1

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.

Adam Ledger
  • 1,240

1 Answers1

1

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.

In the first case you could argue:

  • The arithmetical function $\pi(x)$ counts the number of primes less or equal to $x$. Since this is the same as the index of the greatest prime of the numbers $1,2,\dots,x$, the claim \begin{align*} \pi(x)=\omega(\mathrm{lcm}(1,2,3,...,\lfloor x\rfloor)) \end{align*} follows.

The second case is not valid:

  • We have as counterexample for instance $x=5$. We obtain \begin{align*} \mathrm{lcm}(1,2,3,4,5)&=\mathrm{lcm}(1,2,3,2^2,5)\tag{1}\\ &=2^2\cdot3\cdot 5 \end{align*}

    We see the multiplicity of $2$ in (1) is 2, but $\left\lfloor\frac{\ln(5)}{2}\right\rfloor=0$.

Markus Scheuer
  • 108,315