Relation between The Euler Totient, the counting prime formula and the prime generating Functions [on hold]












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](https://i.stack.imgur.com/lKFS0.png)
Relation between
The Euler Totient,
the counting prime formula
and the prime generating Functions
There is a formula for the ivisor sum hiih is one of the most useful
propertes of the Euler Formula to fond a relations with prime counting Formula ans sum of primes.
Proof : http://vixra.org/abs/1901.0046






number-theory prime-numbers prime-factorization divisor-counting-function







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put on hold as unclear what you're asking by amWhy, Gottfried Helms, Yanko, pre-kidney, metamorphy 2 days ago


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.




















    -1














    ](https://i.stack.imgur.com/lKFS0.png)
    Relation between
    The Euler Totient,
    the counting prime formula
    and the prime generating Functions
    There is a formula for the ivisor sum hiih is one of the most useful
    propertes of the Euler Formula to fond a relations with prime counting Formula ans sum of primes.
    Proof : http://vixra.org/abs/1901.0046






    number-theory prime-numbers prime-factorization divisor-counting-function







    share|cite|improve this question







    New contributor




    Na Zih is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.











    put on hold as unclear what you're asking by amWhy, Gottfried Helms, Yanko, pre-kidney, metamorphy 2 days ago


    Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.


















      -1












      -1








      -1


      1





      ](https://i.stack.imgur.com/lKFS0.png)
      Relation between
      The Euler Totient,
      the counting prime formula
      and the prime generating Functions
      There is a formula for the ivisor sum hiih is one of the most useful
      propertes of the Euler Formula to fond a relations with prime counting Formula ans sum of primes.
      Proof : http://vixra.org/abs/1901.0046






      number-theory prime-numbers prime-factorization divisor-counting-function







      share|cite|improve this question







      New contributor




      Na Zih is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      ](https://i.stack.imgur.com/lKFS0.png)
      Relation between
      The Euler Totient,
      the counting prime formula
      and the prime generating Functions
      There is a formula for the ivisor sum hiih is one of the most useful
      propertes of the Euler Formula to fond a relations with prime counting Formula ans sum of primes.
      Proof : http://vixra.org/abs/1901.0046






      number-theory prime-numbers prime-factorization divisor-counting-function




      number-theory prime-numbers prime-factorization divisor-counting-function






      share|cite|improve this question







      New contributor




      Na Zih is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question







      New contributor




      Na Zih is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question






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      asked 2 days ago









      Na ZihNa Zih

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      New contributor





      Na Zih is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      put on hold as unclear what you're asking by amWhy, Gottfried Helms, Yanko, pre-kidney, metamorphy 2 days ago


      Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.






      put on hold as unclear what you're asking by amWhy, Gottfried Helms, Yanko, pre-kidney, metamorphy 2 days ago


      Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
























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          You are not really linking $varphi(n)$ and $pi(n)$. You are just adding things that cancel out.



          $$pi(n)=sumlimits_{pmid n}p+sumlimits_{pleq n, p,nmid ,n}1+sumlimits_{dneq pmid n}varphi(d)-n$$
          $$pi(n)=sumlimits_{pmid n}p+sumlimits_{pleq n, p,nmid, n}1+sumlimits_{dmid n}varphi(d)-sumlimits_{pmid n}varphi(p)-n$$
          $$pi(n)=sumlimits_{pmid n}p+sumlimits_{pleq n, p,nmid, n}1+n-sumlimits_{pmid n}(p-1)-n$$
          $$pi(n)=sumlimits_{pmid n}p-sumlimits_{pmid n}p+sumlimits_{pleq n, p,nmid, n}1+sumlimits_{pmid n}1+n-n$$
          $$pi(n)=sumlimits_{pleq n}1$$
          $pi(n)$ is hidden in the second term
          $$sumlimits_{pleq n, p,nmid, n}1$$
          You could do that with any formula. Was there any reason to split it that way?






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            1 Answer
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            1 Answer
            1






            active

            oldest

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            active

            oldest

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            active

            oldest

            votes









            1














            You are not really linking $varphi(n)$ and $pi(n)$. You are just adding things that cancel out.



            $$pi(n)=sumlimits_{pmid n}p+sumlimits_{pleq n, p,nmid ,n}1+sumlimits_{dneq pmid n}varphi(d)-n$$
            $$pi(n)=sumlimits_{pmid n}p+sumlimits_{pleq n, p,nmid, n}1+sumlimits_{dmid n}varphi(d)-sumlimits_{pmid n}varphi(p)-n$$
            $$pi(n)=sumlimits_{pmid n}p+sumlimits_{pleq n, p,nmid, n}1+n-sumlimits_{pmid n}(p-1)-n$$
            $$pi(n)=sumlimits_{pmid n}p-sumlimits_{pmid n}p+sumlimits_{pleq n, p,nmid, n}1+sumlimits_{pmid n}1+n-n$$
            $$pi(n)=sumlimits_{pleq n}1$$
            $pi(n)$ is hidden in the second term
            $$sumlimits_{pleq n, p,nmid, n}1$$
            You could do that with any formula. Was there any reason to split it that way?






            share|cite|improve this answer




























              1














              You are not really linking $varphi(n)$ and $pi(n)$. You are just adding things that cancel out.



              $$pi(n)=sumlimits_{pmid n}p+sumlimits_{pleq n, p,nmid ,n}1+sumlimits_{dneq pmid n}varphi(d)-n$$
              $$pi(n)=sumlimits_{pmid n}p+sumlimits_{pleq n, p,nmid, n}1+sumlimits_{dmid n}varphi(d)-sumlimits_{pmid n}varphi(p)-n$$
              $$pi(n)=sumlimits_{pmid n}p+sumlimits_{pleq n, p,nmid, n}1+n-sumlimits_{pmid n}(p-1)-n$$
              $$pi(n)=sumlimits_{pmid n}p-sumlimits_{pmid n}p+sumlimits_{pleq n, p,nmid, n}1+sumlimits_{pmid n}1+n-n$$
              $$pi(n)=sumlimits_{pleq n}1$$
              $pi(n)$ is hidden in the second term
              $$sumlimits_{pleq n, p,nmid, n}1$$
              You could do that with any formula. Was there any reason to split it that way?






              share|cite|improve this answer


























                1












                1








                1






                You are not really linking $varphi(n)$ and $pi(n)$. You are just adding things that cancel out.



                $$pi(n)=sumlimits_{pmid n}p+sumlimits_{pleq n, p,nmid ,n}1+sumlimits_{dneq pmid n}varphi(d)-n$$
                $$pi(n)=sumlimits_{pmid n}p+sumlimits_{pleq n, p,nmid, n}1+sumlimits_{dmid n}varphi(d)-sumlimits_{pmid n}varphi(p)-n$$
                $$pi(n)=sumlimits_{pmid n}p+sumlimits_{pleq n, p,nmid, n}1+n-sumlimits_{pmid n}(p-1)-n$$
                $$pi(n)=sumlimits_{pmid n}p-sumlimits_{pmid n}p+sumlimits_{pleq n, p,nmid, n}1+sumlimits_{pmid n}1+n-n$$
                $$pi(n)=sumlimits_{pleq n}1$$
                $pi(n)$ is hidden in the second term
                $$sumlimits_{pleq n, p,nmid, n}1$$
                You could do that with any formula. Was there any reason to split it that way?






                share|cite|improve this answer














                You are not really linking $varphi(n)$ and $pi(n)$. You are just adding things that cancel out.



                $$pi(n)=sumlimits_{pmid n}p+sumlimits_{pleq n, p,nmid ,n}1+sumlimits_{dneq pmid n}varphi(d)-n$$
                $$pi(n)=sumlimits_{pmid n}p+sumlimits_{pleq n, p,nmid, n}1+sumlimits_{dmid n}varphi(d)-sumlimits_{pmid n}varphi(p)-n$$
                $$pi(n)=sumlimits_{pmid n}p+sumlimits_{pleq n, p,nmid, n}1+n-sumlimits_{pmid n}(p-1)-n$$
                $$pi(n)=sumlimits_{pmid n}p-sumlimits_{pmid n}p+sumlimits_{pleq n, p,nmid, n}1+sumlimits_{pmid n}1+n-n$$
                $$pi(n)=sumlimits_{pleq n}1$$
                $pi(n)$ is hidden in the second term
                $$sumlimits_{pleq n, p,nmid, n}1$$
                You could do that with any formula. Was there any reason to split it that way?







                share|cite|improve this answer














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                share|cite|improve this answer








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