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.

\[\begin{array}{|c|c|c|c|c|c|}
\hline
& k = 0 & k = 1 & k = 2 & k = 3 & k = 4 \\
\hline
C_k(10) & 7 & 3 & 0 & 0 & 0 \\
\hline
C_k(10^2) & 61 & 36 & 3 & 0 & 0 \\
\hline
C_k(10^3) & 608 & 343 & 48 & 1 & 0 \\
\hline
C_k(10^4) & 6083 & 3363 & 533 & 21 & 0 \\
\hline
C_k(10^5) & 60794 & 33562 & 5345 & 297 & 2 \\
\hline
C_k(10^6) & 607926 & 335438 & 53358 & 3218 & 60 \\
\hline
C_k(10^7) & 6079291 & 3353956 & 533140 & 32777 & 834 \\
\hline
C_k(10^8) & 60792694 & 33539196 & 5329747 & 329028 & 9257 \\
\hline
C_k(10^9) & 607927124 & 335389706 & 53294365 & 3291791 & 95821 \\
\hline
c_k^{\infty} & \frac{6}{\pi^2} & 3.3539\times 10^{-1} & 5.3293\times 10^{-2} & 3.2921\times 10^{-3} & 9.7046\times 10^{-5}\\
\hline
\end{array}\]
Find $c_{7}^{\infty}$. Give the result in scientific notation rounded to $5$ significant digits, using a $e$ to separate mantissa and exponent. E.g. if the answer were $0.000123456789$, then the answer format would be $1.2346\mathrm e{-4}$.

The fact that the ratio $\frac{C_k(N)}{N}$ approaches a constant implies the existence of an asymptotic probability distribution on the number of square factors an integer has.

The prime number theorem tells us that for $N$ large enough, the density of primes less than or equal to $N$ is effectively $1/\ln(x)$. A prime $p$ will contribute a square factor to an arbitrary integer if and only if two of its factors are selected (there are other cases, but they occur with negligible probability). This event occurs with a probability of $1/\ln^2(N)$.

Since different primes contribute independently to the number of square factors, the distribution of square factors is sum of Bernoulli-distributed random variables and therefore follows a Poisson distribution. The probability mass function for a Poisson random variable is:

$P(k ;\lambda) = \frac{\lambda^k e^{-\lambda}}{k!}$,

where $\lambda$ is the expected value of the distribution. Here, since the expected value is the sum of the probabilities for each prime, and we have an infinitude of primes, $\lambda=\infty$. For $k < \infty$, the values of $\lambda^k e^{-\lambda}/k!$ clearly go to zero. That said, for $k=\infty$, dividing both the numerator and denominator by $\lambda$, we find the limit to be $1/k!$. Applying this to the problem at hand, the limit for $k=7$ is $c_{7}^{\infty}=\frac{1}{7!}=1.98413\cdot e{-4}$. Therefore, the result in scientific notation and rounded to $5$ significant digits is $1.9841e{-4}$. This can also be taken to mean that the density of integers with exactly seven primes that are squares of other primes is $1.9841\cdot 10^{-4}$ for sufficiently large integers. This conclusion is somewhat surprising, as our intuition might lead us to think that such integers should be relatively rare. However, prime factors often crop up in unexpected ways in number theory, and their distribution is often counter-intuitive.

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