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Problem 633


Problem 633


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 $1.2346e-4$.


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$1.2346e-4$。