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pseudo-free numbers

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  • leavemsg1
    classes of pseudo-free numbers... Let R = k*b^n + (b-1) be just such a number; the multiplier k is an odd natural number; and base b is a prime number,
    Message 1 of 5 , Jul 20, 2009
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      classes of pseudo-free numbers...

      Let R = k*b^n + (b-1) be just such a number;
      the multiplier 'k' is an odd natural number; and
      base 'b' is a prime number, preferably small; and
      exponent 'n' is a prime number, preferably large;

      if and only if... b^(R-1) == 1 (mod R), then 'R' is
      a prime number, and the primality test with respect
      to base 'b' will be pseudo-free.

      does anyone agree with me or can someone find an ex-
      plicit counter-example to refute the idea ???

      the first instance is a Proth number, base 'b' = 2

      Bill
    • Bill Bouris
      no takers ???  the formula s a great find... I invite you to search for a counter-example! ... From: leavemsg1 Subject: [PrimeNumbers]
      Message 2 of 5 , Jul 22, 2009
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        no takers ???  the formula's a great find... I invite you to search for a counter-example!

        --- On Mon, 7/20/09, leavemsg1 <leavemsg1@...> wrote:


        From: leavemsg1 <leavemsg1@...>
        Subject: [PrimeNumbers] pseudo-free numbers
        To: primenumbers@yahoogroups.com
        Date: Monday, July 20, 2009, 11:22 AM


         



        classes of pseudo-free numbers...

        Let R = k*b^n + (b-1) be just such a number;
        the multiplier 'k' is an odd natural number; and
        base 'b' is a prime number, preferably small; and
        exponent 'n' is a prime number, preferably large;

        if and only if... b^(R-1) == 1 (mod R), then 'R' is
        a prime number, and the primality test with respect
        to base 'b' will be pseudo-free.

        does anyone agree with me or can someone find an ex-
        plicit counter-example to refute the idea ???

        the first instance is a Proth number, base 'b' = 2

        Bill



















        [Non-text portions of this message have been removed]
      • richard_in_reading
        ... Unproved assertions of the lack of pseudoprimes for numbers of a special form that get big fast are not particularly interesting. ... Counterexample with
        Message 3 of 5 , Jul 22, 2009
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          > no takers ???  the formula's a great find... I invite you to search for a counter-example!

          Unproved assertions of the lack of pseudoprimes for numbers of a special form that get big fast are not particularly interesting.

          > the first instance is a Proth number, base 'b' = 2

          Counterexample with largest prime exponent and smallish multiplier I've found

          801*2^17+(2-1)=104988673=73.673.2137 is a 2-psp

          also 262145*2^19+1 is a 2-psp

          What's the reason for restricting to prime exponents?

          Richard Heylen
        • Bill Bouris
          sorry the conjecture was missing one component... let R = k * b^n +(b-1) where k
          Message 4 of 5 , Jul 23, 2009
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            sorry the conjecture was missing one component...

            let R = k * b^n +(b-1) where k <= b^n +(b-1), k odd; b is a preferably
            small prime and n is prime, but larger than 3; now, the restriction on
            k is compatible with the Proth definition.
            are there still counter-examples ??? I don't have the computer power
            to verify it; Bill

            --- On Thu, 7/23/09, Maximilian Hasler <maximilian.hasler@...> wrote:

            > From: Maximilian Hasler <maximilian.hasler@...>
            > Subject: Re: [PrimeNumbers] Re: pseudo-free numbers
            > To: "richard_in_reading" <richard_in_reading@...>, "Bill Bouris" <leavemsg1@...>
            > Date: Thursday, July 23, 2009, 4:38 AM
            > Counter-examples [k,n, b^(R-1) mod
            > R]
            > for b=2 :
            > ? bbcheck(2)
            > [85, 2, Mod(1, 341)]
            > [161, 2, Mod(1, 645)]
            > [675, 2, Mod(1, 2701)]
            > [705, 2, Mod(1, 2821)]
            > [819, 2, Mod(1, 3277)]
            > [585, 3, Mod(1, 4681)]
            > [825, 3, Mod(1, 6601)]
            > [77, 5, Mod(1, 2465)]
            > [265, 5, Mod(1, 8481)]
            > [495, 5, Mod(1, 15841)]
            > [585, 5, Mod(1, 18721)]
            > [805, 5, Mod(1, 25761)]
            > [65, 7, Mod(1, 8321)]
            > [259, 7, Mod(1, 33153)]
            > [273, 7, Mod(1, 34945)]
            > [801, 17, Mod(1, 104988673)]
            >
            > ? bbcheck(3) \\ counter-examples for b=3
            > [171, 2, Mod(1, 1541)]
            > [551, 2, Mod(1, 4961)]
            > [57, 3, Mod(1, 1541)]
            > [563,  Mod(1, 15203)]
            >
            > (18:00) gp > bbcheck(5)\\ counter-examples for b=5
            > [69, 2, Mod(1, 1729)]
            > [321, 2, Mod(1, 8029)]
            > [93, 5, Mod(1, 290629)]
            >
            > (18:01) gp > bbcheck(7)\\ counter-examples for b=7
            > [235, 2, Mod(1, 11521)]
            > [527, 2, Mod(1, 25829)]
            > [521, 3, Mod(1, 178709)]
            >
            > (18:02) gp > bbcheck(11)\\ counter-examples for b=11
            > [463, 2, Mod(1, 56033)]
            >
            > Maximilian
            >
            > On 7/23/09, richard_in_reading <richard_in_reading@...>
            > wrote:
            > >
            > >  > no takers ???  the formula's a great
            > find... I invite you to search for a counter-example!
            > >
            > >
            > > Unproved assertions of the lack of pseudoprimes for
            > numbers of a special form that get big fast are not
            > particularly interesting.
            > >
            > >
            > >  > the first instance is a Proth number, base
            > 'b' = 2
            > >
            > >
            > > Counterexample with largest prime exponent and
            > smallish multiplier I've found
            > >
            > >  801*2^17+(2-1)=104988673=73.673.2137 is a 2-psp
            > >
            > >  also 262145*2^19+1 is a 2-psp
            > >
            > >  What's the reason for restricting to prime
            > exponents?
            > >
            > >  Richard Heylen
            > >
            > >
            > >
            > >
            > >  ------------------------------------
            > >
            > >  Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
            > >  The Prime Pages : http://www.primepages.org/
            > >
            > >  Yahoo! Groups Links
            > >
            > >
            > >
            > >
            > >
            > >
            > >
            > >
            >
          • Bill Bouris
            I believe that the following class of numbers is pseudo-free... Let R = k * b^n +(b-1) where k is odd positive, b is a small prime base, n is a larger prime
            Message 5 of 5 , Aug 1, 2009
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              I believe that the following class of numbers is pseudo-free...

              Let R = k * b^n +(b-1) where k is odd positive, b is a small prime base,
              n is a larger prime number and k is restricted to be <= b^n +(b-1).

              when tested against the base 'b', iff b^(R-1) ==1 (mod R), then 'R' is
              prime, never a pseudo-prime number.

              Bill

              can anyone refute this ???
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