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What is the percentage of odd composite positive numbers divisible by 3?

In that same vein, what is the percentage of odd composite positive numbers divisible by 5?

And, for the future, what is the percentage of odd composite positive numbers divisible by n where n is some prime greater than 2?

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  • $\begingroup$ Two questios, do the odd numbers divisible by n occur regularly? How frequently? The answers to those question should be obvious and they should make the answer to your question obvious. $\endgroup$
    – fleablood
    Jul 10, 2016 at 20:56

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"Percentage" here is not a very precise term. So we give a precise version of your question. Let $D(n)$ be the number of odd composites $\le n$ that are divisible by $3$. Let $OC(n)$ be the number of odd composites $\le n$. We want to find $$\lim_{n\to\infty} \frac{D(n)}{OC(n)}.$$ This is $$\lim_{n\to\infty} \frac{D(n)}{n}\cdot \frac{n}{OC(n)}.$$ Every sixth number after $3$ is an odd composite divisible by $3$, so $$\lim_{n\to\infty}\frac{D(n)}{n}=\frac{1}{6}.$$ By the Prime Number Theorem, $$\lim_{n\to\infty}\frac{n}{OC(n)}=2$$ (after a while, primes are sparse). Thus the required limit is $\frac{1}{3}$.

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  • $\begingroup$ This looks like a good answer to the first part of my question, but how about the percentage of composites that are divisible by five, and the rest of the primes? $\endgroup$
    – Linus Rastegar
    Jul 10, 2016 at 21:26
  • $\begingroup$ By $5$, exactly $1/5$, same reasoning. For an odd prime $p$, $\frac{1}{p}$, same reasoning. There are other related problems, such as by $3$ or $5$ (including the possibility of both). Similar reasoning gives $1/3+1/5-\frac{1}{15}=\frac{7}{15}$. $\endgroup$
    – André Nicolas
    Jul 10, 2016 at 21:51
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$$\begin{align} &\lim_{n\to\infty}\frac{\text{#\{odd composites less than $n$ and divisible by $3$\}}}{\text{#\{odd composites less than $n$\}}}\\ &=\lim_{n\to\infty}\frac{\text{#\{odd numbers less than $n$ and divisible by $3$\}-1}}{\text{#\{odd numbers less than $n$\}}-\text{#\{odd primes less than $n$\}}-1}\\ &=\lim_{n\to\infty}\frac{\left\lfloor\frac{n+3}{6}\right\rfloor-1}{\left\lfloor\frac{n}{2}\right\rfloor-O\left(\frac{n}{\log(n)}\right)-1}\\ &=\lim_{n\to\infty}\frac{\left\lfloor\frac{n+3}{6}\right\rfloor}{\left\lfloor\frac{n}{2}\right\rfloor}=\frac{1}{3} \end{align}$$

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  • $\begingroup$ What is the O function in this context? $\endgroup$
    – Linus Rastegar
    Jul 12, 2016 at 19:19
  • $\begingroup$ If you say $O(\frac{n}{\log(n)})$, you mean some function $f(n)$ such that there is a positive number $M$ with $f(n)\leq M\frac{n}{\log(n)}$. So here I'm saying that I don't have an explicit formula for $\#\{\text{odd primes less than n}\}$, but I know from the prime density theorem that this function grows no faster than some constant multiple of $\frac{n}{\log(n)}$ That in turn grows much smaller than $\lfloor{n/2}\rfloor$ that it can be ignored in the limit in the next line. $\endgroup$
    – 2'5 9'2
    Jul 12, 2016 at 23:31

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