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Math Help - Divisibility of central binomial coefficient

  1. #16
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    Yes, you are right.
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  2. #17
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    Quote Originally Posted by chiph588@ View Post
    I was hoping with the addition of  \left\{\frac np\right\}=0 and a constraint on the size of  p , this was an if and only if statement, i.e. "if  0<\left\{\frac np\right\}<\frac12 , then  p\not|\binom{2n}{n} .

    This turns out to be false as  3\mid\binom{14}{7} and  \left\{\frac73\right\}=\frac13 .
    Hello chiph588@, I think I've found what you're looking for, namely a sufficient and necessary condition of the divisibility of  \binom{2n}{n} by a prime p in general!

    A prime p divides  \binom{2n}{n} if and only if the fractional part \left \{ \frac{n}{p^k} \right \}\geq\frac{1}{2} for at least one k\in \mathbb{N}. In particular, when k=1 .

    For example, 3\not|\binom{8}{4} and indeed \left \{ \frac{4}{3^1} \right \}<\frac{1}{2}; \left \{ \frac{4}{3^2} \right \}<\frac{1}{2}; and so on...

    Another example, the one that you gave here is 3\mid\binom{14}{7}<br />
and as you noted \left\{\frac73\right\}=\frac13, but if you continue \left\{\frac{7}{3^2}\right\}\geq\frac12.
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  3. #18
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by melese View Post
    Hello chiph588@, I think I've found what you're looking for, namely a sufficient and necessary condition of the divisibility of  \binom{2n}{n} by a prime p in general!

    A prime p divides  \binom{2n}{n} if and only if the fractional part \left \{ \frac{n}{p^k} \right \}\geq\frac{1}{2} for at least one k\in \mathbb{N}. In particular, when k=1 .

    For example, 3\not|\binom{8}{4} and indeed \left \{ \frac{4}{3^1} \right \}<\frac{1}{2}; \left \{ \frac{4}{3^2} \right \}<\frac{1}{2}; and so on...

    Another example, the one that you gave here is 3\mid\binom{14}{7}<br />
and as you noted \left\{\frac73\right\}=\frac13, but if you continue \left\{\frac{7}{3^2}\right\}\geq\frac12.
    Yes, I came to the same conclusion last night too!
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  4. #19
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    Me too.
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