Let $n$ be an integer more than $9$. Find the number of positive divisors of $n$ lets call it $d(n)$ and take the difference from $n$. So the new number is $n-d(n)$. If we continue this process, we always end up with $2$. I don't know anybody worked this question before but somehow this question reminded me Collatz Conjecture. Anybody can give me an idea how to approach this problem ? Many thanks..

  • $\begingroup$ Interesting ! Did you have the idea for this sequence ? $\endgroup$ – Peter Jun 20 at 8:10
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    $\begingroup$ Yes and no. I have been working on Collatz type series and one of my friend came to me and asked this question. You know collatz always ends up with 1 no matter where you start and this one always ends up with 2. I have a feeling that for n>10, we always end up with 2. $\endgroup$ – Turker Jun 20 at 8:19
  • $\begingroup$ I could not solve the question by induction method. If somebody can solve it and post it , I will be more than happy. $\endgroup$ – Turker Jun 20 at 8:22
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    $\begingroup$ Strictly spekaing, noone knows whether the Collatz-conjecture is true. In the case of this conjecture, I am optimistic that it can be proven. Brute force reveals that it is true upto $n=10^5$ $\endgroup$ – Peter Jun 20 at 8:31
  • $\begingroup$ Seems that Henry already solved it ... $\endgroup$ – Peter Jun 20 at 8:33

If $n < (k+1)^2$, there's at most $2k$ divisors here.

That's because if $x$ is a divisor of $n$ and greater than $\sqrt{n}$, the number $\frac{n}{x}$ must be an integer less than $\sqrt{n}$.

So there are only about $\sqrt{n}$ candidates for $x$, and it implies that the number of divisors is at most about $2\sqrt{n}$. More technically, $d(n)$ is at most $2k$.

The smallest $k$ here is exactly $\lfloor \sqrt{n} \rfloor$. So the upper bound of number of divisors is $2 \times \lfloor \sqrt{n} \rfloor$.

Now, let's check about when will $n - d(n) \geq 9$ hold.
Here, $n - d(n) \geq n - 2 \times \lfloor \sqrt{n} \rfloor \geq n - 2\sqrt{n} \geq 9$. The range of $n - 2 \sqrt{n} \geq 9$ can be solved by translating to quadratic equation.
Let $N = \sqrt{n}$. The inequality will be $N^2 - 2N \geq 9$, and it will be always true when $N \geq 1+\sqrt{10}$.

Now, we can say that $n - d(n) \geq 9$ when $n \geq 18 \geq (1+\sqrt{10})^2$.

The final winning run is to check validity for $n=9, 10, 11, 12, 13, 14, 15, 16, 17$ and we can do inductive proof for $n \geq 18$ because we proved that the next number will always be at least $9$.



  • You can check it is true for $n=6,9,10,11,12$
  • For $n \gt 12$ you could show $9 \le n-d(n) \lt n$
  • Then use induction

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