Goldbach's conjecture and convergence of a Dirichlet series
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Assuming Goldbach's conjecture, let's denote by $ r_{0}(n) : =inf{r>0, (n-r,n+r)inmathbb{P}^{2}} $. The assumption of GC implies $ r_{0}(n)<n $.
Let's now consider the series $ G(s) : =sum_{n>0}frac{a_{n}}{n^s} $ with $ a_{n}=r_{0}(n) $ if $ n $ is composite and $ a_{n}=1 $ otherwise.
Unconditionnally, $ G(s) $ converges for $Re(s)>2 $. What would be the infimum of the exponents $beta $ such that $ r_{0}(n)ll n^{beta} $ if we could prove the abscissa of convergence of $ G(s) $ is $ 1 $?
number-theory prime-numbers analytic-number-theory goldbachs-conjecture
asked Jan 10 at 15:42
Sylvain JulienSylvain Julien
1,135918
$endgroup$
add a comment |
$begingroup$
Assuming Goldbach's conjecture, let's denote by $ r_{0}(n) : =inf{r>0, (n-r,n+r)inmathbb{P}^{2}} $. The assumption of GC implies $ r_{0}(n)<n $.
Let's now consider the series $ G(s) : =sum_{n>0}frac{a_{n}}{n^s} $ with $ a_{n}=r_{0}(n) $ if $ n $ is composite and $ a_{n}=1 $ otherwise.
Unconditionnally, $ G(s) $ converges for $Re(s)>2 $. What would be the infimum of the exponents $beta $ such that $ r_{0}(n)ll n^{beta} $ if we could prove the abscissa of convergence of $ G(s) $ is $ 1 $?
number-theory prime-numbers analytic-number-theory goldbachs-conjecture
asked Jan 10 at 15:42
Sylvain JulienSylvain Julien
1,135918
$endgroup$
add a comment |
$begingroup$
Assuming Goldbach's conjecture, let's denote by $ r_{0}(n) : =inf{r>0, (n-r,n+r)inmathbb{P}^{2}} $. The assumption of GC implies $ r_{0}(n)<n $.
Let's now consider the series $ G(s) : =sum_{n>0}frac{a_{n}}{n^s} $ with $ a_{n}=r_{0}(n) $ if $ n $ is composite and $ a_{n}=1 $ otherwise.
Unconditionnally, $ G(s) $ converges for $Re(s)>2 $. What would be the infimum of the exponents $beta $ such that $ r_{0}(n)ll n^{beta} $ if we could prove the abscissa of convergence of $ G(s) $ is $ 1 $?
number-theory prime-numbers analytic-number-theory goldbachs-conjecture
asked Jan 10 at 15:42
Sylvain JulienSylvain Julien
1,135918
$endgroup$
Assuming Goldbach's conjecture, let's denote by $ r_{0}(n) : =inf{r>0, (n-r,n+r)inmathbb{P}^{2}} $. The assumption of GC implies $ r_{0}(n)<n $.
Let's now consider the series $ G(s) : =sum_{n>0}frac{a_{n}}{n^s} $ with $ a_{n}=r_{0}(n) $ if $ n $ is composite and $ a_{n}=1 $ otherwise.
Unconditionnally, $ G(s) $ converges for $Re(s)>2 $. What would be the infimum of the exponents $beta $ such that $ r_{0}(n)ll n^{beta} $ if we could prove the abscissa of convergence of $ G(s) $ is $ 1 $?
number-theory prime-numbers analytic-number-theory goldbachs-conjecture
number-theory prime-numbers analytic-number-theory goldbachs-conjecture
asked Jan 10 at 15:42
Sylvain JulienSylvain Julien
1,135918
asked Jan 10 at 15:42
Sylvain JulienSylvain Julien
1,135918
asked Jan 10 at 15:42
Sylvain JulienSylvain Julien
1,135918
asked Jan 10 at 15:42
Sylvain JulienSylvain Julien
1,135918
asked Jan 10 at 15:42
Sylvain JulienSylvain Julien
1,135918
1,135918
add a comment |
add a comment |
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