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Re: Carmichael numbers of order 2

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  • Robert Gerbicz
    I was lazy and did not use binary indexing for products of primes. Yes, I used long long int (8bytes) to store the residue (L
    Message 1 of 12 , Apr 2, 2009
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      "I was lazy and did not use binary indexing for products of primes."

      Yes, I used long long int (8bytes) to store the residue (L<2^63), and int (4
      bytes) to store the offset of the prime divisors (there were 26 primes in P
      so it was enough).

      Searching for Carmichael number of order 3 would be very hard:
      by the following quick code:
      generatesmooth(n)=T=1;forprime(p=2,n,q=p;while(q*p<=n,q*=p);T*=q);return(T)
      this is generating super smooth numbers, but for example for n=1000 this is
      giving only P(3,T)~225 numbers
      , the last 3 of them in order what I have found are:
      153572717
      217815289
      264982051
      So there are very few larger primes in P(3,T) set. Say there are another 25,
      then the expected number of solutions are
      2^250/eulerphi(generatesmooth(1000))=3*10^(-357), and it would require about
      2^200 multiplications/divisions (using man-in-the middle algorithm #P=25,
      #Q=200). Obviously using smaller n gives larger but still very small
      probability to find a 3 order number. So we have to increase the value of n:

      For example: L=generatesmooth(20000) gives P(L,3)>30000 primes and this
      means that it generates many Carmichael numbers of order 3:
      2^30000/eulerphi(L)>>1. But to find even one solution using this L value is
      still very hard.


      [Non-text portions of this message have been removed]
    • David Broadhurst
      In a talk http://alumnus.caltech.edu/%7Ehowever/talks/FortCollins.pdf given in December 2006, Everett Howe posed this Open Problem : What are the first 3
      Message 2 of 12 , Apr 2, 2009
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        In a talk
        http://alumnus.caltech.edu/%7Ehowever/talks/FortCollins.pdf
        given in December 2006, Everett Howe posed this "Open Problem":

        "What are the first 3 Carmichael numbers of order 2?"

        Here is the answer:

        443372888629441
        39671149333495681
        842526563598720001

        David Broadhurst
      • David Broadhurst
        ... Richard Finch calls these unusually strong Lucas-Carmichael-minus (uLC-) numbers, with p^2-1|N-1, for every prime p|N. See:
        Message 3 of 12 , Apr 3, 2009
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          --- In primenumbers@yahoogroups.com,
          "David Broadhurst" <d.broadhurst@...> wrote:

          > "What are the first 3 Carmichael numbers of order 2?"
          >
          > 443372888629441
          > 39671149333495681
          > 842526563598720001

          Richard Finch calls these
          "unusually strong Lucas-Carmichael-minus" (uLC-) numbers,
          with p^2-1|N-1, for every prime p|N. See:
          http://www.chalcedon.demon.co.uk/rgep/p20.pdf

          I found the third of these by mining Richard's file of
          Carmichael numbers between 10^17 than 10^18.
          The first two had already been noted in
          http://www.chalcedon.demon.co.uk/rgep/cartable.html

          Richard classified
          582920080863121 = 41 * 53 * 79 * 103 * 239 * 271 * 509
          as a "strong", but not "unusually strong",
          Lucas-Carmichael-minus number
          since in this case both p-1 and p+1
          divide N-1, for each prime p|N, but p^2-1
          does not divide N-1 in the cases p = 79, 239, 271,
          where p^2 = 1 mod 2^5
          whilst N = 17 mod 2^5.

          If we ask merely that p-1 and (p+1)/2 divide N-1,
          then the following 3 numbers also occur, for N < 10^18:
          28295303263921
          894221105778001
          2013745337604001
          making 7 in all, of which only

          842526563598720001
          = 17 * 61 * 71 * 89 * 197 * 311 * 769 * 2729

          occurs for 10^18 > N > 10^17.

          Finally, I remark that Richard found precisely one
          "unusually strong Lucas-Carmichael-plus" (uLC+) number,
          with p^2-1|N+1, for prime p|N and N < 10^13, namely

          79397009999 = 23 * 29 * 41 * 43 * 251 * 269.

          David
        • David Broadhurst
          Oh dear, another silly typo: I meant Richard Pinch, of course.
          Message 4 of 12 , Apr 3, 2009
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            Oh dear, another silly typo: I meant Richard Pinch, of course.
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