Question links to paper which states: $$ \sigma(n)= \frac{6}{n^2(n-1)}\sum_{k=1}^{n-1}(3n^2-10k^2)\sigma(k)\sigma(n-k) \qquad (1) $$
where $\sigma(n)$ is the sum of divisors of $n$.
Another similar result from OEIS about Euler totient function: $$ \phi(n)=\sum_{i=1}^n |k(n,i)| \qquad (2) $$
where $k(n,i)$ is the Kronecker symbol.
And
$$ \phi(n) = \operatorname{binomial}(n+1, 2) - \sum_{i=1}^{n-1} \phi(i)\lfloor n/i\rfloor \qquad (3) $$
Observe that the RHS of all of the above doesn't require the factorization of $n$ and in addition (2) doesn't require factorization at all.
We find the (1) and (3) factorization of $n$-free identities counter-intuitive.
Q1 Is there any intuition why such identities exist?
Q2 Are there similar factorization of $n$-free identities for $\omega(n)$ or $\Omega(n)$ (the number of prime factors of $n$) (not counting the generating function)?