1

I was wondering if there is a nice formula for the number of divisors $d$ of a perfect square ($n^2$), such that $\frac{n}{2} <d < n$. For example, for $n = 12$, the divisors $d$ of $12^2=144$ such that $12/2=6<d<12$ consist of $8$ and $9$, so $f(12) = 2$. $$$$ Another way to express this is that $12 = 2^2 * 3^1$, so $144=2^4*3^2$. Now we must find the number of tuples $(e_1, e_2)$, such that $6 \lt 2^{e_1} * 3^{e_2} \lt 12$ $$$$ Is there a closed formula for this or at least another way to express this?

Varun Vejalla
  • 8,873
  • $f(12)=2$ is correct. This takes the $144=2^43^2$ and looks for how many $\lambda, \mu :0\leq\lambda\leq 4, 0\leq \mu \leq 2$ satisfy $6<2^\lambda3^\mu<12$. In this case, we get two: $(\lambda, \mu)= (3,0)\ ,\ (0,2)$. OP, you should edit your question to reflect this and avoid confusing readers. – Rhys Hughes Apr 18 '19 at 23:55
  • 1
    The number of divisor function $\tau(n)$ is erratic but at least it is multiplicative, the same holds for $\tau(n^2)$, when removing the divisors $< n/2$ you are loosing that. In those cases to hope interesting properties such as asymptotic estimates we look at $\sum_{n \le x} f(n)$. Also why do you care of $f(n)= \sum_{d | n, n/2<d < n}1$, why cutting at $n/2$, do you have any application ? – reuns Apr 19 '19 at 00:52

2 Answers2

2

I don't think there is a nice formula. There might be an upper and lower bound, but I don't even think there is an accurate asymptotic functional approximation. Here is why:
This is the graph of the number of solutions for each $x^2$.

You might want to take a look at $\sigma(n)$, the divisor counting function, $\Omega(n)$, the prime factorization counting function, and $\omega(n)$, the unique prime factor counting function, and note the discontinuity.
Edit
This being said, I did a little more researching and I was able to graph the average solutions up to $x$.
This smoothes the graph and gives it a clear function approximation. Empirically, I guessed that there exist constants $$a_1, a_2, b_1, b_2 \neq 0 : a_1\ln(x)+b_1 \leq f(x) \leq a_2\ln(x)+b_2$$ Then the following follows by definition. $$\frac{1}{x}\sum_{i=1}^x f(i) \asymp \ln(x)$$ Note that this is a guess. Hopefully this will steer you in the right direction

Ryan Shesler
  • 1,498
  • $f(12)=2$ is correct. This takes the $144=2^43^2$ and looks for how many $\lambda, \mu :0\leq\lambda\leq 4, 0\leq \mu \leq 2$ satisfy $6<2^\lambda3^\mu<12$. In this case, we get two: $(\lambda, \mu)= (3,0)\ ,\ (0,2)$ – Rhys Hughes Apr 18 '19 at 23:51
  • 1
    Oh ok, that is more clear. You should suggest an edit and include that. @RhysHughes – Ryan Shesler Apr 18 '19 at 23:54
  • Thank you for the suggestion. I added that to my comment at the top. – Rhys Hughes Apr 18 '19 at 23:59
0

What I'm about to show isn't a formula but rules on how to quickly find the divisors ($d$) between $n/2$ and $n$. For each $d$ $\exists a,b\in\Bbb{N}|2a>b>a,gcd(a,b)=1,a|n,b|n$

$d=\frac{n*a}{b}$

I will show three examples below.

If $n=12$ then $n$'s factors are $1,2,3,4,6, \text{ and } 12$. All $a$ and $b$ must be one of these numbers. All $(a,b)$ pairs that satisfy the inequality ($2a>b>a$) are $(2,3),(3,4),(4,6)$. The gcd of the third pair isn't one, so it is removed. $$8=\frac{12*2}{3}\quad 9=\frac{12*3}{4}$$

If $n=45$ then $n$'s factors are $1,3,5,9,15,$ and $45$. The three pairs $(3,5),(5,9),(9,15)$ satisfy the inequality. The third pair isn't coprime so it is removed.

$$27=\frac{3*45}{5}\quad 25=\frac{5*45}{9}$$

If $n=60$ Factors of $n$: $1,2,3,4,5,6,10,12,15,20,30,60$

Coprime pairs: $(2,3),(3,4),(3,5),(4,5),(5,6)$

$$d\in (40,45,36,48,50)$$

While I don't have a general formula for $f(n)$ there is a formula for a specific case. If all factors of $n^2$ less than $n/2$ divides $n$ then the formula for $f(n^2)$ is $$\lceil\frac{\sigma_0(n^2)}{2}\rceil -\sigma_0(n)$$ where $\sigma_0(x)$ is the number of factors of $x$

This formula does work for $12$ and $45$ but not $60$.

quantus14
  • 2,624