This may be a very naive question, but I've noticed something about the factors of most composite numbers. When listing out the factors of a number (excluding the number itself), the factors seem to get arbitrarily farther apart from each other as they approach, $\frac{c}{2}$, c being the number. Is there any reason for this, and does it have any consequences in other parts of number theory/mathematics?

Some of my reasoning:

I tried looking for some patterns, and all I've seen is that the sum of differences between the factors add up to the (largest factor -1), whereas the product of all the differences between the factors add up to $2^f$, f being the number of factors. Perhaps this could relate to the fact that the number of factors for a number $c\quad =\quad { p }_{ 1 }^{ { e }_{ 1 } }*{ p }_{ 2 }^{ { e }_{ 2 } }*{ p }_{ 3 }^{ { e }_{ 3 } }...{ p }_{ n }^{ { e }_{ n } }\\ f\quad =\quad ({ e }_{ n }+1)({ e }_{ n }+1)...({ e }_{ n }+1)$.

For example, for the number 48, the product of all the differences between factors (excluding 48 itself), is 512, which is $2^9$, where 9 is the number of factors.

  • 1
    $\begingroup$ Yes, for the factors to be close together, they need to be close to $\sqrt[k]{c}$, if there are $k$ factors for the number $c$. For any of them that is lowered by say $\Delta_f$ from that $\sqrt[k]{c}$, we can investigate how much one of the other needs to grow by to compensate. $\endgroup$ Nov 18, 2017 at 21:13
  • $\begingroup$ Here is one idea: take your number $n$ and two factors $a,b$. Then $a\cdot \frac na=b\cdot \frac nb$. You could now use this to try to study $b-a$ compared to $\frac na-\frac nb$. For simplicity, say that $a<b<\frac na<\frac nb$. $\endgroup$
    – Arthur
    Nov 18, 2017 at 21:14
  • 2
    $\begingroup$ Write $\log n = \sum_{i=1}^j a_i$ (where $a_i = \log p_i^{e_i}$) Then the logarithm of the divisors of $n$ are sums of subsets of the $a_i$. They are more or less linearly spaced in $[0,\log n]$. Thus the divisors of $n$ are more or less exponentially spaced in $[1,n]$ $\endgroup$
    – reuns
    Nov 18, 2017 at 21:15
  • $\begingroup$ Are you conjecturing that $\prod_{d,e | n; d,e \neq n} (d - e) = \# \{ d | n \}$? $\endgroup$ Nov 18, 2017 at 21:15
  • $\begingroup$ @EnjoysMath No, he says "sum of differences", which means that he has discovered that $\sum_{a,b\text{ are consecutive divisors of } n}(b-a)$ telescopes. $\endgroup$
    – Arthur
    Nov 18, 2017 at 21:18


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