I recently asked what percentage of odd composite integers was divisible by three here. The best answer proved that since the frequency in the amount of primes goes to $0$ as they approach infinity, $\frac{1}{3}$ of all odd composites are divisible by three. This proved very helpful to me, but I am now looking for an upper bound for the amount of odd composites divisible by $3$. I am hoping for an upper bound less than $100$% of all odd composites.

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    $\begingroup$ I'n not sure why you keep stressing "composites". Other than $3$, any number divisible by $3$ is composite. $\endgroup$ – lulu Mar 14 '17 at 21:32
  • $\begingroup$ @lulu Well, $-3$ is also prime, but I doubt Linus was thinking about it. So alright, I'm just as puzzled as you are. $\endgroup$ – Mr. Brooks Mar 14 '17 at 22:16
  • $\begingroup$ Also...in the header you ask for odd composites but in the body of the question you leave off the "odd". I think that's what causes the confusion with the posted solution below. The same method works, of course. You want numbers less than $n$ of the form $6k+3$, excluding $3$. $\endgroup$ – lulu Mar 14 '17 at 23:16
  • $\begingroup$ @lulu Yes, but I am looking for the amount of composites divisible by 3. For instance, $25$ is composite, but not divisible by 3. $\endgroup$ – Linus Rastegar Mar 15 '17 at 1:48
  • $\begingroup$ Well, I don't get it. The posted solution (from @vrugtehagel) is correct other than for the confusion in your original post about "odd" or not. If you define $f(n)$ to be the number of odd multiples of $3$, other than $3$ itself, which are less than $n$ then $f(n)=\lfloor \frac {n-3}6\rfloor$ by a simple variation on the argument given in that solution. Thus $f(16)=\lfloor \frac {13}6\rfloor=2$ and we can check that by inspection ($9,15$ are the cases). Is that what you wanted? $\endgroup$ – lulu Mar 15 '17 at 11:02

Exactly $n-1$ numbers below $3n$ are divisible by three, in other words, there are exactly $\lfloor \frac n3\rfloor-1$ numbers below $n$ are divisible by $3$. This is because $3$ is the only prime divisible by three and so the rest of the numbers divisible by three are composite.

Exactly a third of the numbers $1,2,3,4,\cdots,3n$ is divisible by $3$. Only one is not composite. Thus, $n-1$ numbers are divisible by $3$. We can tweak this to say $\lfloor \frac n3\rfloor-1$ numbers less than $n$ are divisible by $3$.

Since this is the exact amount, it is both an upperbound and a lowerbound simultaneously.

  • $\begingroup$ I agree with your result on the amount of numbers that are divisible by $3$. I do not agree that the amount of composites less than $n$ that are divisible by $3$ is the same amount. In your sequence $1, 2, 3, 4, ... , 3n$, $2$ is prime, $5$, $7$, $11$, $13$ ... are all prime. This leaves you only with $1, 4, 6, 9, 15 ... $, of which $2$ are odd composites divisible by $3$, which is definitely not $\frac{n}{3} - 1$ $\endgroup$ – Linus Rastegar Mar 14 '17 at 22:14
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    $\begingroup$ I'm not saying it's the same, it's one less. $3$ is the only prime divisibly by three. Of the numbers $1,2,3,\cdots,3n$, the ones divisible by three are $3,6,9,\cdots,3n$, and of those, only $3$ is prime and thus the rest is composite. $\endgroup$ – vrugtehagel Mar 14 '17 at 22:21
  • $\begingroup$ I've been hesitant to accept this answer because I was convinced that the answer was more complicated. Your reasoning makes sense to me now. Can I apply this same logic to the amount of number divisible by $5$ or $7$? $\endgroup$ – Linus Rastegar Apr 27 '17 at 13:40
  • $\begingroup$ Yes you can. You can apply the same exact reasoning to find the amount of numbers divisible by any prime $p$. $\endgroup$ – vrugtehagel Apr 27 '17 at 20:23

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