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Math Help - Sequence and divisibility

  1. #1
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    Sequence and divisibility

    We have recurrent sequence of integer number a1,a2,... a1=1, a2=2 an=3a(n-1)+5a(n-2) for n=3,4,5,... Is integer number k>=2, that a(k+1)*a(k+2) mod ak = 0 ? Please for quick help
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by terafull View Post
    We have recurrent sequence of integer number a1,a2,... a1=1, a2=2 an=3a(n-1)+5a(n-2) for n=3,4,5,... Is integer number k>=2, that a(k+1)*a(k+2) mod ak = 0 ? Please for quick help
    Lets just translate this into good English.

    \{a_n\} is a sequence definde by the recurrence:

    a_1=1,\ a_2=2,\ a_n=3a_{n-1}+5a_{n-2} \mbox{ for }n>2

    Is there a k \ge 2 such that:

    <br />
a_{k+1}\times a_{k+2} \equiv 0 \mbox{ modulo } a_k?

    RonL
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  3. #3
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    Exactly, thx
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    Lets just translate this into good English.

    \{a_n\} is a sequence definde by the recurrence:

    a_1=1,\ a_2=2,\ a_n=3a_{n-1}+5a_{n-2} \mbox{ for }n>2

    Is there a k \ge 2 such that:

    <br />
a_{k+1}\times a_{k+2} \equiv 0 \mbox{ modulo } a_k?

    RonL
    Well, let's start by writing out an explicit representation of the series:
    a_n = \left ( \frac{-29 + 13\sqrt{29}}{290} \right ) \left ( \frac{3 + \sqrt{29}}{2} \right ) ^n + \left ( \frac{-29 - 13\sqrt{29}}{290} \right ) \left ( \frac{3 - \sqrt{29}}{2} \right ) ^n

    You wish to know if
    a_{n + 2} a_{n + 1} \equiv 0 \text{ mod }a_n

    a_{n + 2} = 3a_{n + 1} + 5a_n

    So this amounts to wanting to know if
    (3a_{n + 1} + 5a_n ) a_{n + 1} \equiv 0 \text{mod }a_n

    Which amounts to questioning if a_n ever divides a_{n + 1} evenly.

    -Dan

    By the way, I've been practicing my difference equations:
    a_n = \left ( \frac{-29 + 13\sqrt{29}}{290} \right ) \left ( \frac{3 + \sqrt{29}}{2} \right ) ^n + \left ( \frac{-29 - 13\sqrt{29}}{290} \right ) \left ( \frac{3 - \sqrt{29}}{2} \right ) ^n
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  5. #5
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    I had done what you wrote, but I dont know what next... I dont see proof for: a_n ever divides a_{n + 1} od doesnt divide.
    Last edited by terafull; October 8th 2007 at 05:45 AM.
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  6. #6
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    We will show that \gcd(a_n,a_{n+1}) = 1 for all n this is true for n=1 now accept induction. Say d=\gcd(a_{n+1},a_{n+2}) then d|a_{n+1} and d|a_{n+2}\implies d|(5a_{n+1}+3a_n)\implies d|(3a_n). Now if d>1 (by contradiction) then d\not | a_n\implies d|3 thus d=3. It remains to show that none of the numbers in this sequence ever are divisible by 3. Again accept induction. Say a_{n+1} = 5a_n+3a_{n-1} is divisible by 3 this means 3|(5a_n)\implies 3|5 by induction. Which is an impossibility.
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