Problem 861

Products of Bi-Unitary Divisors

A unitary divisor of a positive integer $n$ is a divisor $d$ of $n$ such that $gcd (d, n / d) = 1$ .

A bi-unitary divisor of $n$ is a divisor $d$ for which $1$ is the only unitary divisor of $d$ that is also a unitary divisor of $n / d$ .

For example, $2$ is a bi-unitary divisor of $8$ , because the unitary divisors of $2$ are ${1, 2}$ , and the unitary divisors of $8 / 2$ are ${1, 4}$ , with $1$ being the only unitary divisor in common.

The bi-unitary divisors of $240$ are ${1, 2, 3, 5, 6, 8, 10, 15, 16, 24, 30, 40, 48, 80, 120, 240}$ .

Let $P (n)$ be the product of all bi-unitary divisors of $n$ . Define $Q_{k} (N)$ as the number of positive integers $1 < n \leq N$ such that $P (n) = n^{k}$ . For example, $Q_{2} (10^{2}) = 51$ and $Q_{6} (10^{6}) = 6189$ .

Find $\sum_{k = 2}^{10} Q_{k} (10^{12})$ .

双元因数的乘积

若正整数 $n$ 的因数 $d$ 满足 $gcd (d, n / d) = 1$ ，则称 $d$ 为 $n$ 的元因数。

若 $1$ 是 $d$ 和 $n / d$ 唯一的公共元因数，则称 $d$ 为 $n$ 的双元因数。

例如， $2$ 是 $8$ 的双元因数，因为 $2$ 的元因数有 ${1, 2}$ ，而 $8 / 2$ 的元因数有 ${1, 4}$ ，只有 $1$ 是唯一的公共元因数。

$240$ 的双元因数有 ${1, 2, 3, 5, 6, 8, 10, 15, 16, 24, 30, 40, 48, 80, 120, 240}$ 。

令 $P (n)$ 为 $n$ 的所有双元因数之积。定义 $Q_{k} (N)$ 为满足 $1 < n \leq N$ 和 $P (n) = n^{k}$ 的正整数 $n$ 的数量。例如， $Q_{2} (10^{2}) = 51$ ， $Q_{6} (10^{6}) = 6189$ 。

求 $\sum_{k = 2}^{10} Q_{k} (10^{12})$ 。