Relation between The Euler Totient, the counting prime formula and the prime generating Functions [on hold]
](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
asked 2 days ago
Na ZihNa Zih
11
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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.
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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
asked 2 days ago
Na ZihNa Zih
11
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.
add a comment |
](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
asked 2 days ago
Na ZihNa Zih
11
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
asked 2 days ago
Na ZihNa Zih
11
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.
asked 2 days ago
Na ZihNa Zih
11
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.
asked 2 days ago
Na ZihNa Zih
11
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.
asked 2 days ago
Na ZihNa Zih
11
asked 2 days ago
Na ZihNa Zih
11
11
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.
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.
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.
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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1 Answer
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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?
edited 2 days ago
reuns
19.7k21046
answered 2 days ago
Collag3nCollag3n
669211
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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?
edited 2 days ago
reuns
19.7k21046
answered 2 days ago
Collag3nCollag3n
669211
add a comment |
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?
edited 2 days ago
reuns
19.7k21046
answered 2 days ago
Collag3nCollag3n
669211
add a comment |
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?
edited 2 days ago
reuns
19.7k21046
answered 2 days ago
Collag3nCollag3n
669211
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?
edited 2 days ago
reuns
19.7k21046
answered 2 days ago
Collag3nCollag3n
669211
edited 2 days ago
reuns
19.7k21046
edited 2 days ago
reuns
19.7k21046
edited 2 days ago
reuns
19.7k21046
19.7k21046
answered 2 days ago
Collag3nCollag3n
669211
answered 2 days ago
Collag3nCollag3n
669211
answered 2 days ago
Collag3nCollag3n
669211
669211
add a comment |
add a comment |
