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Thread: Carmichael numbers

  1. #16
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    A new direction...

    Iíve got you to $\displaystyle lcm(a_1,a_2,a_3)|gcd(a_2a_3+a_1a_3+a_1a_2 , a_1+a_2+a_3)$

    Now, letís keep going. Define $\displaystyle g=gcd(a_1,a_2,a_3)$ and let $\displaystyle u_1=a_1/g$ , $\displaystyle u_2=a_2/g$ , $\displaystyle u_3=a_3/g$ .

    Then $\displaystyle u_1u_2u_3g|gcd((u_2u_3+u_1u_3+u_1u_2 )g^2, (u_1+u_2+u_3)g)$

    Or, alternatively:

    $\displaystyle u_1|u_2u_3(gk)+(u_2+u_3)$
    $\displaystyle u_2|u_1u_3(gk)+(u_1+u_3)$
    $\displaystyle u_3|u_2u_1(gk)+(u_2+u_1)$

    In order to satisfy this for all $\displaystyle k$, $\displaystyle u_1u_2u_3|g$ and $\displaystyle u_1u_2u_3|u_1+u_2+u_3$ . This calls for three relatively prime integers whose product divides their sum. I believe $\displaystyle 1,2,3$ is the only such arrangement, leading us into a dead end.

    However, after looking at some raw data, it looks like $\displaystyle gcd(p_1-1,p_2-1,p_3-1)$ and subsequent $\displaystyle p_1/g, p_2/g, p_3/g$ are all over the map. This leads me to pose another conjecture:

    Weak: For any three relatively prime $\displaystyle u_1<u_2<u_3$, there exists a $\displaystyle k$ such that $\displaystyle (u_1k+1) (u_2k+1) (u_3k+1)$ is a three-factor Carmichael Number.

    Strong: For any three relatively prime $\displaystyle u_1<u_2<u_3$, there exists an $\displaystyle a,b$ such that $\displaystyle (u_1k+1) (u_2k+1) (u_3k+1)$ is a three-factor Carmichael Number iff $\displaystyle k \equiv b (\bmod a)$ and $\displaystyle u_ik+1$ is prime for all $\displaystyle i$.

    Examples:
    For $\displaystyle u=(1,2,3), a=6, b=0$ generates Carmichaels 1729, 294409, 56052361, 118901521 Ö
    For $\displaystyle u=(1,3,5), a=30, b=12$ generates Carmichaels 29341, 1152271, 64377911, 775368901 Ö
    For $\displaystyle u=(1,4,7), a=28, b=4$ generates Carmichaels 2465, 6189121, 19384289, 34153717249 Ö
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  2. #17
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    Quote Originally Posted by Media_Man View Post
    Iíve got you to $\displaystyle lcm(a_1,a_2,a_3)|gcd(a_2a_3+a_1a_3+a_1a_2 , a_1+a_2+a_3)$

    Now, letís keep going. Define $\displaystyle g=gcd(a_1,a_2,a_3)$ and let $\displaystyle u_1=a_1/g$ , $\displaystyle u_2=a_2/g$ , $\displaystyle u_3=a_3/g$ .

    Then $\displaystyle u_1u_2u_3g|gcd((u_2u_3+u_1u_3+u_1u_2 )g^2, (u_1+u_2+u_3)g)$
    I'll have to take some time to think about your other conjectures, however you should know that we do not in general have the equality $\displaystyle lcm(a_1,a_2,a_3)=u_1u_2u_3g$, where $\displaystyle g=gcd(a_1,a_2,a_3)$. Notice that having the property $\displaystyle gcd(u_1,u_2,u_3)=1$ does not necessarily imply that $\displaystyle gcd(u_i,u_j)=1$ for each distinct $\displaystyle i,j$.
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  3. #18
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    Quote Originally Posted by Media_Man View Post

    However, after looking at some raw data, it looks like $\displaystyle gcd(p_1-1,p_2-1,p_3-1)$ and subsequent $\displaystyle p_1/g, p_2/g, p_3/g$ are all over the map.
    Suppose $\displaystyle n=p_1p_2p_3$ is a Carmichael and let $\displaystyle D =gcd(p_1-1,p_2-1,p_3-1)$. Try finding out what the number of (approximately) distinct values the ratio $\displaystyle n/D$ can take for all Carmichaels $\displaystyle n$ bounded above by some integer N (ex N=10,000, N=100,000 etc).

    What are the smallest and/or average values of $\displaystyle n/D$ for various N?
    Last edited by jamix; Apr 21st 2009 at 09:56 PM.
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