Mixing
Interesting missing proof (Prime Number Theorem)
$begingroup$
Let $n in mathbb{N}$, a number large enough.
Let $q(n)$ be the smallest prime number verify $n< displaystyle{small prod_{substack{a leq q(n) \ text{a prime}}} {normalsize a}}$
If $I_n$ denote the number of elements less than $n$ and coprime to $displaystyle{small left( prod_{substack{a leq q(n) \ text{a prime}}} {normalsize a} right)}$, then : $$I_n sim dfrac{n}{lnln(n)} , e^{-gamma}$$
Using prime number theorem : $dfrac{n}{ln(ln(n))} , e^{-gamma} = dfrac{n}{ln(n)} dfrac{ln(n)}{ln(ln(n))} , e^{-gamma} sim pi(n) big( pi(q(n)) e^{-gamma} big)$
My Question is : there is an other proof that $I_n sim pi(n) big( pi(q(n)) e^{-gamma} big)$ ? (not using the proof above! and just this formula without details about $pi(n)$ or $pi(q(n))$)
$I_n$ denote the number of elements less than $n$ and coprime to $displaystyle{small left( prod_{substack{a leq q(n) \ text{a prime}}} {normalsize a} right)}$
number-theory prime-numbers
asked Jan 23 at 14:48
LAGRIDALAGRIDA
100110
$endgroup$
add a comment |
$begingroup$
Let $n in mathbb{N}$, a number large enough.
Let $q(n)$ be the smallest prime number verify $n< displaystyle{small prod_{substack{a leq q(n) \ text{a prime}}} {normalsize a}}$
If $I_n$ denote the number of elements less than $n$ and coprime to $displaystyle{small left( prod_{substack{a leq q(n) \ text{a prime}}} {normalsize a} right)}$, then : $$I_n sim dfrac{n}{lnln(n)} , e^{-gamma}$$
Using prime number theorem : $dfrac{n}{ln(ln(n))} , e^{-gamma} = dfrac{n}{ln(n)} dfrac{ln(n)}{ln(ln(n))} , e^{-gamma} sim pi(n) big( pi(q(n)) e^{-gamma} big)$
My Question is : there is an other proof that $I_n sim pi(n) big( pi(q(n)) e^{-gamma} big)$ ? (not using the proof above! and just this formula without details about $pi(n)$ or $pi(q(n))$)
$I_n$ denote the number of elements less than $n$ and coprime to $displaystyle{small left( prod_{substack{a leq q(n) \ text{a prime}}} {normalsize a} right)}$
number-theory prime-numbers
asked Jan 23 at 14:48
LAGRIDALAGRIDA
100110
$endgroup$
$begingroup$
I don't see a proof in what you wrote.
$endgroup$
– reuns
Jan 23 at 15:34
$begingroup$
From the definition of $q(n)$ we have $displaystyle{small left( prod_{substack{a leq p(n) \ text{a prime}}} {normalsize a} right)} leq n < {small left( prod_{substack{a leq q(n) \ text{a prime}}} {normalsize a} right)}$ with $q(n)$ the next prime to $p(n)$. And Mertens 3rd theorem give $displaystyle{small prod_{substack{a leq q(n) \ text{a prime}}} left({normalsize 1-frac{1}{a}}right)} sim frac{e^{-gamma}}{ln(q(n))}$.
$endgroup$
– LAGRIDA
Jan 23 at 15:42
$begingroup$
And the prime number theorem give $displaystyle ln {small left( prod_{substack{a leq q(n) \ text{a prime}}} {normalsize a} right)} sim q(n)$ and $ln(p(n)) sim ln(q(n)) sim ln(ln(n))$ and that give $displaystyle I_n sim dfrac{n}{lnln(n)} , e^{-gamma}$
$endgroup$
– LAGRIDA
Jan 23 at 15:44
$begingroup$
The question is how you evaluate $I_n$. The answer is in my post
$endgroup$
– reuns
Jan 23 at 15:56
add a comment |
$begingroup$
Let $n in mathbb{N}$, a number large enough.
Let $q(n)$ be the smallest prime number verify $n< displaystyle{small prod_{substack{a leq q(n) \ text{a prime}}} {normalsize a}}$
If $I_n$ denote the number of elements less than $n$ and coprime to $displaystyle{small left( prod_{substack{a leq q(n) \ text{a prime}}} {normalsize a} right)}$, then : $$I_n sim dfrac{n}{lnln(n)} , e^{-gamma}$$
Using prime number theorem : $dfrac{n}{ln(ln(n))} , e^{-gamma} = dfrac{n}{ln(n)} dfrac{ln(n)}{ln(ln(n))} , e^{-gamma} sim pi(n) big( pi(q(n)) e^{-gamma} big)$
My Question is : there is an other proof that $I_n sim pi(n) big( pi(q(n)) e^{-gamma} big)$ ? (not using the proof above! and just this formula without details about $pi(n)$ or $pi(q(n))$)
$I_n$ denote the number of elements less than $n$ and coprime to $displaystyle{small left( prod_{substack{a leq q(n) \ text{a prime}}} {normalsize a} right)}$
number-theory prime-numbers
asked Jan 23 at 14:48
LAGRIDALAGRIDA
100110
$endgroup$
Let $n in mathbb{N}$, a number large enough.
Let $q(n)$ be the smallest prime number verify $n< displaystyle{small prod_{substack{a leq q(n) \ text{a prime}}} {normalsize a}}$
If $I_n$ denote the number of elements less than $n$ and coprime to $displaystyle{small left( prod_{substack{a leq q(n) \ text{a prime}}} {normalsize a} right)}$, then : $$I_n sim dfrac{n}{lnln(n)} , e^{-gamma}$$
Using prime number theorem : $dfrac{n}{ln(ln(n))} , e^{-gamma} = dfrac{n}{ln(n)} dfrac{ln(n)}{ln(ln(n))} , e^{-gamma} sim pi(n) big( pi(q(n)) e^{-gamma} big)$
My Question is : there is an other proof that $I_n sim pi(n) big( pi(q(n)) e^{-gamma} big)$ ? (not using the proof above! and just this formula without details about $pi(n)$ or $pi(q(n))$)
$I_n$ denote the number of elements less than $n$ and coprime to $displaystyle{small left( prod_{substack{a leq q(n) \ text{a prime}}} {normalsize a} right)}$
number-theory prime-numbers
number-theory prime-numbers
asked Jan 23 at 14:48
LAGRIDALAGRIDA
100110
asked Jan 23 at 14:48
LAGRIDALAGRIDA
100110
asked Jan 23 at 14:48
LAGRIDALAGRIDA
100110
asked Jan 23 at 14:48
LAGRIDALAGRIDA
100110
asked Jan 23 at 14:48
LAGRIDALAGRIDA
100110
100110
$begingroup$
I don't see a proof in what you wrote.
$endgroup$
– reuns
Jan 23 at 15:34
$begingroup$
From the definition of $q(n)$ we have $displaystyle{small left( prod_{substack{a leq p(n) \ text{a prime}}} {normalsize a} right)} leq n < {small left( prod_{substack{a leq q(n) \ text{a prime}}} {normalsize a} right)}$ with $q(n)$ the next prime to $p(n)$. And Mertens 3rd theorem give $displaystyle{small prod_{substack{a leq q(n) \ text{a prime}}} left({normalsize 1-frac{1}{a}}right)} sim frac{e^{-gamma}}{ln(q(n))}$.
$endgroup$
– LAGRIDA
Jan 23 at 15:42
$begingroup$
And the prime number theorem give $displaystyle ln {small left( prod_{substack{a leq q(n) \ text{a prime}}} {normalsize a} right)} sim q(n)$ and $ln(p(n)) sim ln(q(n)) sim ln(ln(n))$ and that give $displaystyle I_n sim dfrac{n}{lnln(n)} , e^{-gamma}$
$endgroup$
– LAGRIDA
Jan 23 at 15:44
$begingroup$
The question is how you evaluate $I_n$. The answer is in my post
$endgroup$
– reuns
Jan 23 at 15:56
add a comment |
$begingroup$
I don't see a proof in what you wrote.
$endgroup$
– reuns
Jan 23 at 15:34
$begingroup$
From the definition of $q(n)$ we have $displaystyle{small left( prod_{substack{a leq p(n) \ text{a prime}}} {normalsize a} right)} leq n < {small left( prod_{substack{a leq q(n) \ text{a prime}}} {normalsize a} right)}$ with $q(n)$ the next prime to $p(n)$. And Mertens 3rd theorem give $displaystyle{small prod_{substack{a leq q(n) \ text{a prime}}} left({normalsize 1-frac{1}{a}}right)} sim frac{e^{-gamma}}{ln(q(n))}$.
$endgroup$
– LAGRIDA
Jan 23 at 15:42
$begingroup$
And the prime number theorem give $displaystyle ln {small left( prod_{substack{a leq q(n) \ text{a prime}}} {normalsize a} right)} sim q(n)$ and $ln(p(n)) sim ln(q(n)) sim ln(ln(n))$ and that give $displaystyle I_n sim dfrac{n}{lnln(n)} , e^{-gamma}$
$endgroup$
– LAGRIDA
Jan 23 at 15:44
$begingroup$
The question is how you evaluate $I_n$. The answer is in my post
$endgroup$
– reuns
Jan 23 at 15:56
$begingroup$
I don't see a proof in what you wrote.
$endgroup$
– reuns
Jan 23 at 15:34
$begingroup$
I don't see a proof in what you wrote.
$endgroup$
– reuns
Jan 23 at 15:34
$begingroup$
From the definition of $q(n)$ we have $displaystyle{small left( prod_{substack{a leq p(n) \ text{a prime}}} {normalsize a} right)} leq n < {small left( prod_{substack{a leq q(n) \ text{a prime}}} {normalsize a} right)}$ with $q(n)$ the next prime to $p(n)$. And Mertens 3rd theorem give $displaystyle{small prod_{substack{a leq q(n) \ text{a prime}}} left({normalsize 1-frac{1}{a}}right)} sim frac{e^{-gamma}}{ln(q(n))}$.
$endgroup$
– LAGRIDA
Jan 23 at 15:42
$begingroup$
From the definition of $q(n)$ we have $displaystyle{small left( prod_{substack{a leq p(n) \ text{a prime}}} {normalsize a} right)} leq n < {small left( prod_{substack{a leq q(n) \ text{a prime}}} {normalsize a} right)}$ with $q(n)$ the next prime to $p(n)$. And Mertens 3rd theorem give $displaystyle{small prod_{substack{a leq q(n) \ text{a prime}}} left({normalsize 1-frac{1}{a}}right)} sim frac{e^{-gamma}}{ln(q(n))}$.
$endgroup$
– LAGRIDA
Jan 23 at 15:42
$begingroup$
And the prime number theorem give $displaystyle ln {small left( prod_{substack{a leq q(n) \ text{a prime}}} {normalsize a} right)} sim q(n)$ and $ln(p(n)) sim ln(q(n)) sim ln(ln(n))$ and that give $displaystyle I_n sim dfrac{n}{lnln(n)} , e^{-gamma}$
$endgroup$
– LAGRIDA
Jan 23 at 15:44
$begingroup$
And the prime number theorem give $displaystyle ln {small left( prod_{substack{a leq q(n) \ text{a prime}}} {normalsize a} right)} sim q(n)$ and $ln(p(n)) sim ln(q(n)) sim ln(ln(n))$ and that give $displaystyle I_n sim dfrac{n}{lnln(n)} , e^{-gamma}$
$endgroup$
– LAGRIDA
Jan 23 at 15:44
$begingroup$
The question is how you evaluate $I_n$. The answer is in my post
$endgroup$
– reuns
Jan 23 at 15:56
$begingroup$
The question is how you evaluate $I_n$. The answer is in my post
$endgroup$
– reuns
Jan 23 at 15:56
add a comment |
1 Answer
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For $c> 0$ fixed, $P_k= prod_{p le k}p ,k to infty$ then $$sum_{m le c P_k} 1_{gcd(m,P_k) = 1} sim c, varphi(P_k)sim c P_k frac{e^{-gamma}}{log k}$$
The proof shows it is also valid for $c$ depending on $k$ not decreasing too fast.
$$sum_{m le c P_k} 1_{gcd(m,P_k) = 1} = sum_{d | P_k} mu(d)sum_{m le c P_k} 1_{d | m} = sum_{d | P_k} mu(d) lfloor frac{c P_k}{d}rfloor = sum_{d | P_k} mu(d) (frac{c P_k}{d}+O(1)) \ = c prod_{p le k} p (1-p^{-1}) + O(tau(P_k))= c varphi(P_k)+O(P_k^epsilon)=c P_k frac{e^{-gamma}}{log k}+O(P_k^epsilon)$$
Where the last step is Mertens third theorem and $tau(m) = sum_{d | m} 1 = O(m^epsilon)$
The $O$ constant is uniform in $ c$.
In your question $q(n) = k$ and $P_k$ is the least primorial above $n$ so $ n = c_n P_k, c_n in (frac{1}{k},1]$ then $ I_n = sum_{m le c_n P_k} 1_{gcd(m,P_k) = 1}$, $frac1{c_n} = O(k) = O(log P_k)$ so you get the result
$$I_n = sum_{m le c_n P_k} 1_{gcd(m,P_k) = 1}= c_n P_k frac{e^{-gamma}}{log k}+O(P_k^epsilon) = n frac{e^{-gamma}}{log k}+O((kn)^epsilon)= e^{-gamma}frac{n}{log log n}+O(n^{epsilon'})$$
The PNT gives $q(n) sim log(P_k) sim log n$ and $pi(n) sim frac{n}{log n}$ so $I_n sim e^{-gamma}frac{pi(n) q(n)}{log log n} sim e^{-gamma}frac{pi(n) q(n)}{log q(n)}sim e^{-gamma}pi(n) pi(q(n))$ but this is just an obfuscation of the result, to be avoided at all cost.
answered Jan 23 at 15:33
reunsreuns
21.2k21351
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$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
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– Aloizio Macedo♦
Jan 26 at 3:13
$begingroup$
We need the quantitative missing proof $displaystyle I_n sim pi(n) big( pi(q(n)) e^{-gamma} big)$, and that will give us the proof $displaystyle I_{q(n), m}(n) sim pi_m(n) big( pi(q(n)) , e^{- gamma} big)^2$ and that proove imediately Hardy-LittleWood conjecture : $displaystyle pi_m(n) sim displaystyle{small left( prod_{substack{a | m \ text{a prime}}} {normalsize frac{a-1}{a-2}} right)} 2 , C_2 ; dfrac{n}{ln(n)^2}$
$endgroup$
– LAGRIDA
Feb 13 at 15:38
add a comment |
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$begingroup$
For $c> 0$ fixed, $P_k= prod_{p le k}p ,k to infty$ then $$sum_{m le c P_k} 1_{gcd(m,P_k) = 1} sim c, varphi(P_k)sim c P_k frac{e^{-gamma}}{log k}$$
The proof shows it is also valid for $c$ depending on $k$ not decreasing too fast.
$$sum_{m le c P_k} 1_{gcd(m,P_k) = 1} = sum_{d | P_k} mu(d)sum_{m le c P_k} 1_{d | m} = sum_{d | P_k} mu(d) lfloor frac{c P_k}{d}rfloor = sum_{d | P_k} mu(d) (frac{c P_k}{d}+O(1)) \ = c prod_{p le k} p (1-p^{-1}) + O(tau(P_k))= c varphi(P_k)+O(P_k^epsilon)=c P_k frac{e^{-gamma}}{log k}+O(P_k^epsilon)$$
Where the last step is Mertens third theorem and $tau(m) = sum_{d | m} 1 = O(m^epsilon)$
The $O$ constant is uniform in $ c$.
In your question $q(n) = k$ and $P_k$ is the least primorial above $n$ so $ n = c_n P_k, c_n in (frac{1}{k},1]$ then $ I_n = sum_{m le c_n P_k} 1_{gcd(m,P_k) = 1}$, $frac1{c_n} = O(k) = O(log P_k)$ so you get the result
$$I_n = sum_{m le c_n P_k} 1_{gcd(m,P_k) = 1}= c_n P_k frac{e^{-gamma}}{log k}+O(P_k^epsilon) = n frac{e^{-gamma}}{log k}+O((kn)^epsilon)= e^{-gamma}frac{n}{log log n}+O(n^{epsilon'})$$
The PNT gives $q(n) sim log(P_k) sim log n$ and $pi(n) sim frac{n}{log n}$ so $I_n sim e^{-gamma}frac{pi(n) q(n)}{log log n} sim e^{-gamma}frac{pi(n) q(n)}{log q(n)}sim e^{-gamma}pi(n) pi(q(n))$ but this is just an obfuscation of the result, to be avoided at all cost.
answered Jan 23 at 15:33
reunsreuns
21.2k21351
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 26 at 3:13
$begingroup$
We need the quantitative missing proof $displaystyle I_n sim pi(n) big( pi(q(n)) e^{-gamma} big)$, and that will give us the proof $displaystyle I_{q(n), m}(n) sim pi_m(n) big( pi(q(n)) , e^{- gamma} big)^2$ and that proove imediately Hardy-LittleWood conjecture : $displaystyle pi_m(n) sim displaystyle{small left( prod_{substack{a | m \ text{a prime}}} {normalsize frac{a-1}{a-2}} right)} 2 , C_2 ; dfrac{n}{ln(n)^2}$
$endgroup$
– LAGRIDA
Feb 13 at 15:38
add a comment |
$begingroup$
For $c> 0$ fixed, $P_k= prod_{p le k}p ,k to infty$ then $$sum_{m le c P_k} 1_{gcd(m,P_k) = 1} sim c, varphi(P_k)sim c P_k frac{e^{-gamma}}{log k}$$
The proof shows it is also valid for $c$ depending on $k$ not decreasing too fast.
$$sum_{m le c P_k} 1_{gcd(m,P_k) = 1} = sum_{d | P_k} mu(d)sum_{m le c P_k} 1_{d | m} = sum_{d | P_k} mu(d) lfloor frac{c P_k}{d}rfloor = sum_{d | P_k} mu(d) (frac{c P_k}{d}+O(1)) \ = c prod_{p le k} p (1-p^{-1}) + O(tau(P_k))= c varphi(P_k)+O(P_k^epsilon)=c P_k frac{e^{-gamma}}{log k}+O(P_k^epsilon)$$
Where the last step is Mertens third theorem and $tau(m) = sum_{d | m} 1 = O(m^epsilon)$
The $O$ constant is uniform in $ c$.
In your question $q(n) = k$ and $P_k$ is the least primorial above $n$ so $ n = c_n P_k, c_n in (frac{1}{k},1]$ then $ I_n = sum_{m le c_n P_k} 1_{gcd(m,P_k) = 1}$, $frac1{c_n} = O(k) = O(log P_k)$ so you get the result
$$I_n = sum_{m le c_n P_k} 1_{gcd(m,P_k) = 1}= c_n P_k frac{e^{-gamma}}{log k}+O(P_k^epsilon) = n frac{e^{-gamma}}{log k}+O((kn)^epsilon)= e^{-gamma}frac{n}{log log n}+O(n^{epsilon'})$$
The PNT gives $q(n) sim log(P_k) sim log n$ and $pi(n) sim frac{n}{log n}$ so $I_n sim e^{-gamma}frac{pi(n) q(n)}{log log n} sim e^{-gamma}frac{pi(n) q(n)}{log q(n)}sim e^{-gamma}pi(n) pi(q(n))$ but this is just an obfuscation of the result, to be avoided at all cost.
answered Jan 23 at 15:33
reunsreuns
21.2k21351
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 26 at 3:13
$begingroup$
We need the quantitative missing proof $displaystyle I_n sim pi(n) big( pi(q(n)) e^{-gamma} big)$, and that will give us the proof $displaystyle I_{q(n), m}(n) sim pi_m(n) big( pi(q(n)) , e^{- gamma} big)^2$ and that proove imediately Hardy-LittleWood conjecture : $displaystyle pi_m(n) sim displaystyle{small left( prod_{substack{a | m \ text{a prime}}} {normalsize frac{a-1}{a-2}} right)} 2 , C_2 ; dfrac{n}{ln(n)^2}$
$endgroup$
– LAGRIDA
Feb 13 at 15:38
add a comment |
$begingroup$
For $c> 0$ fixed, $P_k= prod_{p le k}p ,k to infty$ then $$sum_{m le c P_k} 1_{gcd(m,P_k) = 1} sim c, varphi(P_k)sim c P_k frac{e^{-gamma}}{log k}$$
The proof shows it is also valid for $c$ depending on $k$ not decreasing too fast.
$$sum_{m le c P_k} 1_{gcd(m,P_k) = 1} = sum_{d | P_k} mu(d)sum_{m le c P_k} 1_{d | m} = sum_{d | P_k} mu(d) lfloor frac{c P_k}{d}rfloor = sum_{d | P_k} mu(d) (frac{c P_k}{d}+O(1)) \ = c prod_{p le k} p (1-p^{-1}) + O(tau(P_k))= c varphi(P_k)+O(P_k^epsilon)=c P_k frac{e^{-gamma}}{log k}+O(P_k^epsilon)$$
Where the last step is Mertens third theorem and $tau(m) = sum_{d | m} 1 = O(m^epsilon)$
The $O$ constant is uniform in $ c$.
In your question $q(n) = k$ and $P_k$ is the least primorial above $n$ so $ n = c_n P_k, c_n in (frac{1}{k},1]$ then $ I_n = sum_{m le c_n P_k} 1_{gcd(m,P_k) = 1}$, $frac1{c_n} = O(k) = O(log P_k)$ so you get the result
$$I_n = sum_{m le c_n P_k} 1_{gcd(m,P_k) = 1}= c_n P_k frac{e^{-gamma}}{log k}+O(P_k^epsilon) = n frac{e^{-gamma}}{log k}+O((kn)^epsilon)= e^{-gamma}frac{n}{log log n}+O(n^{epsilon'})$$
The PNT gives $q(n) sim log(P_k) sim log n$ and $pi(n) sim frac{n}{log n}$ so $I_n sim e^{-gamma}frac{pi(n) q(n)}{log log n} sim e^{-gamma}frac{pi(n) q(n)}{log q(n)}sim e^{-gamma}pi(n) pi(q(n))$ but this is just an obfuscation of the result, to be avoided at all cost.
answered Jan 23 at 15:33
reunsreuns
21.2k21351
$endgroup$
For $c> 0$ fixed, $P_k= prod_{p le k}p ,k to infty$ then $$sum_{m le c P_k} 1_{gcd(m,P_k) = 1} sim c, varphi(P_k)sim c P_k frac{e^{-gamma}}{log k}$$
The proof shows it is also valid for $c$ depending on $k$ not decreasing too fast.
$$sum_{m le c P_k} 1_{gcd(m,P_k) = 1} = sum_{d | P_k} mu(d)sum_{m le c P_k} 1_{d | m} = sum_{d | P_k} mu(d) lfloor frac{c P_k}{d}rfloor = sum_{d | P_k} mu(d) (frac{c P_k}{d}+O(1)) \ = c prod_{p le k} p (1-p^{-1}) + O(tau(P_k))= c varphi(P_k)+O(P_k^epsilon)=c P_k frac{e^{-gamma}}{log k}+O(P_k^epsilon)$$
Where the last step is Mertens third theorem and $tau(m) = sum_{d | m} 1 = O(m^epsilon)$
The $O$ constant is uniform in $ c$.
In your question $q(n) = k$ and $P_k$ is the least primorial above $n$ so $ n = c_n P_k, c_n in (frac{1}{k},1]$ then $ I_n = sum_{m le c_n P_k} 1_{gcd(m,P_k) = 1}$, $frac1{c_n} = O(k) = O(log P_k)$ so you get the result
$$I_n = sum_{m le c_n P_k} 1_{gcd(m,P_k) = 1}= c_n P_k frac{e^{-gamma}}{log k}+O(P_k^epsilon) = n frac{e^{-gamma}}{log k}+O((kn)^epsilon)= e^{-gamma}frac{n}{log log n}+O(n^{epsilon'})$$
The PNT gives $q(n) sim log(P_k) sim log n$ and $pi(n) sim frac{n}{log n}$ so $I_n sim e^{-gamma}frac{pi(n) q(n)}{log log n} sim e^{-gamma}frac{pi(n) q(n)}{log q(n)}sim e^{-gamma}pi(n) pi(q(n))$ but this is just an obfuscation of the result, to be avoided at all cost.
answered Jan 23 at 15:33
reunsreuns
21.2k21351
edited Jan 25 at 11:31
answered Jan 23 at 15:33
reunsreuns
21.2k21351
answered Jan 23 at 15:33
reunsreuns
21.2k21351
answered Jan 23 at 15:33
reunsreuns
21.2k21351
21.2k21351
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– Aloizio Macedo♦
Jan 26 at 3:13
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We need the quantitative missing proof $displaystyle I_n sim pi(n) big( pi(q(n)) e^{-gamma} big)$, and that will give us the proof $displaystyle I_{q(n), m}(n) sim pi_m(n) big( pi(q(n)) , e^{- gamma} big)^2$ and that proove imediately Hardy-LittleWood conjecture : $displaystyle pi_m(n) sim displaystyle{small left( prod_{substack{a | m \ text{a prime}}} {normalsize frac{a-1}{a-2}} right)} 2 , C_2 ; dfrac{n}{ln(n)^2}$
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– LAGRIDA
Feb 13 at 15:38
add a comment |
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 26 at 3:13
$begingroup$
We need the quantitative missing proof $displaystyle I_n sim pi(n) big( pi(q(n)) e^{-gamma} big)$, and that will give us the proof $displaystyle I_{q(n), m}(n) sim pi_m(n) big( pi(q(n)) , e^{- gamma} big)^2$ and that proove imediately Hardy-LittleWood conjecture : $displaystyle pi_m(n) sim displaystyle{small left( prod_{substack{a | m \ text{a prime}}} {normalsize frac{a-1}{a-2}} right)} 2 , C_2 ; dfrac{n}{ln(n)^2}$
$endgroup$
– LAGRIDA
Feb 13 at 15:38
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 26 at 3:13
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 26 at 3:13
$begingroup$
We need the quantitative missing proof $displaystyle I_n sim pi(n) big( pi(q(n)) e^{-gamma} big)$, and that will give us the proof $displaystyle I_{q(n), m}(n) sim pi_m(n) big( pi(q(n)) , e^{- gamma} big)^2$ and that proove imediately Hardy-LittleWood conjecture : $displaystyle pi_m(n) sim displaystyle{small left( prod_{substack{a | m \ text{a prime}}} {normalsize frac{a-1}{a-2}} right)} 2 , C_2 ; dfrac{n}{ln(n)^2}$
$endgroup$
– LAGRIDA
Feb 13 at 15:38
$begingroup$
We need the quantitative missing proof $displaystyle I_n sim pi(n) big( pi(q(n)) e^{-gamma} big)$, and that will give us the proof $displaystyle I_{q(n), m}(n) sim pi_m(n) big( pi(q(n)) , e^{- gamma} big)^2$ and that proove imediately Hardy-LittleWood conjecture : $displaystyle pi_m(n) sim displaystyle{small left( prod_{substack{a | m \ text{a prime}}} {normalsize frac{a-1}{a-2}} right)} 2 , C_2 ; dfrac{n}{ln(n)^2}$
$endgroup$
– LAGRIDA
Feb 13 at 15:38
add a comment |
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I don't see a proof in what you wrote.
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– reuns
Jan 23 at 15:34
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From the definition of $q(n)$ we have $displaystyle{small left( prod_{substack{a leq p(n) \ text{a prime}}} {normalsize a} right)} leq n < {small left( prod_{substack{a leq q(n) \ text{a prime}}} {normalsize a} right)}$ with $q(n)$ the next prime to $p(n)$. And Mertens 3rd theorem give $displaystyle{small prod_{substack{a leq q(n) \ text{a prime}}} left({normalsize 1-frac{1}{a}}right)} sim frac{e^{-gamma}}{ln(q(n))}$.
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– LAGRIDA
Jan 23 at 15:42
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And the prime number theorem give $displaystyle ln {small left( prod_{substack{a leq q(n) \ text{a prime}}} {normalsize a} right)} sim q(n)$ and $ln(p(n)) sim ln(q(n)) sim ln(ln(n))$ and that give $displaystyle I_n sim dfrac{n}{lnln(n)} , e^{-gamma}$
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– LAGRIDA
Jan 23 at 15:44
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The question is how you evaluate $I_n$. The answer is in my post
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– reuns
Jan 23 at 15:56