Second Chebyshev function and Unique prime factorization












1












$begingroup$


the identity for the second Chebyshev function:
$$sum _{j=1}^{pi left( x right) } Biggllfloor {frac {ln
left( x right) }{ln left( p_{{j}} right) }} Biggrrfloor ln
left( p_{{j}} right)=ln(operatorname{lcm}(1,2,3,...,lfloor x rfloor))quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(0)
$$



Any natural number $N$ has a unique prime factorization product:
$$N=p_{1,N}^{v_{1,N}}p_{2,N}^{v_{2,N}}p_{3,N}^{v_{3,N}}cdotcdotcdot p_{omega(N),N}^{v_{omega(N),N}}quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(1)$$



Allows us to make the following assertion about the natural logarithm of any such natural:
$$ln left( N right) =sum _{j=1}^{omega left( N right) }v_{{j,N}}
ln left( p_{{j,N}} right)
quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(1')$$



My question is as to whether or not stating the above is sufficient justification for stating the following to be true:



$pi(x)=omega(operatorname{lcm}(1,2,3,...,lfloor xrfloor))quad quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(a)$



the largest multiplicity any factor of $operatorname{lcm}(1,2,3,...,lfloor xrfloor)$ may have is $ lfloorfrac{ln left( x right) }{2}rfloor quadquadquadquadquadquadquad,,,(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.






number-theory







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    the identity for the second Chebyshev function:
    $$sum _{j=1}^{pi left( x right) } Biggllfloor {frac {ln
    left( x right) }{ln left( p_{{j}} right) }} Biggrrfloor ln
    left( p_{{j}} right)=ln(operatorname{lcm}(1,2,3,...,lfloor x rfloor))quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(0)
    $$



    Any natural number $N$ has a unique prime factorization product:
    $$N=p_{1,N}^{v_{1,N}}p_{2,N}^{v_{2,N}}p_{3,N}^{v_{3,N}}cdotcdotcdot p_{omega(N),N}^{v_{omega(N),N}}quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(1)$$



    Allows us to make the following assertion about the natural logarithm of any such natural:
    $$ln left( N right) =sum _{j=1}^{omega left( N right) }v_{{j,N}}
    ln left( p_{{j,N}} right)
    quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(1')$$



    My question is as to whether or not stating the above is sufficient justification for stating the following to be true:



    $pi(x)=omega(operatorname{lcm}(1,2,3,...,lfloor xrfloor))quad quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(a)$



    the largest multiplicity any factor of $operatorname{lcm}(1,2,3,...,lfloor xrfloor)$ may have is $ lfloorfrac{ln left( x right) }{2}rfloor quadquadquadquadquadquadquad,,,(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.






    number-theory







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      0



      $begingroup$


      the identity for the second Chebyshev function:
      $$sum _{j=1}^{pi left( x right) } Biggllfloor {frac {ln
      left( x right) }{ln left( p_{{j}} right) }} Biggrrfloor ln
      left( p_{{j}} right)=ln(operatorname{lcm}(1,2,3,...,lfloor x rfloor))quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(0)
      $$



      Any natural number $N$ has a unique prime factorization product:
      $$N=p_{1,N}^{v_{1,N}}p_{2,N}^{v_{2,N}}p_{3,N}^{v_{3,N}}cdotcdotcdot p_{omega(N),N}^{v_{omega(N),N}}quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(1)$$



      Allows us to make the following assertion about the natural logarithm of any such natural:
      $$ln left( N right) =sum _{j=1}^{omega left( N right) }v_{{j,N}}
      ln left( p_{{j,N}} right)
      quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(1')$$



      My question is as to whether or not stating the above is sufficient justification for stating the following to be true:



      $pi(x)=omega(operatorname{lcm}(1,2,3,...,lfloor xrfloor))quad quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(a)$



      the largest multiplicity any factor of $operatorname{lcm}(1,2,3,...,lfloor xrfloor)$ may have is $ lfloorfrac{ln left( x right) }{2}rfloor quadquadquadquadquadquadquad,,,(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.






      number-theory







      share|cite|improve this question











      $endgroup$




      the identity for the second Chebyshev function:
      $$sum _{j=1}^{pi left( x right) } Biggllfloor {frac {ln
      left( x right) }{ln left( p_{{j}} right) }} Biggrrfloor ln
      left( p_{{j}} right)=ln(operatorname{lcm}(1,2,3,...,lfloor x rfloor))quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(0)
      $$



      Any natural number $N$ has a unique prime factorization product:
      $$N=p_{1,N}^{v_{1,N}}p_{2,N}^{v_{2,N}}p_{3,N}^{v_{3,N}}cdotcdotcdot p_{omega(N),N}^{v_{omega(N),N}}quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(1)$$



      Allows us to make the following assertion about the natural logarithm of any such natural:
      $$ln left( N right) =sum _{j=1}^{omega left( N right) }v_{{j,N}}
      ln left( p_{{j,N}} right)
      quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(1')$$



      My question is as to whether or not stating the above is sufficient justification for stating the following to be true:



      $pi(x)=omega(operatorname{lcm}(1,2,3,...,lfloor xrfloor))quad quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(a)$



      the largest multiplicity any factor of $operatorname{lcm}(1,2,3,...,lfloor xrfloor)$ may have is $ lfloorfrac{ln left( x right) }{2}rfloor quadquadquadquadquadquadquad,,,(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.






      number-theory




      number-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 15 '18 at 16:50







      Adam

















      asked Aug 7 '18 at 19:42









      AdamAdam

      54114




      54114






















          1 Answer
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          1












          $begingroup$

          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 xrfloor))
            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}\
            &=2^2cdot3cdot 5
            end{align*}



            We see the multiplicity of $2$ in (1) is 2, but $leftlfloorfrac{ln(5)}{2}rightrfloor=0$.









          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Much appreciated Markus I really am not sure now why I thought $(b)$ was true to be honest
            $endgroup$
            – Adam
            Dec 3 '18 at 3:12






          • 1




            $begingroup$
            @Adam: You're welcome.
            $endgroup$
            – Markus Scheuer
            Dec 3 '18 at 5:38











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          1 Answer
          1






          active

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          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          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 xrfloor))
            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}\
            &=2^2cdot3cdot 5
            end{align*}



            We see the multiplicity of $2$ in (1) is 2, but $leftlfloorfrac{ln(5)}{2}rightrfloor=0$.









          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Much appreciated Markus I really am not sure now why I thought $(b)$ was true to be honest
            $endgroup$
            – Adam
            Dec 3 '18 at 3:12






          • 1




            $begingroup$
            @Adam: You're welcome.
            $endgroup$
            – Markus Scheuer
            Dec 3 '18 at 5:38
















          1












          $begingroup$

          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 xrfloor))
            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}\
            &=2^2cdot3cdot 5
            end{align*}



            We see the multiplicity of $2$ in (1) is 2, but $leftlfloorfrac{ln(5)}{2}rightrfloor=0$.









          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Much appreciated Markus I really am not sure now why I thought $(b)$ was true to be honest
            $endgroup$
            – Adam
            Dec 3 '18 at 3:12






          • 1




            $begingroup$
            @Adam: You're welcome.
            $endgroup$
            – Markus Scheuer
            Dec 3 '18 at 5:38














          1












          1








          1





          $begingroup$

          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 xrfloor))
            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}\
            &=2^2cdot3cdot 5
            end{align*}



            We see the multiplicity of $2$ in (1) is 2, but $leftlfloorfrac{ln(5)}{2}rightrfloor=0$.









          share|cite|improve this answer









          $endgroup$



          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 xrfloor))
            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}\
            &=2^2cdot3cdot 5
            end{align*}



            We see the multiplicity of $2$ in (1) is 2, but $leftlfloorfrac{ln(5)}{2}rightrfloor=0$.










          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 1 '18 at 22:59









          Markus ScheuerMarkus Scheuer

          60.4k455144




          60.4k455144












          • $begingroup$
            Much appreciated Markus I really am not sure now why I thought $(b)$ was true to be honest
            $endgroup$
            – Adam
            Dec 3 '18 at 3:12






          • 1




            $begingroup$
            @Adam: You're welcome.
            $endgroup$
            – Markus Scheuer
            Dec 3 '18 at 5:38


















          • $begingroup$
            Much appreciated Markus I really am not sure now why I thought $(b)$ was true to be honest
            $endgroup$
            – Adam
            Dec 3 '18 at 3:12






          • 1




            $begingroup$
            @Adam: You're welcome.
            $endgroup$
            – Markus Scheuer
            Dec 3 '18 at 5:38
















          $begingroup$
          Much appreciated Markus I really am not sure now why I thought $(b)$ was true to be honest
          $endgroup$
          – Adam
          Dec 3 '18 at 3:12




          $begingroup$
          Much appreciated Markus I really am not sure now why I thought $(b)$ was true to be honest
          $endgroup$
          – Adam
          Dec 3 '18 at 3:12




          1




          1




          $begingroup$
          @Adam: You're welcome.
          $endgroup$
          – Markus Scheuer
          Dec 3 '18 at 5:38




          $begingroup$
          @Adam: You're welcome.
          $endgroup$
          – Markus Scheuer
          Dec 3 '18 at 5:38


















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