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Problem 633

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Square prime factors II

For an integer $n$, we define the square prime factors of $n$ to be the primes whose square divides $n$. For example, the square prime factors of $1500 = 2^2 \times 3 \times 5^3$ are $2$ and $5$.

Let $C_k(N)$ be the number of integers between $1$ and $N$ inclusive with exactly $k$ square prime factors. It can be shown that with growing $N$ the ratio $\frac{C_k(N)}{N}$ gets arbitrarily close to a constant $c_k^\infty$, as suggested by the table below.

	$k=0$	$k=1$	$k=2$	$k=3$	$k=4$
$C_k(10)$	$7$	$3$	$0$	$0$	$0$
$C_k(10^2)$	$61$	$36$	$3$	$0$	$0$
$C_k(10^3)$	$608$	$343$	$48$	$1$	$0$
$C_k(10^4)$	$6083$	$3363$	$533$	$21$	$0$
$C_k(10^5)$	$60794$	$33562$	$5345$	$297$	$2$
$C_k(10^6)$	$607926$	$335438$	$53358$	$3218$	$60$
$C_k(10^7)$	$6079291$	$3353956$	$533140$	$32777$	$834$
$C_k(10^8)$	$60792694$	$33539196$	$5329747$	$329028$	$9257$
$C_k(10^9)$	$607927124$	$335389706$	$53294365$	$3291791$	$95821$
$c_k^\infty$	$\frac{6}{\pi^2}$	$3.3539\times 10^{-1}$	$5.3293\times 10^{-2}$	$3.2921\times 10^{-3}$	$9.7046\times 10^{-5}

Find $c_7^\infty$. Give the result in scientific notation rounded to $5$ significant digits, using a $e$ to seperate mantissa and exponent. E.g., if the answer were $0.000123456789$, then the answer format would be $1.2346e-4$.

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平方质因数II

对于整数$n$，称平方后能整除$n$的质数为$n$的平方质因数。例如，$1500 = 2^2 \times 3 \times 5^3$的平方质因数包括$2$和$5$。

记$C_k(N)$为$1$与$N$之间（含$1$和$N$）恰好有$k$个平方质因数的整数之和。可以发现，随着$N$不断增大，比值$\frac{C_k(N)}{N}$逐渐趋向于一个常数$c_k^\infty$，如下表所示：

	$k=0$	$k=1$	$k=2$	$k=3$	$k=4$	$k=5$
$N=10$	$7$	$3$	$0$	$0$	$0$	$0$
$N=10^2$	$61$	$36$	$3$	$0$	$0$	$0$
$N=10^3$	$608$	$343$	$48$	$1$	$0$	$0$
$N=10^4$	$6083$	$3363$	$533$	$21$	$0$	$0$
$N=10^5$	$60794$	$33562$	$5345$	$297$	$2$	$0$
$N=10^6$	$607926$	$335438$	$53358$	$3218$	$60$	$0$
$N=10^7$	$6079291$	$3353956$	$533140$	$32777$	$834$	$2$
$N=10^8$	$60792694$	$33539196$	$5329747$	$329028$	$9257$	$78$

求$c_7^\infty$，并将结果用科学计数法表示，保留5位有效数字，尾数和指数间用$e$分隔。例如，如果结果是$0.000123456789$，则应当回答$1.2346e-4$。

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