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Re: Carmichael question

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  • mistermac39
    Here is a little contribution about handwavey methods. It does not specifically deal with Carmichaels, although my first examinations are coming up with some
    Message 1 of 77 , Nov 2, 2002
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      Here is a little contribution about handwavey methods. It does not
      specifically deal with Carmichaels, although my first examinations
      are coming up with some interesting material on them as well.
      Let us consider f(a,b) = b^2-a^2-a*b and f(c,d) = d^2-c^2-c*d
      a,b,c,d all integers. Actually it works for all a,b,c,d.

      Take
      a,c
      b,d
      a+b,c+d

      Multiply each line and get
      a*c (L1)
      b*d (L2)
      a*c+b*c+b*c+b*d (L3)

      Take (L1)+(L2) (L4)
      and (L2)+(L3) (L5)

      Take (L5)-(L4) (L6)

      You will find that (L6)^2 - (L4)*(L5) = +/-f(a,b)*f(b,c)

      Do the same for
      a,-d
      b,c
      a+b,c-d

      and you will get a new L6,L5,L4 which gives +/-f(a,b)*f(b,c)

      Thus you have two new functions which give the product ot the first 2

      A worked example

      31 = 7^2-2*9 and 11 = 4^2-1*5
      2*1=2
      7*4=28
      9*5=45

      2+28=30
      28+45=73
      73-30=43

      43^2-30*73=-341=-31*11

      2*(-4)=-8
      7*1=7
      9*(-3)=-27

      -8+7=-1
      7-27=-20
      -20-(-1)=-19

      (-19)^2 -(-1)(-19) = 341

      This method will work for all Lucas sequences, with appropriate
      changes in the original polynomial coefficients.

      John McNamara
    • Phil Carmody
      Apologies for the aborted attempt at a post. Tab followed by either space or return (too quick to know what happened) seems to be a lethal key
      Message 77 of 77 , Apr 26, 2005
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        Apologies for the aborted attempt at a post. 'Tab' followed by either
        'space' or 'return' (too quick to know what happened) seems to be a
        lethal key combination.

        From: "mcnamara_gio" <mcnamara_gio@...>
        >
        > What do you think about this sequence a(n)=n^2+7n-1. Its terms are
        > usually prime. I have calculated that 72% of a(n) so that n<500 is
        > prime. 81% of a(n) is prime when n<5000. 85.6% of a(n) is prime when
        > n<50000 and 88.5% of a(n) is prime when n<500000. I am going to find
        > more prime terms in this sequence. What do you think about it?

        2, 3, and 5 can never be factors. This boosts density by a factor of
        (2/1)*(3/2)*(5/4) over arbitrary ranges. However, 7,11, 13 and 17 both
        divide 2 of the p possible residues. This decreases density by a factor
        of (5/6)*(9/10)*(11/12)*(15/16) over arbitrary ranges.

        Looking at primes up to 10000, the density boost is almost exactly 2.75.
        This is pretty feeble compared with Euler's famous trinomials.

        Run this script in Pari/GP:

        rnorm=1.0
        rthis=1.0
        forprime(p=2,10000,roots=polrootsmod(x^2+7*x-1,p)~;rnorm*=(p-1)/p;rthis*=(p-#roots)/p;print(p"
        "rthis" "rnorm" "roots))
        print(rthis/rnorm);

        Research the Euler trinomials, and try the above script on them too, to see why
        I say 2.75 is pretty feeble.

        Phil


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