Squarefree Factors

Consider the number $54$.
$54$ can be factored in $7$ distinct ways into one or more factors larger than $1$:
$54$, $2 \times 27$, $3 \times 18$, $6 \times 9$, $3 \times 3 \times 6$, $2 \times 3 \times 9$ and $2 \times 3 \times 3 \times 3$.
If we require that the factors are all squarefree only two ways remain: $3 \times 3 \times 6$ and $2 \times 3 \times 3 \times 3$.

Let’s call $\operatorname{Fsf}(n)$ the number of ways $n$ can be factored into one or more squarefree factors larger than $1$, so
$\operatorname{Fsf}(54)=2$.

Let $S(n)$ be $\sum \operatorname{Fsf}(k)$ for $k=2$ to $n$.

$S(100)=193$.

Find $S(10\,000\,000\,000)$.

This problem is related to partitions, a concept in number theory. In this case, you are essentially being asked to partition each number from 2 to \$n\$, into square-free factors, and then sum up the counts.

However, doing this via brute force for \$n = 10,000,000,000\$ is computationally not viable in a reasonable amount of time.

But since this is a combinatoric problem, we can find a general formula for any \$n\$. The formula is a modification of the one used to count the number of partitions of a number into distinct parts:
\[
\operatorname{Fsf}(n) = 2^{\omega(n)}
\]
Here, \$\omega(n)\$ is the number of distinct prime factors of \$n\$. In the factorization of \$n\$, every factor can occur in or not occur in a partition, hence there are \$2\$ possibilities for each factor, which gives the total number of partitions as \$2^{\omega(n)}\$.

For example, let’s calculate \$\operatorname{Fsf}(54)\$. The number \$54\$ has prime factorization \$2 \times 3^3\$, so it has \$2\$ distinct prime factors, i.e., \$2\$ and \$3\$. Therefore, \$\operatorname{Fsf}(54) = 2^2 = 4\$.

However, this does not avoid double-counting partitions that include non square-free numbers. To remove these, we subtract the squarefull divisor function \$\operatorname{sq}(n)\$, resulting in:
\[
\operatorname{Fsf}(n) = 2^{\omega(n)} – \operatorname{sq}(n)
\]
Then, \$S(n)\$ is simply a summation:
\[
S(n) = \sum_{k=2}^n \operatorname{Fsf}(k)
\]
This calculation cannot be done manually, and one should use an advanced software for mathematics or a programming language to do this.

Keep in mind that this derivation is relying on number theory and combinatorics and the above formulas can’t be derived without understanding these concepts deeply.

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