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A VERY interesting Fermat-like sequence: A_n=4^3^n+2^3^n+1

  1. May 7, 2005 #1
    Hi,
    In 1995, Yannick Saouter produced the study of a family of numbers close to the Fermat numbers: [tex]A_n=4^{3^n}+2^{3^n}+1[/tex] .
    (See: http://www.inria.fr/rrrt/rr-2728.html)

    Saouter proved that this A_n serie shares many properties with the Fermat numbers:

    3.4 A_n numbers are pairwise relatively primes

    3.3 [tex]A_n \text{ is prime iff } 5^{(A_n-1)/2} \equiv -1 \pmod{A_n}[/tex]

    3.5 [tex]p | A_n ==> p = 1 \pmod{2.3^{n+1}}[/tex]


    I've also discovered (and checked with PARI) that the following property is true:
    [tex]A_n = 3+2(2^{3^{\scriptstyle n}-1}+1)\prod_{i=0}^{n-1}A_i[/tex]

    (I've summarized the properties at: http://tony.reix.free.fr/Mersenne/PropertiesOfFermatLikeTNumbers.pdf)

    A VERY interesting thing is that the primality of these numbers can be checked with the Pépin's test, with 5 instead of 3, like Fermat numbers.

    Saouter provides the divisors of several of these numbers (n up to 39).
    It appears:
    - that 10 of them have no divisors known
    - only the 3 first Saouter numbes are prime (same as Fermat numbers).

    Is someone interested in studying these numbers in more details ?

    Is it possible to adapt some Pépin's test code to this kind of numbers ?

    Regards,
    Tony
     
  2. jcsd
  3. May 7, 2005 #2

    Zurtex

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    Erm interesting, but due to their size it's a bit computationally awkward to calculate isn't it? I mean I'm sure there are quite a few sequences that behave in similar ways but of such great size that they can't be calculated in any short amount of time.

    Not criticizing you at all, it's something I couldn't do, just wondering.
     
  4. May 8, 2005 #3
    You're right. These numbers grow awfully fast, much faster than Fermat numbers.
    What I think really interesting is that they share several properties with Fermat numbers. Do they also seem to have a finite number of primes ? What about the equivalent of Sierpinski's problem Saouter has studied ? Maybe a proof for Saouter numbers could help for Fermat numbers ?
    Tony
     
  5. May 8, 2005 #4
    Hi Zurtex,
    I've updated the paper with some proof and with the number of digits of A_n.
    For n=16, A_n has ~26 millions of digits. Very big !
    Tony
     
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