the identity for the second Chebyshev function:
$$\sum _{j=1}^{\pi \left( x \right) } \Biggl\lfloor {\frac {\ln
\left( x \right) }{\ln \left( p_{{j}} \right) }} \Biggr\rfloor \ln
\left( p_{{j}} \right)=\ln(\operatorname{lcm}(1,2,3,…,\lfloor x \rfloor))\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(0)
$$
Any natural number $N$ has a unique prime factorization product:
$$N=p_{1,N}^{v_{1,N}}p_{2,N}^{v_{2,N}}p_{3,N}^{v_{3,N}}\cdot\cdot\cdot p_{\omega(N),N}^{v_{\omega(N),N}}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(1)$$
Allows us to make the following assertion about the natural logarithm of any such natural:
$$\ln \left( N \right) =\sum _{j=1}^{\omega \left( N \right) }v_{{j,N}}
\ln \left( p_{{j,N}} \right)
\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(1')$$
My question is as to whether or not stating the above is sufficient justification for stating the following to be true:
$\pi(x)=\omega(\operatorname{lcm}(1,2,3,…,\lfloor x\rfloor))\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(a)$
the largest multiplicity any factor of $\operatorname{lcm}(1,2,3,…,\lfloor x\rfloor)$ may have is $ \lfloor\frac{\ln \left( x \right) }{2}\rfloor \quad\quad\quad\quad\quad\quad\quad\,\,\,(b)$
Both (a) and (b) I concluded from (0) & ('1), I just feel as if I am missing something that is necessary in a direct proof of these statements.
Best Answer
Hint: If you are not sure whether the stated justification is sufficient or not, this is an indication that you should add some more information.
The second case is not valid: