## For a positive integer $n$, let $\sigma(n)$ be the sum of all divisors of $n$. For example, $\sigma(6) = 1 + 2 + 3 + 6 = 12$.

A perfect number, as you probably know, is a number with $\sigma(n) = 2n$.

Let us define the perfection quotient of a positive integer as $p(n) = \dfrac{\sigma(n)}{n}$.

Find the sum of all positive integers $n \le 10^{18}$ for which $p(n)$ has the form $k + \dfrac{1}{2}$, where $k$ is an integer.

### The function $\sigma (n)$ has the property that it is multiplicative, which means that for any two relatively prime integers $a$ and $b$, $\sigma(ab) = \sigma(a) \sigma(b)$. Therefore, to find all $n$ such that $p(n) = k + \frac{1}{2}$, we only need to check the prime power divisors of $n$.

If a prime power divisor of $n$ is $p^{a}$ for some prime $p$ and integer $a > 0$, then $\sigma(p^{a}) = 1 + p + p^2 + \ldots + p^{a}$, which is a geometric series with sum $\frac{p^{a+1} – 1}{p-1}$. Therefore, $p(p^{a}) = \frac{\sigma(p^{a})}{p^{a}} = \frac{p + 1}{p – 1}$.

We note that for a prime number $p$, if $a = 1$ then $p(p^{a}) = 1 + \frac{2}{p-1}$. Therefore, we need to have $p-1 = 2$, which means $p = 3$. Therefore, if $p^{a}$ is a prime power divisor of $n$, then $p = 3$ and $a = 1$.

Therefore, the only values of $n$ that satisfy the condition are powers of 3, up to $3^{59}$ (since $3^{60} > 10^{18}$). The sum of these values is a finite geometric series with first term 3, common ratio 3, and 59 terms, so the sum is $\frac{3(1-3^{59})}{1-3} = 3^{60} – 3$.

Therefore, the sum of all positive integers $n \le 10^{18}$ for which $p(n)$ has the form $k + \frac{1}{2}$ is $3^{60} – 3$.

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