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Let $q(n)$ be the number of integers $m<n$ such that $m$ is square-free. Let $p(n)$ be the number of integers $m<n$ such that the sum of the prime factors of $m$ is square-free. And let $s(n)$ be the number of integers $m<n$ such that the sum of the divisors of $m$ is square-free. The table shows values of $q$, $p$ and $s$ for some $n$:

      n       q       p      s
    100      60      68     24
   1000     607     660    157
  10000    6082    6343   1090
 100000   60793   62352   8097
1000000  607925  618969  64306

It's known that the distribution is asymptotically equivalent with a straight line and that $\frac{6}{\pi^2}\cdot 100\approx 60.79$ percent of all integers are square-free. So how to explain this anomaly?

Why is the distribution of square-free sums of divisors so low?

Erick Wong
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Lehs
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    I'm not sure I grasp what you feel is anomalous. Your table shows $q(n)$ approaching the prescribed ratio to $n$. Your final sentence suggests the anomaly is connected to sequence $s(n)$. But clearly the sums of divisors are not distributed like arbitrary integers in relation to the property of being square-free. Have I misunderstood your concern? – hardmath Nov 19 '18 at 21:04
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    @hardmath: the word anomaly is maybe not the right word, but there should be an explanation why $\sigma(x)$ tends to be a non square-free number. – Lehs Nov 19 '18 at 21:14
  • Do you know how to apply the standard analytic-number-theory tools to your sequences ? – reuns Nov 19 '18 at 21:22
  • @reuns: probably not. – Lehs Nov 19 '18 at 21:29
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    The basic fact about the sum of divisors function is that it is multiplicative, i.e. $\sigma(mn) = \sigma(m) \sigma(n)$ when $m,n$ are coprime. It follows that the prime power factors of $n$ determine the factors of $\sigma(n)$, and thus a necessary condition for $\sigma(n)$ to be square-free is that each $\sigma(p^k)$ is square-free (with $p^k$ the generic prime power in a unique prime factorization of $n$). – hardmath Nov 20 '18 at 00:45

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Expanding on hardmath's insightful comment, another necessary condition for $\sigma(n)$ to be square-free is that the values of $\sigma(p^k)$ are pairwise relatively prime. For instance, almost all of the prime power factors of $n$ would need to have even exponents in order for $\sigma(n)$ to avoid being divisible by $4$. Consider also that the majority of $n$ have many prime divisors (about $\log \log n$ typically). So $n$ needs to be "almost" a perfect square (except for one allowable exceptional odd prime and the prime $2$).

This is really quite rare and it would be unreasonable to expect any positive proportion of $n$ to satisfy such constraints. It’s fairly easy to prove that the density of such $n$ is only $O(x/\log x)$ up to $x$, similar to that of the primes (but with a larger constant in front).

Erick Wong
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