I recently had an idea and derived some formulas through elementary means. Let, $F(x)$ be an function which returns the sum of the unique prime factors of $x$. $\newcommand{\floor}[1]{\lfloor #1 \rfloor}$. $\lambda(x)$ return the number of unique primes.
For example,
$$ F(12) = 2 + 3 = 5$$ $$ \lambda(12) = 1+ 1 =2$$
Let, $p_i$ be the $i$'th prime and $\floor{x} $ be the floor function. Then,
$$ \sum_{p_k < p_1p_2\dots p_n} p_k \floor{\frac{\prod_{i=1}^n p_i}{p_k}} = \sum_{r=2}^{p_1 p_2\dots p_n} F(r)$$
Using similar means:
$$ \sum_{p_k < (p_1p_2\dots p_n)^2} \floor{\frac{(\prod_{i=1}^n p_i)^2}{p_k}} = \sum_{r=2}^{(p_1 p_2\dots p_n)^2} \lambda(r)$$
Question
Is it possible to improve upon the formulas by introducing some error?