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

4^8230-3 is prime!

Expand Messages
  • Wojciech.Florek@amu.edu.pl
    Hi, John Renze s PARI chg.gp script has just finished a test of 4955-digit number 4^8230-3 = 2^16460-2-1 (as some people prefer). It is the largest prime
    Message 1 of 30 , Mar 2, 2006
    • 0 Attachment
      Hi,

      John Renze's PARI chg.gp script has just finished a test of 4955-digit
      number 4^8230-3 = 2^16460-2-1 (as some people prefer). It is the largest
      prime number b^n-(b-1) I know.


      I have no verifying program, so I'd be very much obliged if someone would
      provide me with such program or would do an independet test. I'll provide
      her/him with n, F, and G numbers or/and with the output of chg.gp. The
      test has taken some days on my laptop since (from the output):
      Number to be tested has 4955 digits.
      Modulus has 1321 digits.
      Modulus is 26.64255144% of n

      Regards to all, especially to
      \\ [...] John Renze
      \\ [...] David Broadhurst, Greg Childers and the PrimeForm community.


      Wojtek

      PS
      This result will be included in my page
      http://perta.fizyka.amu.edu.pl/pnq/prp1.html

      WsF

      ===============================================
      Wojciech Florek (WsF)
      Adam Mickiewicz University, Faculty of Physics
      ul. Umultowska 85, 61-614 Poznan, Poland

      Phone: (++48-61) 8295033 fax: (++48-61) 8257758
      email: Wojciech.Florek@...
    • Devaraj Kandadai
      Sometime ago I had conjectured that all the prime factors of a Carmichael number cannot be Mersenne. For that I need the following lemma: *Thread*:
      Message 2 of 30 , May 28 9:05 PM
      • 0 Attachment
        Sometime ago I had conjectured that all the prime factors of a Carmichael
        number cannot be Mersenne. For that I need the following lemma: *Thread*:
        Programming a Conjecture<http://www.mersenneforum.org/showthread.php?p=55271#post55271>
        View Single Post [image: Old] 01 Jun 05, 07:39 PM
        #*14*<http://www.mersenneforum.org/showpost.php?p=55271&postcount=14>
        maxal <http://www.mersenneforum.org/member.php?u=1643>

        [image: maxal's Avatar] <http://www.mersenneforum.org/member.php?u=1643>

        Posts: 145
        Join Date: Feb 2005
        [image: Default] *Pomerance's proof*
        ------------------------------
        I've got a permission from Carl Pomerance to reproduce his proof that every
        Carmichael number is Devaraj number. I present an extended version of the
        proof which applies to r-factor Carmichael numbers for arbitrary r (the
        original proof is for r=3).

        Recall that n is Carmichael number iff (p-1) divide (n-1) for every prime p
        dividing n.

        Let n=p1*p2*...*pr be Carmichael number, where p1,p2,...,pr are prime.
        To show that n is Devaraj number it's enough to prove that
        (p1-1)*(n-1)^(r-2)/(p2-1)...(pr-1) is integer.

        (p1-1)*(n-1)^(r-2)/(p2-1)...(pr-1) is integer iff for every prime q its
        maximum degree dividing the numerator is greater or equal than its maximum
        degree dividing the denominator.

        Let q be any prime number and let a,a1,a2,...,ar be maximum non-negative
        integers such that q^a divides n-1, q^a1 divides (p1-1), ..., q^ar divides
        (pr-1). Since n is Carmichael number, a1<=a, a2<=a, ..., ar<=a. (*)
        Then the maximum degree of q dividing the numerator (p1-1)*(n-1)^(r-2) is
        a1+(r-2)*a while the maximum degree of q dividing the numerator
        (p2-1)...(pr-1) is a2+a3+...+ar.
        We will show that a1 >= min(a2,a3,...,ar).

        Assume that a1 < min(a2,a3,...,ar).
        Let p1=k1*q^a1 + 1, p2 = k2*q^a2 + 1, ..., pr=kr*q^ar + 1 where k1, k2, ...,
        kr are not divisible by q. Then
        n = p1*p2*...*pr = (k1*q^a1 + 1)*(k2*q^a2 + 1)*...*(kr*q^ar + 1) = 1 +
        k1*q^a1 + k2*q^a2 + ... + kr*q^ar + C where C includes all summands
        divisible by at least q^(a1+min(a2,a3,...,ar)).
        Hence, n-1 = k1*q^a1 + k2*q^a2 + ... + kr*q^ar + C is divisible by at most
        q^a1 meanning a=a1, a contradiction to (*). Therefore, a1 >=
        min(a2,a3,...,ar).

        It is easy to see that a1 >= min(a2,a3,...,ar) together with (*) implies
        a1+(r-2)*a >= a2+a3+...+ar.
        Q.E.D.
        ------------------------------
        *Last fiddled with by maxal on 01 Jun 05 at 07:41 PM. *
        [image: maxal is offline] [image: Report
        Post]<http://www.mersenneforum.org/report.php?p=55271> [image:
        Reply With Quote]<http://www.mersenneforum.org/newreply.php?do=newreply&p=55271>
        maxal View Public Profile <http://www.mersenneforum.org/member.php?u=1643> Send
        a private message to
        maxal<http://www.mersenneforum.org/private.php?do=newpm&u=1643> Find
        More Posts by maxal<http://www.mersenneforum.org/search.php?do=finduser&u=1643>
        Add
        maxal to Your Buddy
        List<http://www.mersenneforum.org/profile.php?do=addlist&userlist=buddy&u=1643>
        Delete
        This User<http://www.mersenneforum.org/login.php?do=logout&logouthash=d855c8cb50c936b8fa20e6caf3e6eb19>
        Mersenne. I hope to prove it in the near future; for this I need
        the above lemma :

        A.K.Devaraj


        [Non-text portions of this message have been removed]
      • Devaraj Kandadai
        A CONJECTURE Let f(x) =a^b^x + c where a, b,c & x belong to N ( a, b and c fixed.). Then a^(b^ (x+k* Eulerphi(Eulerphi(f(x)) + c is congruent to 0 (mod f(x)).
        Message 3 of 30 , Jun 11, 2009
        • 0 Attachment
          A CONJECTURE



          Let f(x) =a^b^x + c where a, b,c & x belong to N ( a, b and c fixed.).


          Then a^(b^ (x+k* Eulerphi(Eulerphi(f(x)) + c is congruent to 0 (mod f(x)).
          Here k belongs to N.


          Sketch proof: This is a corollary of �Euler's Generalisation of Fermat's
          Theorem- a further generalisation. (Hawaii International Conference,
          2004-ISSN 1550 3747).



          A.K. Devaraj


          [Non-text portions of this message have been removed]
        • Phil Carmody
          ... Nesting brackets isn t that hard a skill. Please acquire it. ... A conjecture has no proof, sketchy or otherwise. A theorem has a proof. ... The
          Message 4 of 30 , Jun 11, 2009
          • 0 Attachment
            --- On Fri, 6/12/09, Devaraj Kandadai <dkandadai@...> wrote:
            > A CONJECTURE
            >
            > Let f(x) =a^b^x + c where a, b,c & x belong to N ( a, b
            > and c fixed.).
            >
            > Then a^(b^ (x+k* Eulerphi(Eulerphi(f(x)) + c is congruent
            > to 0 (mod f(x)).

            Nesting brackets isn't that hard a skill. Please acquire it.

            > Here k belongs to N.
            >
            > Sketch proof:

            A conjecture has no proof, sketchy or otherwise. A theorem has a proof.

            > This is a corollary of “Euler's Generalisation of Fermat's
            > Theorem- a further generalisation. (Hawaii International
            > Conference, 2004-ISSN 1550 3747).

            The International Conference of People Who Want to Pay Large Amounts of Money in Order to Claim That They've Presented a Paper at a Conference?

            However, I'm pleased to see that at the above "conference" someone presented a paper on "Factorization of Intergers[sic]"; it would be a shame to see such a wonderfully illucid concept as intergers fade away. (Or maybe it was just a typo, that would be a shame.)

            Phil
          • James Merickel
            Dear Group Members,   The following is an extension of ideas recently presented for research.  Part of the story of this particular sequence, also (I might
            Message 5 of 30 , Sep 16, 2012
            • 0 Attachment
              Dear Group Members,
               
              The following is an extension of ideas recently presented for research.  Part of the story of this particular sequence, also (I might add), has something to do with the plausibility of Cancer VZ as a signal source.  Cancer VZ is a Scuti-alpha (formerly, 'dwarf Cepheid') type of star.
               
              Conjecture:  The number of values of the form 2^a*3^b for which the number of times each digit from 0 through 9 appears exactly a prime number of times is 40. 
               
              The largest I have so far is 2^1997*3^135, with an exceptionally long wait.  However, this conjecture a) is in a computing environment I am not sure I should trust, b) is not based upon any real heuristic reasoning of finitude, and c) has not even been discussed with anybody else.  It took me a while to even remember or figure out what I had programmed. 
               
              The computing environment is probably sound to data but possibly not to the question of whether certain programs have been asked to cease, and even the former is in question.  So, if anybody would like to tackle a check, some real reasoning, and/or an attempt at continuing this sequence, it would be welcome.  I do have data, so if a choice is made to submit data to OEIS and the 40th term is right, it would be appropriate to join me as co-author or for me to add a confirmer when I submit.
               
              Yours Truly, James G. Merickel

              [Non-text portions of this message have been removed]
            • djbroadhurst
              ... http://www.micosmos.com/geos/nc/NC_0857.pdf I do not think that variable stars care much about the number of human fingers. David
              Message 6 of 30 , Sep 16, 2012
              • 0 Attachment
                --- In primenumbers@yahoogroups.com,
                James Merickel <moralforce120@...> wrote:

                > Conjecture: The number of values of the form 2^a*3^b for
                > which the number of times each digit from 0 through 9
                > appears exactly a prime number of times is 40.

                > something to do with the plausibility of Cancer VZ
                > as a signal source

                http://www.micosmos.com/geos/nc/NC_0857.pdf

                I do not think that variable stars care much
                about the number of human fingers.

                David
              • Robert Gerbicz
                The largest I have so far is 2^1997*3^135, with an exceptionally long wait. n=2^1742*3^4350 is a larger solution, the distribution of the digits: [241, 251,
                Message 7 of 30 , Sep 16, 2012
                • 0 Attachment
                  "The largest I have so far is 2^1997*3^135, with an exceptionally long
                  wait."

                  n=2^1742*3^4350 is a larger solution, the distribution of the digits: [241,
                  251, 257, 269, 251, 257, 241, 281, 281, 271]

                  And it is very likely that your conjecture is false, for e2,e3<N the
                  expected number of solutions is at least c*N^2/log(N)^10, for c>0. Moreover
                  it should be false even for the sequence of 2^e2.


                  [Non-text portions of this message have been removed]
                • James Merickel
                  Dr. Broadhurst, There is a physics text in a Springer Verlag Series on plausible physics for manipulating the Sun (I do not own this particular book or know
                  Message 8 of 30 , Sep 16, 2012
                  • 0 Attachment
                    Dr. Broadhurst,
                    There is a physics text in a Springer Verlag Series on plausible physics for manipulating the Sun (I do not own this particular book or know its title offhand, but it does exist; so what a star cares about is not what I was thinking (and what I was thinking was digressive, but I'll use your response for my placement of a specific correction: The type of star is called 'delta-Scuti', not 'alpha-')).  Your focus on my digressive note is ... noted.  It was not the primary topic. 
                    Jim Merickel

                    --- On Sun, 9/16/12, djbroadhurst <d.broadhurst@...> wrote:


                    From: djbroadhurst <d.broadhurst@...>
                    Subject: [PrimeNumbers] Re: Conjecture
                    To: primenumbers@yahoogroups.com
                    Date: Sunday, September 16, 2012, 10:33 AM



                     



                    --- In primenumbers@yahoogroups.com,
                    James Merickel <moralforce120@...> wrote:

                    > Conjecture: The number of values of the form 2^a*3^b for
                    > which the number of times each digit from 0 through 9
                    > appears exactly a prime number of times is 40.

                    > something to do with the plausibility of Cancer VZ
                    > as a signal source

                    http://www.micosmos.com/geos/nc/NC_0857.pdf

                    I do not think that variable stars care much
                    about the number of human fingers.

                    David








                    [Non-text portions of this message have been removed]
                  • James Merickel
                    Okay.    Thanks, Mr. Gerbicz.  I retract as a reasonable conjecture, without checking.  This is somewhere  There was or was not a check of term?  This
                    Message 9 of 30 , Sep 16, 2012
                    • 0 Attachment
                      Okay. 
                       
                      Thanks, Mr. Gerbicz.  I retract as a reasonable conjecture, without checking.  This is somewhere  There was or was not a check of term?  This one is the following term, as far as your computation goes?  This sequence, as I have it, is coincidence-laden from any perspective and more strongly so from my own.  And my highschool class number was the count of 0s and 7s of the term you have added.    
                       
                      JGM

                      --- On Sun, 9/16/12, Robert Gerbicz <robert.gerbicz@...> wrote:


                      From: Robert Gerbicz <robert.gerbicz@...>
                      Subject: [PrimeNumbers] Re: Conjecture
                      To: primenumbers@yahoogroups.com
                      Date: Sunday, September 16, 2012, 11:26 AM



                       



                      "The largest I have so far is 2^1997*3^135, with an exceptionally long
                      wait."

                      n=2^1742*3^4350 is a larger solution, the distribution of the digits: [241,
                      251, 257, 269, 251, 257, 241, 281, 281, 271]

                      And it is very likely that your conjecture is false, for e2,e3<N the
                      expected number of solutions is at least c*N^2/log(N)^10, for c>0. Moreover
                      it should be false even for the sequence of 2^e2.

                      [Non-text portions of this message have been removed]








                      [Non-text portions of this message have been removed]
                    • djbroadhurst
                      -- In primenumbers@yahoogroups.com, ... I don t know that book, but I have read about the IRAS upper limit on nearby Dyson spheres:
                      Message 10 of 30 , Sep 16, 2012
                      • 0 Attachment
                        -- In primenumbers@yahoogroups.com,
                        James Merickel <moralforce120@...> wrote:

                        > There is a physics text in a Springer Verlag Series on plausible
                        > physics for manipulating the Sun

                        I don't know that book, but I have read about the
                        IRAS upper limit on nearby Dyson spheres:

                        http://arxiv.org/ftp/arxiv/papers/0811/0811.2376.pdf

                        >> This Dyson Sphere search has looked at a significant fraction of the IRAS LRS sources with temperatures under 600 ºK. Since IRAS covered 96% of the sky this is essentially a whole-sky search. Indeed this search may be one of the only SETI/cosmic archaeology whole-sky searches conducted so far. Unlike many radio and optical SETI searches this one does not require purposeful intent to communicate on the part of the originating source of the signature of intelligence. <<

                        In any case, it still seems to me to be silly to link
                        such ideas to base-10 representations of integers,
                        as seemed to be implied by James' obiter dictum.

                        David
                      • James Merickel
                        Dr. Broadhurst: It was not meant to be the main topic, but since you have continued on it...the assumption of the use of Dyson spheres is an assumption of
                        Message 11 of 30 , Sep 16, 2012
                        • 0 Attachment
                          Dr. Broadhurst:
                          It was not meant to be the main topic, but since you have continued on it...the assumption of the use of Dyson spheres is an assumption of energy greediness in a near-zone.  We should not assume too much about intelligence beyond Earth.  We only know us.  I am grappling with three possibilities for life's origination: 1) Pre-bigbang, 2) Pre-metalization, and 3) post-metalization but before and affecting Earth's life.  These base-10 coincidences are the foundation of this.  You have the right to your opinion.  It's more my area than yours, though.
                          Jim Merickel
                          P.S. I will present the data on the sequence in question as I have it tomorrow, along with the code I use to generate it a second time.  Right now I cannot include the term given by Mr. (or Dr.) Gentzen as the 41st term, though I assume that's what was meant right now.  I will test it from my computer, but cannot get that far quickly to rule out an intermediate term.

                          --- On Sun, 9/16/12, djbroadhurst <d.broadhurst@...> wrote:


                          From: djbroadhurst <d.broadhurst@...>
                          Subject: [PrimeNumbers] Re: Conjecture
                          To: primenumbers@yahoogroups.com
                          Date: Sunday, September 16, 2012, 4:26 PM



                           



                          -- In primenumbers@yahoogroups.com,
                          James Merickel <moralforce120@...> wrote:

                          > There is a physics text in a Springer Verlag Series on plausible
                          > physics for manipulating the Sun

                          I don't know that book, but I have read about the
                          IRAS upper limit on nearby Dyson spheres:

                          http://arxiv.org/ftp/arxiv/papers/0811/0811.2376.pdf

                          >> This Dyson Sphere search has looked at a significant fraction of the IRAS LRS sources with temperatures under 600 ºK. Since IRAS covered 96% of the sky this is essentially a whole-sky search. Indeed this search may be one of the only SETI/cosmic archaeology whole-sky searches conducted so far. Unlike many radio and optical SETI searches this one does not require purposeful intent to communicate on the part of the originating source of the signature of intelligence. <<

                          In any case, it still seems to me to be silly to link
                          such ideas to base-10 representations of integers,
                          as seemed to be implied by James' obiter dictum.

                          David








                          [Non-text portions of this message have been removed]
                        • Robert Gerbicz
                          Three more terms, and no more for e2,e3
                          Message 12 of 30 , Sep 17, 2012
                          • 0 Attachment
                            Three more terms, and no more for e2,e3<40000 (if we write n=2^e2*3^e3)
                            2^26887*3^13438
                            2^35399*3^26448
                            2^18350*3^38015


                            [Non-text portions of this message have been removed]
                          • maximilian_hasler
                            ... At least this definitively rules out the Conjecture in its initial form. (To leave it open, change 40 to a finite number .) Anyway, as would say a
                            Message 13 of 30 , Sep 17, 2012
                            • 0 Attachment
                              --- In primenumbers@yahoogroups.com, Robert Gerbicz <robert.gerbicz@...> wrote:
                              >
                              > Three more terms, and no more for e2,e3<40000 (if we write n=2^e2*3^e3)
                              > 2^26887*3^13438
                              > 2^35399*3^26448
                              > 2^18350*3^38015


                              At least this definitively rules out the Conjecture in its initial form.
                              (To leave it open, change "40" to "a finite number".)
                              Anyway, as would say a Russian colleague of mine, 40 is not a number(*).
                              More surprisingly, it seems that 42 isn't the answer, either.

                              (* Although I vaguely recall some people stating that 10^40 was a "fundamental" ratio occuring in several instances when looking at sizes and/or masses in the known universe... no link to dwarf Cepheid type stars though.)

                              For the records, I tentatively submitted the draft http://oeis.org/draft/A216854 to store these results.
                              (Not sure other editors of OEIS will not delete rather than "approve" & publish this...)

                              Regards,
                              Maximilian
                            • James Merickel
                              No, Max.  I have already (in another mailing on this topic) conceded it should not be finite.  It would be nice to have that there are 44 values with
                              Message 14 of 30 , Sep 18, 2012
                              • 0 Attachment
                                No, Max.  I have already (in another mailing on this topic) conceded it should not be finite.  It would be nice to have that there are 44 values with exponents under 44000 (or 44444), though.  It's just cutesy stuff related to coincidences though. 
                                 
                                Robert, I have checked these terms.
                                 
                                I will not have time for this or anything else here.  I slept when I got home last night.  It's more important to change that home than do this. 

                                --- On Mon, 9/17/12, maximilian_hasler <maximilian.hasler@...> wrote:


                                From: maximilian_hasler <maximilian.hasler@...>
                                Subject: [PrimeNumbers] Re: Conjecture
                                To: primenumbers@yahoogroups.com
                                Date: Monday, September 17, 2012, 3:34 PM



                                 



                                --- In primenumbers@yahoogroups.com, Robert Gerbicz <robert.gerbicz@...> wrote:
                                >
                                > Three more terms, and no more for e2,e3<40000 (if we write n=2^e2*3^e3)
                                > 2^26887*3^13438
                                > 2^35399*3^26448
                                > 2^18350*3^38015

                                At least this definitively rules out the Conjecture in its initial form.
                                (To leave it open, change "40" to "a finite number".)
                                Anyway, as would say a Russian colleague of mine, 40 is not a number(*).
                                More surprisingly, it seems that 42 isn't the answer, either.

                                (* Although I vaguely recall some people stating that 10^40 was a "fundamental" ratio occuring in several instances when looking at sizes and/or masses in the known universe... no link to dwarf Cepheid type stars though.)

                                For the records, I tentatively submitted the draft http://oeis.org/draft/A216854 to store these results.
                                (Not sure other editors of OEIS will not delete rather than "approve" & publish this...)

                                Regards,
                                Maximilian








                                [Non-text portions of this message have been removed]
                              • James Merickel
                                Got it, Max, but just focused on the early part of it to the exclusion of the end.  Sorry I did not read more carefully.  Your sequence is only the beginning
                                Message 15 of 30 , Sep 18, 2012
                                • 0 Attachment
                                  Got it, Max, but just focused on the early part of it to the exclusion of the end.  Sorry I did not read more carefully.  Your sequence is only the beginning of what's available.  Mine are the exponents.  It seems that what is best is to keep mine and possibly also yours (which cannot have 44 numbers in the visible 'front part' ofr the sequence at OEIS, but actually link full data.  If Robert has that, that would be good.  I cannot move my data in such a large way to anywhere on the web right now.  Incompetence is part, but only actually a rather small cause of that. 
                                  JimMe

                                  --- On Mon, 9/17/12, maximilian_hasler <maximilian.hasler@...> wrote:


                                  From: maximilian_hasler <maximilian.hasler@...>
                                  Subject: [PrimeNumbers] Re: Conjecture
                                  To: primenumbers@yahoogroups.com
                                  Date: Monday, September 17, 2012, 3:34 PM



                                   



                                  --- In primenumbers@yahoogroups.com, Robert Gerbicz <robert.gerbicz@...> wrote:
                                  >
                                  > Three more terms, and no more for e2,e3<40000 (if we write n=2^e2*3^e3)
                                  > 2^26887*3^13438
                                  > 2^35399*3^26448
                                  > 2^18350*3^38015

                                  At least this definitively rules out the Conjecture in its initial form.
                                  (To leave it open, change "40" to "a finite number".)
                                  Anyway, as would say a Russian colleague of mine, 40 is not a number(*).
                                  More surprisingly, it seems that 42 isn't the answer, either.

                                  (* Although I vaguely recall some people stating that 10^40 was a "fundamental" ratio occuring in several instances when looking at sizes and/or masses in the known universe... no link to dwarf Cepheid type stars though.)

                                  For the records, I tentatively submitted the draft http://oeis.org/draft/A216854 to store these results.
                                  (Not sure other editors of OEIS will not delete rather than "approve" & publish this...)

                                  Regards,
                                  Maximilian








                                  [Non-text portions of this message have been removed]
                                • James Merickel
                                  Very well.  Do that.  My reading has been that all my terms for exponents are correct and in order through the 44th terms.  It may only be the case for
                                  Message 16 of 30 , Sep 19, 2012
                                  • 0 Attachment
                                    Very well.  Do that.  My reading has been that all my terms for exponents are correct and in order through the 44th terms.  It may only be the case for (at least) the 41st term--you may wish to consider 42nd through 44th tentatively correct or likely wrong.  I'm in no position to check those (next several days).  Just add them in a note or hold off until they are checked.  The problem: It appears that Robert is saying that he is checking in sequence through a maximum exponent value.  The first 40 were found 'correctly', and it shouldn't matter for at least the 41st.
                                    --- On Wed, 9/19/12, Maximilian Hasler <maximilian.hasler@...> wrote:


                                    From: Maximilian Hasler <maximilian.hasler@...>
                                    Subject: Re: [PrimeNumbers] Re: Conjecture
                                    To: "James Merickel" <moralforce120@...>
                                    Date: Wednesday, September 19, 2012, 12:00 PM


                                    I can put your name as co-author in the three sequences already published since Monday
                                     (see FORMULA).


                                    Maximilian




                                    On Wed, Sep 19, 2012 at 2:52 PM, James Merickel <moralforce120@...> wrote:






                                    Max:
                                    I'm changing the titles to include your A-number.  It is a sequence I had been planning to do myself.  So, all three should link across to two others (to start with, but there's that skew-coincidence also that is going to be expanded upon as well, and probably other stuff).

                                    JimMe

                                    --- On Mon, 9/17/12, maximilian_hasler <maximilian.hasler@...> wrote:



                                    From: maximilian_hasler <maximilian.hasler@...>
                                    Subject: [PrimeNumbers] Re: Conjecture
                                    To: primenumbers@yahoogroups.com
                                    Date: Monday, September 17, 2012, 3:34 PM



                                     





                                    --- In primenumbers@yahoogroups.com, Robert Gerbicz <robert.gerbicz@...> wrote:
                                    >
                                    > Three more terms, and no more for e2,e3<40000 (if we write n=2^e2*3^e3)
                                    > 2^26887*3^13438
                                    > 2^35399*3^26448
                                    > 2^18350*3^38015

                                    At least this definitively rules out the Conjecture in its initial form.
                                    (To leave it open, change "40" to "a finite number".)
                                    Anyway, as would say a Russian colleague of mine, 40 is not a number(*).
                                    More surprisingly, it seems that 42 isn't the answer, either.

                                    (* Although I vaguely recall some people stating that 10^40 was a "fundamental" ratio occuring in several instances when looking at sizes and/or masses in the known universe... no link to dwarf Cepheid type stars though.)

                                    For the records, I tentatively submitted the draft http://oeis.org/draft/A216854 to store these results.
                                    (Not sure other editors of OEIS will not delete rather than "approve" & publish this...)

                                    Regards,
                                    Maximilian





                                    [Non-text portions of this message have been removed]
                                  • ronhallam@lineone.net
                                    The other day I was browsing through Riesel and opened it at Appendix 3, Legendre’s symbol, x^2 º a mod p; a thought came into my mind from where who
                                    Message 17 of 30 , Jul 6, 2014
                                    • 0 Attachment
                                      The other day I was browsing through Riesel and opened it at Appendix
                                      3, Legendre’s symbol,
                                      x^2 º a mod p; a thought came into my mind from where who knows!!
                                      It was just silly, that it was equivalent to N = pq.



                                      Conjecture

                                      An odd p < s (s the integer square root) will be a factor of N a
                                      composite odd number if and only if



                                      s^2 = - (N - s^2) mod p .





                                      I can prove the main part but not that p can be either a prime or a
                                      composite.

                                      Proof

                                      N = pq

                                      s = flr(sqr( N))

                                      Redefine p and q as follows:-



                                      p = s - t ( t is an integer - 0 < t < s )

                                      q = s + t + k ( k is an integer and will always be of the form of
                                      either (0 mod 4) or (2 mod 4) depending on N



                                      Substituting the redefined p and q , back into N = pq



                                      N = ( s - t)( s + t + k)

                                      N = s^2 + ks - kt - t^2 (the st values cancel out)

                                      Solve for t^2



                                      t^2 = k(s - t) - (N - s^2)

                                      Taking the modulus of both sides using (s - t) gives



                                      t^2 = 0 - ( N - s^2) (mod ( s - t))



                                      t^2 mod(s - t) = s^2 mod ( s- t) (any difference is a multiple of ( s
                                      - t))



                                      This gives



                                      s^2 = - ( N - s^2) mod p.

                                      QED



                                      Example

                                      I will use a number from Riesel (page 147 of my edition)

                                      N = 13199

                                      s = 114

                                      p = 67



                                      (114^2) mod 67 = 65

                                      (N - s^2) = 203 => 203 mod 67 = 2

                                      This is 67 - 2 = 65



                                      I have looked at some of the RSA numbers that have been solved and they
                                      also confirm the above.



                                      A number that shows that p can be composite is 1617

                                      N = 1617

                                      s = 40

                                      p = 33

                                      (40^2) mod 33 = 16

                                      N - s^2 = 1617 - 1600 = 17

                                      p - 17 = 33 - 17 = 16



                                      If the residues to a number N, are found using the N and then (N-s^2)
                                      they give 2 different sets; example using 12007001



                                      Residues for N =? (-1 2 3 5 7 11 13 23 29 37 43 71 73 89 97)



                                      Residues for (N - s^2) => (-1 2 5 23 31 43 53 59 61 67 71 89 97)



                                      Intersection of the 2 sets gives (-1 2 5 23 43 71 89 97)



                                      I am not sure that this will be of any real benefit, but who knows.





                                      Ron
                                    • djbroadhurst
                                      ... This is trivially equivalent to N = 0 mod p, whatever s might be. So the conjecture is that N = 0 mod p if and only if p divides N. This is true, but
                                      Message 18 of 30 , Jul 7, 2014
                                      • 0 Attachment
                                        ---In primenumbers@yahoogroups.com, <ronhallam@...> wrote :

                                        > s^2 = - (N - s^2) mod p .

                                        This is trivially equivalent to N = 0 mod p, whatever s might be.
                                        So the "conjecture" is that N = 0 mod p if and only if p
                                        divides N. This is true, but hardly noteworthy.

                                        David 






                                      Your message has been successfully submitted and would be delivered to recipients shortly.