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Re: Is phi(p^2-1)/(p^2-1) bounded?

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  • David Broadhurst <d.broadhurst@open.ac.u
    Jon Perry mistakenly claimed that ... Invoke it with -p10000000 to precompute primes to 10M. Limit is about 430M, as I recall, but then you need to allocate
    Message 1 of 20 , Jan 3, 2003
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      Jon Perry mistakenly claimed that

      > Pari chokes on primes>1000000

      Invoke it with -p10000000 to precompute primes to 10M.
      Limit is about 430M, as I recall, but then you
      need to allocate core with the -s<size> modifier.
    • David Broadhurst <d.broadhurst@open.ac.u
      ... .......................? No. p=3 gives 1/2 and p=2 gives 2/3 which is maximal, for prime p.
      Message 2 of 20 , Jan 3, 2003
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        Phil:
        > Therefore the highest value you will find will be
        > 1/2*2/3 = 1/3 from p=3,5,17
        .......................?
        No. p=3 gives 1/2
        and p=2 gives 2/3 which is maximal, for prime p.
      • Phil Carmody
        ... Deliberate mistake, to see if Jon was paying attention! ;-) Phil (lying through his teeth!) ===== The answer to life s mystery is simple and direct: Sex
        Message 3 of 20 , Jan 3, 2003
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          --- "David Broadhurst <d.broadhurst@...>" <d.broadhurst@...> wrote:
          > Phil:
          > > Therefore the highest value you will find will be
          > > 1/2*2/3 = 1/3 from p=3,5,17
          > .......................?
          > No. p=3 gives 1/2
          > and p=2 gives 2/3 which is maximal, for prime p.

          Deliberate mistake, to see if Jon was paying attention! ;-)

          Phil
          (lying through his teeth!)


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        • David Broadhurst <d.broadhurst@open.ac.u
          ... 0. We believe (but cannot prove) that there are an infinite number of primes of the form primorial+1. That would be enough to make f(p)=phi(p^2-1)/(p^2-1)
          Message 4 of 20 , Jan 3, 2003
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            Phil:
            > Lower bound - anyone care for a stab?

            0.

            We believe (but cannot prove)
            that there are an infinite number of primes of
            the form primorial+1. That would be enough
            to make f(p)=phi(p^2-1)/(p^2-1)
            as close to zero as one likes.

            At present we know that
            p=392113#+1 is prime,
            giving (a la Mertens)
            f(p) < 0.0436

            Can anyone get lower than that?

            David
          • David Broadhurst <d.broadhurst@open.ac.u
            Let f(p)=phi(p^2-1)/(p^2-1). Say a prime p is lowest yet if there is no prime q
            Message 5 of 20 , Jan 3, 2003
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              Let f(p)=phi(p^2-1)/(p^2-1).
              Say a prime p is "lowest yet" if there is
              no prime q<p with f(q)<f(p).
              The "lowest yet" sequence begins
              2, 3, 5, 11, 29, 131, 139, 181, 419, 1429, 17291, 23561,
              23869, 188189, 315589, 483209, 614041, 1624349, 1729001,
              8242961, 15431989, 22486309, 27033161, 36058021, 57762431,
              61577671, 117048931, ...

              (117048931^2-1)/4=
              2*3*5*7*11*13*19*23*29*31*41*73*97

              What comes next?
            • Phil Carmody
              ... It s what I would have guessed, but my brain has begun to stop working in the last few hours. (e.g. the p=3 - 0.5 line was on my screen when I typed p=3
              Message 6 of 20 , Jan 3, 2003
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                --- "David Broadhurst <d.broadhurst@...>" <d.broadhurst@...> wrote:
                > Phil:
                > > Lower bound - anyone care for a stab?
                >
                > 0.

                It's what I would have guessed, but my brain has begun to stop working in
                the last few hours. (e.g. the p=3 -> 0.5 line was on my screen when I typed
                p=3 -> 1/3, so I'm really not with it!)

                > We believe (but cannot prove)
                > that there are an infinite number of primes of
                > the form primorial+1.

                As many different prime factors as possible, such that
                Product[(p-1)/p] could be over as many terms as possible.

                > That would be enough
                > to make f(p)=phi(p^2-1)/(p^2-1)
                > as close to zero as one likes.

                Of course.

                > At present we know that
                > p=392113#+1 is prime,
                > giving (a la Mertens)
                > f(p) < 0.0436
                >
                > Can anyone get lower than that?

                Not without using a larger number, probably (it's possible, though, as you
                can use fewer factors in p-1, and dope p+1 with them instead - all you
                need's a few coincidences). However, finding primes of that size is not for
                the faint hearted.

                Phil


                =====
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              • Phil Carmody
                ... Nice concrete follow-on from Jon s original. I can t see a clever way to improve on brute-force search without leaving the possibility of missing some.
                Message 7 of 20 , Jan 3, 2003
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                  --- "David Broadhurst <d.broadhurst@...>" <d.broadhurst@...> wrote:
                  > Let f(p)=phi(p^2-1)/(p^2-1).
                  > Say a prime p is "lowest yet" if there is
                  > no prime q<p with f(q)<f(p).
                  > The "lowest yet" sequence begins
                  > 2, 3, 5, 11, 29, 131, 139, 181, 419, 1429, 17291, 23561,
                  > 23869, 188189, 315589, 483209, 614041, 1624349, 1729001,
                  > 8242961, 15431989, 22486309, 27033161, 36058021, 57762431,
                  > 61577671, 117048931, ...
                  >
                  > (117048931^2-1)/4=
                  > 2*3*5*7*11*13*19*23*29*31*41*73*97
                  >
                  > What comes next?

                  Nice concrete follow-on from Jon's original.

                  I can't see a clever way to improve on brute-force search without leaving
                  the possibility of missing some.
                  However, it might be possible to stick some markers in the ground for people
                  to aim at by finding squarefree-smooths that have isprime(sqrt(4s+1)).


                  Phil


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                • richard_heylen <richard_heylen@yahoo.co.
                  ... ... I believe it continues as follows 181333151 267190769 331413809 376754951 636510601 1737265531 3019962791 One can obtain fairly low values of
                  Message 8 of 20 , Jan 3, 2003
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                    --- In primenumbers@yahoogroups.com, "David Broadhurst
                    <d.broadhurst@o...>" <d.broadhurst@o...> wrote:
                    > Let f(p)=phi(p^2-1)/(p^2-1).
                    > Say a prime p is "lowest yet" if there is
                    > no prime q<p with f(q)<f(p).
                    > The "lowest yet" sequence begins

                    <snip>

                    > 61577671, 117048931, ...

                    I believe it continues as follows
                    181333151
                    267190769
                    331413809
                    376754951
                    636510601
                    1737265531
                    3019962791

                    One can obtain fairly low values of f(p) realtively easily. Consider
                    for example
                    p=58531393985146662592474024598667898081212671 prime
                    p-1=2.5.7.13.19.43.53.67.71.73.43520821168673.98287085283258329
                    p+1=2^8.3^3.11.17.23.29.31.37.41.47.59.61.79.83.89.97.101.103.107.109.
                    113.127.131.661

                    So the first 33 primes are factors of p^2-1 and I believe this gives
                    an f(p) around 0.113
                    This is significantly smaller than the f(p) around 0.148 for the best
                    of the minimal examples listed.
                    To break the 0.10 barrier you need primes up to 257.
                    To break the 0.05 barrier you need primes up to 75029
                    By this stage, the numbers are getting rather large.

                    Richard Heylen
                  • David Broadhurst <d.broadhurst@open.ac.u
                    ... I proved the first 4 of your addenda with the unsmart brute-force Pari-GP source m=1;mp=430*10^6; Jon please note
                    Message 9 of 20 , Jan 4, 2003
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                      Richard:

                      > I believe it continues as follows
                      > 181333151
                      > 267190769
                      > 331413809
                      > 376754951
                      > 636510601
                      > 1737265531
                      > 3019962791

                      I proved the first 4 of your addenda with
                      the unsmart brute-force Pari-GP source

                      m=1;mp=430*10^6; \\ Jon please note
                      forprime(p=2,mp,n=p^2-1;s=eulerphi(n)/n;if(s<m,m=s;print(p)))

                      David
                    • Jon Perry
                      m=1;mp=430*10^6; Jon please note forprime(p=2,mp,n=p^2-1;s=eulerphi(n)/n;if(s
                      Message 10 of 20 , Jan 4, 2003
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                        'm=1;mp=430*10^6; \\ Jon please note
                        forprime(p=2,mp,n=p^2-1;s=eulerphi(n)/n;if(s<m,m=s;print(p)))'

                        I'm looking...

                        'Use 'calc' instead. Or bc. Or the other 'calc'. Or use gp and use
                        'p=nextprime(p+1)' rather than 'forprime(p='.'

                        Is this the K.R. Matthews Number Theory calculator 'calc'?

                        Is there such a concept as the 'average value of f(p)'?

                        Jon Perry
                        perry@...
                        http://www.users.globalnet.co.uk/~perry/maths/
                        http://www.users.globalnet.co.uk/~perry/DIVMenu/
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                      • David Broadhurst <d.broadhurst@open.ac.u
                        ... forprime is faster than nextprime if you can afford the memory up to p=430M
                        Message 11 of 20 , Jan 4, 2003
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                          > use 'p=nextprime(p+1)' rather than 'forprime(p='
                          'forprime' is faster than 'nextprime'
                          if you can afford the memory up to p=430M
                        • Phil Carmody
                          ... Possibly. I m using Chongo s calc (Curt Landon Noll, record prime finder 2-3 decades ago), which is the standard GNU utility. The whereabouts of the
                          Message 12 of 20 , Jan 4, 2003
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                            --- Jon Perry <perry@...> wrote:
                            > 'm=1;mp=430*10^6; \\ Jon please note
                            > forprime(p=2,mp,n=p^2-1;s=eulerphi(n)/n;if(s<m,m=s;print(p)))'
                            >
                            > I'm looking...
                            >
                            > 'Use 'calc' instead. Or bc. Or the other 'calc'. Or use gp and use
                            > 'p=nextprime(p+1)' rather than 'forprime(p='.'
                            >
                            > Is this the K.R. Matthews Number Theory calculator 'calc'?

                            Possibly. I'm using Chongo's calc (Curt Landon Noll, record prime finder 2-3
                            decades ago), which is the standard 'GNU' utility. The whereabouts of the
                            other calc is answered in the archives some time around a year back, maybe
                            more.

                            > Is there such a concept as the 'average value of f(p)'?

                            I expect it to drift downards so it's not well-defined.
                            (or maybe it is, maybe it's zero. On average numbers have 1/eps distinct
                            divisors, i.e. a divergent number. That's got to take a toll on the phi
                            value. Any sample up to 300000# is puny compared with the sizes of almost
                            all integers...)

                            Phil


                            =====
                            The answer to life's mystery is simple and direct:
                            Sex and death. -- Ian 'Lemmy' Kilminster

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                          • Jon Perry
                            I expect it to drift downards so it s not well-defined. (or maybe it is, maybe it s zero. On average numbers have 1/eps distinct divisors, i.e. a divergent
                            Message 13 of 20 , Jan 4, 2003
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                              'I expect it to drift downards so it's not well-defined.
                              (or maybe it is, maybe it's zero. On average numbers have 1/eps distinct
                              divisors, i.e. a divergent number. That's got to take a toll on the phi
                              value. Any sample up to 300000# is puny compared with the sizes of almost
                              all integers...)'

                              'Therefore the highest value you will find will be
                              1/2*2/3 = 1/3 from p=3,5,17
                              and 1/3-eps from numbers with a few prime factors larger than 2 or 3 in p+1
                              and p-1.'

                              Cough, cough. You make these up, or do they come naturally?

                              Jon Perry
                              perry@...
                              http://www.users.globalnet.co.uk/~perry/maths/
                              http://www.users.globalnet.co.uk/~perry/DIVMenu/
                              BrainBench MVP for HTML and JavaScript
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                            • Phil Carmody
                              ... Where the 3 was already pointed out as a typo. ... OK John. Which of the two statements do you think is wrong? And why? Come on, show us the flaws, I yearn
                              Message 14 of 20 , Jan 4, 2003
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                                --- Jon Perry <perry@...> wrote:

                                Quoting me:

                                > 'I expect it to drift downards so it's not well-defined.
                                > (or maybe it is, maybe it's zero. On average numbers have 1/eps distinct
                                > divisors, i.e. a divergent number. That's got to take a toll on the phi
                                > value. Any sample up to 300000# is puny compared with the sizes of almost
                                > all integers...)'

                                > 'Therefore the highest value you will find will be
                                > 1/2*2/3 = 1/3 from p=3,5,17
                                > and 1/3-eps from numbers with a few prime factors larger than 2 or 3 in p+1
                                > and p-1.'

                                Where the 3 was already pointed out as a typo.

                                > Cough, cough. You make these up, or do they come naturally?

                                OK John. Which of the two statements do you think is wrong?
                                And why?

                                Come on, show us the flaws, I yearn to be enlightened by your razer-sharp
                                mathematical quill. I'll even fill in the ellipses, if you like, as have a
                                feeling you're getting confused by my elision.

                                �10 to Oxfam for each statement you persuade me to retract. I trust you'll
                                reciprocate?

                                Phil


                                =====
                                The answer to life's mystery is simple and direct:
                                Sex and death. -- Ian 'Lemmy' Kilminster

                                __________________________________________________
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                              • Jon Perry
                                £10 to Oxfam for each statement you persuade me to retract. I trust you ll reciprocate? What... Oxfam will pay me £10 to persuade you to retract them? I
                                Message 15 of 20 , Jan 4, 2003
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                                  '£10 to Oxfam for each statement you persuade me to retract. I trust you'll
                                  reciprocate?'

                                  What... Oxfam will pay me £10 to persuade you to retract them?

                                  I believe they are both correct, hence I will not allow Oxfam to waste their
                                  money on me, and this in turn leads me to believe that f(p) could have an
                                  average value.

                                  Jon Perry
                                  perry@...
                                  http://www.users.globalnet.co.uk/~perry/maths/
                                  http://www.users.globalnet.co.uk/~perry/DIVMenu/
                                  BrainBench MVP for HTML and JavaScript
                                  http://www.brainbench.com
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