEMENT ... I had just noticed that again. P = 1 mod 20 ratio ~ 1/1 P = 11 mod 20 ratio ~ 1/2 P = 9 mod 20
                  Message 8 of 16 , Oct 20, 2002
                  • 0 Attachment
                    Subject: Re: Linear factors of F(p), L(p)


                    ADVERTISEMENT






                    > w.r.t. mod 20, not mod 10.
                    >
                    > Mike


                    I had just noticed that again.

                    P = 1 mod 20
                    ratio ~ 1/1
                    P = 11 mod 20

                    ratio ~ 1/2

                    P = 9 mod 20
                    ratio ~ 1/1
                    P = 19 mod 20


                    Does this continue to hold though?
                  • Shane
                    Mike Oakes wrote, For p = 1, 9 mod 20 there is no finite ... they do or don t]. Isn t that remarkable? There seems to be a certain probability that mirrors
                    Message 9 of 16 , Oct 20, 2002
                    • 0 Attachment
                      Mike Oakes wrote,
                      For p = 1, 9 mod 20 "there is no finite
                      > set of divisibility conditions to describe completely" [whether
                      they do or don't]. Isn't that remarkable?


                      There seems to be a certain probability that mirrors p=11,19 mod 20.
                      Remarks?

                      1<=11
                      9<=19 ?
                    • David Broadhurst
                      Jack Brennen wrote ... For primes
                      Message 10 of 16 , Oct 20, 2002
                      • 0 Attachment
                        Jack Brennen wrote

                        > Computing it for all primes < 10^6,
                        > of which 4878 meet the criterion,
                        > it only reaches a product of approx.
                        > 3.228752.

                        For primes < 10^7 it attains the value

                        3.25453335...

                        I'll try letting the code run up to 10^8.

                        David
                      • David Broadhurst
                        It looks as if the product converges, about as slowly as Brun s twin-prime constant. Here, at random, is a screenful of prime {q,p} with q|F(p), ordered by the
                        Message 11 of 16 , Oct 20, 2002
                        • 0 Attachment
                          It looks as if the product converges, about
                          as slowly as Brun's twin-prime constant.

                          Here, at random, is a screenful
                          of prime {q,p} with q|F(p),
                          ordered by the divisor q:

                          [58470317, 3248351],
                          [58470793, 29235397],
                          [58470869, 14617717],
                          [58471069, 4872589],
                          [58471213, 29235607],
                          [58471513, 29235757],
                          [58473397, 29236699],
                          [58473433, 29236717],
                          [58473977, 9745663],
                          [58474193, 9745699],
                          [58474573, 29237287],
                          [58474777, 29237389],
                          [58475069, 14618767],
                          [58475617, 29237809],
                          [58476521, 1461913],
                          [58477813, 29238907],
                          [58478593, 29239297],
                          [58478677, 29239339],
                          [58479437, 9746573],
                          [58479697, 29239849],

                          At this size, slightly more than a half
                          are Drobot-Cunningham pairs:
                          q=2*p-1 with p=13 or 17 mod 20.

                          If drobots remained a finite fraction, the log of
                          the product would converge in a manner similar
                          to Brun's constant, since
                          integral dx/(x*log(x)^2)
                          =1/1og(x)+const
                          is kosher at large x.

                          David
                        • Shane
                          David: With these linear,primes mod 20, Are the remainder ratios, Jack and I have observed correct? Or is this just a local phenomenon? Shane F.
                          Message 12 of 16 , Oct 20, 2002
                          • 0 Attachment
                            David:
                            With these linear,primes mod 20,
                            Are the remainder ratios, Jack and I have observed correct?
                            Or is this just a local phenomenon?





                            Shane F.
                          • David Broadhurst
                            Primes q
                            Message 13 of 16 , Oct 20, 2002
                            • 0 Attachment
                              Primes q < 10^8 dividing some F(p) with prime p;
                              final digit pattern for the divisor, q:

                              Ending in 1....6835
                              Ending in 2.......1
                              Ending in 3..104149
                              Ending in 5.......1
                              Ending in 7..104217
                              Ending in 9...36781

                              Total....... 251984

                              of which

                              Drobots......152904 (60.7%)

                              prod_{q<10^8} 1/(1-1/q)
                              =3.27365650101...
                            • David Broadhurst
                              Since gcd(F(n),F(m))=F(gcd(m,n)), we may associate to each prime q at most one prime p such that q|F(p). For 10^8 q|F(p), the final-digit p-pattern is
                              Message 14 of 16 , Oct 20, 2002
                              • 0 Attachment
                                Since gcd(F(n),F(m))=F(gcd(m,n)),
                                we may associate to each prime q
                                at most one prime p such that q|F(p).

                                For 10^8>q|F(p), the
                                final-digit p-pattern is
                                ............!

                                Ending in 1...14378
                                Ending in 2........
                                Ending in 3...27047
                                Ending in 5.......1
                                Ending in 7..105132
                                Ending in 9..105426

                                PS: The largest F(p) with
                                prime index p and a prime divisor
                                q<10^8 has more than 10 million decimal digits.
                              • mistermac39
                                Would it not be odd if it reached 3.359885666....? Although, I must confess, it is a long shot. The above is the sum of the reciprocals of the Fibonacci.
                                Message 15 of 16 , Oct 20, 2002
                                • 0 Attachment
                                  Would it not be odd if it reached 3.359885666....?

                                  Although, I must confess, it is a long shot.

                                  The above is the sum of the reciprocals of the Fibonacci.
                                • Shane
                                  ... This number is [veta 1]+1 The answer is no, when [veta s] is small, these two numbers are farthest apart. Conversely when s is large they converge, at a
                                  Message 16 of 16 , Oct 23, 2002
                                  • 0 Attachment
                                    >"mistermac39" wrote:
                                    > Would it not be odd if it reached 3.359885666....?>
                                    > Although, I must confess, it is a long shot.
                                    > The above is the sum of the reciprocals of the Fibonacci.

                                    This number is [veta 1]+1

                                    The answer is no, when [veta s] is small, these two numbers are
                                    farthest apart. Conversely when s is large they converge, at a
                                    wonderful rate.

                                    I believe you were off by 1.



                                    See veta message,
                                    Shane F.
                                  Your message has been successfully submitted and would be delivered to recipients shortly.