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

Euclid primes [Fwd: SEQ FROM Roger L. Bagula]

Expand Messages
  • Roger Bagula
    ... Subject: SEQ FROM Roger L. Bagula Date: Fri, 31 Oct 2003 14:14:46 -0500 (EST) From: Reply-To: tftn@earthlink.net To:
    Message 1 of 8 , Oct 31, 2003
    • 0 Attachment
      -------- Original Message --------
      Subject: SEQ FROM Roger L. Bagula
      Date: Fri, 31 Oct 2003 14:14:46 -0500 (EST)
      From: <njas@...>
      Reply-To: tftn@...
      To: njas@...
      CC: tftn@...



      The following is a copy of the email message that was sent to njas
      containing the sequence you submitted.

      All greater than and less than signs have been replaced by their html
      equivalents. They will be changed back when the message is processed.

      This copy is just for your records. No reply is expected.
      Subject: NEW SEQUENCE FROM Roger L. Bagula


      %I A000001
      %S A000001 3,7,19,31,59,61,73,97,131,139,167,173,181,211,223,277,317,331,347,509,571,953,
      1039,1063,2311,2521,2719,3467,22093,27953,34231,46727,60611,105229,265739,
      303049,450451,598841,676421,703763,2336993,11072701,13848803,1503181961,
      13808181181,19026377261,54730729297,200560490131,2892214489673,18564761860301,
      78339888213593,143581524529603,96888414202798247,525956867082542470777,
      71237436024091007473549,16156160491570418147806951,64225891884294373371806141,
      1004988035964897329167431269
      %N A000001 Unique prime factors from Eculid's proof of infinite primes
      %C A000001 These Euclid primes get very large very fast and represent
      only a very few of the actual Primes.
      %F A000001 Prime Factors of F[n_]=Product[Prime[i],{i,1,n}}+1
      %t A000001 p[n_]=Product[Prime[i],{i,1,n}]
      a=Table[FactorInteger[p[n]+1],{n,1,25}];
      b=Flatten[a];
      c=Delete[Union[b],1]
      %O A000001 1
      %K A000001 ,nonn,
      %A A000001 Roger L. Bagula (tftn@...), Oct 31 2003
      RH
      RA 209.178.146.96
      RU
      RI



      --
      Respectfully, Roger L. Bagula
      tftn@..., 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 :
      URL : http://home.earthlink.net/~tftn
      URL : http://victorian.fortunecity.com/carmelita/435/




      [Non-text portions of this message have been removed]
    • Edwin Clark
      Roger, How do you know that if you increase the range from 25 to to some larger number that you won t get the missing primes: 5, 11, 13, ...? --Edwin
      Message 2 of 8 , Oct 31, 2003
      • 0 Attachment
        Roger,

        How do you know that if you increase the range from 25 to
        to some larger number that you won't get the missing primes:
        5, 11, 13, ...?


        --Edwin


        On Fri, 31 Oct 2003, Roger Bagula wrote:
        >
        > %I A000001
        > %S A000001 3,7,19,31,59,61,73,97,131,139,167,173,181,211,223,277,317,331,347,509,571,953,
        > 1039,1063,2311,2521,2719,3467,22093,27953,34231,46727,60611,105229,265739,
        > 303049,450451,598841,676421,703763,2336993,11072701,13848803,1503181961,
        > 13808181181,19026377261,54730729297,200560490131,2892214489673,18564761860301,
        > 78339888213593,143581524529603,96888414202798247,525956867082542470777,
        > 71237436024091007473549,16156160491570418147806951,64225891884294373371806141,
        > 1004988035964897329167431269
        > %N A000001 Unique prime factors from Eculid's proof of infinite primes
        > %C A000001 These Euclid primes get very large very fast and represent
        > only a very few of the actual Primes.
        > %F A000001 Prime Factors of F[n_]=Product[Prime[i],{i,1,n}}+1
        > %t A000001 p[n_]=Product[Prime[i],{i,1,n}]
        > a=Table[FactorInteger[p[n]+1],{n,1,25}];
        > b=Flatten[a];
        > c=Delete[Union[b],1]
        > %O A000001 1
        > %K A000001 ,nonn,
        > %A A000001 Roger L. Bagula (tftn@...), Oct 31 2003
        > RH
        > RA 209.178.146.96
        > RU
        > RI
      • Décio Luiz Gazzoni Filho
        ... Hash: SHA1 ... Consider a prime p
        Message 3 of 8 , Oct 31, 2003
        • 0 Attachment
          -----BEGIN PGP SIGNED MESSAGE-----
          Hash: SHA1

          On Friday 31 October 2003 17:54, Edwin Clark wrote:
          > Roger,
          >
          > How do you know that if you increase the range from 25 to
          > to some larger number that you won't get the missing primes:
          > 5, 11, 13, ...?

          Consider a prime p' < p. Then p' | p# and thus p' does not divide p#+1. So if
          p' does not divide p#+1 for some p < p', then surely it won't do so for any
          p#+1 with p >= p'.

          Décio
          -----BEGIN PGP SIGNATURE-----
          Version: GnuPG v1.2.3 (GNU/Linux)

          iD8DBQE/osSRce3VljctsGsRAssfAJ4g2mlv18ByUhhqMXgaCt1WogifgwCgtZYv
          w5WUSa5Ht2l0ntJK8PR9lHU=
          =lgWL
          -----END PGP SIGNATURE-----
        • Edwin Clark
          ... Okay, I see that. But how about 101 which is not in his list and he only uses primes up to the 25th prime which is 97? Am I missing something else? Edwin
          Message 4 of 8 , Oct 31, 2003
          • 0 Attachment
            On Fri, 31 Oct 2003, [iso-8859-1] Décio Luiz Gazzoni Filho wrote:
            >
            > Consider a prime p' < p. Then p' | p# and thus p' does not divide p#+1. So if
            > p' does not divide p#+1 for some p < p', then surely it won't do so for any
            > p#+1 with p >= p'.
            >

            Okay, I see that. But how about 101 which is not in his list and he only
            uses primes up to the 25th prime which is 97?

            Am I missing something else?

            Edwin
          • Edwin Clark
            ... Here s an example. Suppose we calculate the sequence going to 13 (the 6th prime) then we get the sequence [3, 7, 31, 59, 211, 509, 2311] Note that 19, 97
            Message 5 of 8 , Oct 31, 2003
            • 0 Attachment
              On Fri, 31 Oct 2003, Edwin Clark wrote:

              > On Fri, 31 Oct 2003, [iso-8859-1] Décio Luiz Gazzoni Filho wrote:
              > >
              > > Consider a prime p' < p. Then p' | p# and thus p' does not divide p#+1. So if
              > > p' does not divide p#+1 for some p < p', then surely it won't do so for any
              > > p#+1 with p >= p'.
              > >
              >
              > Okay, I see that. But how about 101 which is not in his list and he only
              > uses primes up to the 25th prime which is 97?
              >
              > Am I missing something else?
              >
              > Edwin
              >


              Here's an example. Suppose we calculate the sequence going to 13 (the 6th
              prime) then we get the sequence

              [3, 7, 31, 59, 211, 509, 2311]

              Note that 19, 97 and 277 don't appear. But if I go up one more prime
              to 17 (the 7th prime) then we get:


              [3, 7, 19, 31, 59, 97, 211, 277, 509, 2311]


              So Roger's sequence is complete only up to 97 (the 25th prime), n'est
              pas?

              --Edwin
            • Jack Brennen
              ... Yes, that s correct. Here is the actual sequence. You can see that Roger missed some values, starting with 149: 3 7 19 31 59 61 73 97 131 139 149 167 173
              Message 6 of 8 , Oct 31, 2003
              • 0 Attachment
                > So Roger's sequence is complete only up to 97 (the 25th prime), n'est
                > pas?

                Yes, that's correct. Here is the actual sequence. You can see that Roger
                missed some values, starting with 149:

                3
                7
                19
                31
                59
                61
                73
                97
                131
                139
                149
                167
                173
                181
                211
                223
                271
                277
                307
                313
                317
                331
                347
                463
                467
                509
                571
                601
                673
                809
                827
                877
                881
                953
                983
                997
                1031
                1033
                1039
                1051
                1063
                1069
                1109
                1259
                1279
                1283
                1291
                1297
                1361
                1381
                1471
                .
                .
                .


                PARI/GP Code to generate:

                for(z=2,10000,p=prime(z);g=Mod(1,p);for(a=1,z-1,g*=prime(a);if(g+1==0,print(p);break)));
              • Jens Kruse Andersen
                ... for(z=2,10000,p=prime(z);g=Mod(1,p);for(a=1,z-1,g*=prime(a);if(g+1==0,print(p) ;break))); Roger is missing _lots_ of factors below his largest
                Message 7 of 8 , Oct 31, 2003
                • 0 Attachment
                  Jack Brennen wrote:
                  > Yes, that's correct. Here is the actual sequence. You can see that Roger
                  > missed some values, starting with 149:
                  ...
                  > PARI/GP Code to generate:
                  >
                  >
                  for(z=2,10000,p=prime(z);g=Mod(1,p);for(a=1,z-1,g*=prime(a);if(g+1==0,print(p)
                  ;break)));

                  Roger is missing _lots_ of factors below his largest
                  1004988035964897329167431269. Some of them are here:
                  http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/matha102.htm
                  and here:
                  http://web.tiscali.it/lorenzo75/primorials+1.html

                  I wonder whether all sequences posted by Roger Bagula to OEIS are crap.
                  Judged by his recent posts here, I fear so. I am not going into the discussion
                  with him. Life is too short to spend on some people.

                  By coincidence, I have just today searched primorial factors for question 3 in
                  http://www.primepuzzles.net/puzzles/puzz_240.htm

                  --
                  Jens Kruse Andersen
                • ashok
                  hi, I am working on AKS algorithm. I am done with implementation of the algorithm. Can someone give me any idea as a continuation of this. I don t have much
                  Message 8 of 8 , Nov 1, 2003
                  • 0 Attachment
                    hi,
                    I am working on AKS algorithm. I am done with implementation of
                    the algorithm. Can someone give me any idea as a continuation of this. I
                    don't have much knowlege on the developments that have come up on AKS. If
                    anyone has some idea on this, please provide me with the links.


                    thanking in advance,
                    ashok
                  Your message has been successfully submitted and would be delivered to recipients shortly.