A series related to prime numbers Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)A heuristic argument for the Goldbach conjecture?The relation of $zeta$-function and $p^k$ for $Re(s) le 1$?The ordinary generating function for $ζ(s)$what the RH equivalent for Riemann prime formula $Pi(x)$?How prime numbers are related to special functions?Relationship between Riemann Zeta function and Prime zeta functionPrime Zeta without Möbius Function and Prime SummationPrime-counting function: EvaluationRelation of prime indexed series with Riemann zeta functionCan the Riemann Explicit Formula be used to find prime numbers?Sum of almost-prime zeta functions
How can I use the Python library networkx from Mathematica?
Is it fair for a professor to grade us on the possession of past papers?
Why wasn't DOSKEY integrated with COMMAND.COM?
Denied boarding although I have proper visa and documentation. To whom should I make a complaint?
How to Make a Beautiful Stacked 3D Plot
Is there any way for the UK Prime Minister to make a motion directly dependent on Government confidence?
Did MS DOS itself ever use blinking text?
Is it common practice to audition new musicians one-on-one before rehearsing with the entire band?
How to compare two different files line by line in unix?
How does the math work when buying airline miles?
What does "lightly crushed" mean for cardamon pods?
Crossing US/Canada Border for less than 24 hours
Trademark violation for app?
Using audio cues to encourage good posture
Fundamental Solution of the Pell Equation
What causes the direction of lightning flashes?
Fantasy story; one type of magic grows in power with use, but the more powerful they are, they more they are drawn to travel to their source
How to convince students of the implication truth values?
Should I use a zero-interest credit card for a large one-time purchase?
How to find all the available tools in mac terminal?
Can you use the Shield Master feat to shove someone before you make an attack by using a Readied action?
Around usage results
Is it ethical to give a final exam after the professor has quit before teaching the remaining chapters of the course?
What is implied by the word 'Desika'
A series related to prime numbers
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)A heuristic argument for the Goldbach conjecture?The relation of $zeta$-function and $p^k$ for $Re(s) le 1$?The ordinary generating function for $ζ(s)$what the RH equivalent for Riemann prime formula $Pi(x)$?How prime numbers are related to special functions?Relationship between Riemann Zeta function and Prime zeta functionPrime Zeta without Möbius Function and Prime SummationPrime-counting function: EvaluationRelation of prime indexed series with Riemann zeta functionCan the Riemann Explicit Formula be used to find prime numbers?Sum of almost-prime zeta functions
$begingroup$
Let $H(t) = sum_n=1 ^infty pi(n)t^n$ where $pi(n)$ is the prime counting function. This is the Hilbert series of some $mathbbQ$-vector space. By the prime number theorem, the radius of convergence is $1$. Observing $pi(n) = pi(n-1)+1$ if $n$ is prime and $pi(n) = pi(n-1)$ if $n$ is composite, we might rewrite this as $H(t) = f(t)/(1-t)$ where $f(t) =sum_p text prime t^p$. Define the sequence $b_n$ for $n=-1,0,1,2,cdots$ as $ f'(t)/f(t) = sum_n=-1 b_n t^n$. Then we can recover the primes from this sequence:
$$ p = 2 + sum_q <p, q text prime b_p-1-q$$
For example:
For example the first coefficients are given by the series:
$$2*t^(-1) + 1 + (-1)*t + 4*t^2 + (-5)*t^3 + 11*t^4 + (-16)*t^5 + 22*t^6 + (-37)*t^7 + 67*t^8 + (-101)*t^9 + 166*t^10 + (-260)*t^11 + 404*t^12 + (-652)*t^13 + cdots $$
so $b_-1 = 2, b_0 = 1, b_1 = 1 , b_2 = 4$ etc.
We have for example:
$$3 = 2+b_0 = 2+1$$
$$5 = 2+b_2+b_1 = 2+4-1$$
$$7 = 2+b_4+b_3+b_1 = 2+11-5-1$$
Let $a_n,k$ denote the number of ordered ways of writing $n$ as a sum of $k$ primes.
Then after some calculation, one finds that:
$$a_n,k = frackn-2k sum_v=0^n-1 a_v,k b_n-1-v$$
which is a recurence relation.
Furthermore if $alpha_n$ $n=0,1,2,3,cdots$ are all roots of $f(t)$ not equal to zero, then for $nge 0$
$$ b_n = - sum_k=0 ^ infty frac1alpha_k^n+1$$
The numbers $b_n$ might be computed inductively using:
$$ n a_n,1 = sum_v=0^n b_v-1 a_n-v,1$$
from which one sees, that $b_n in mathbbZ$.
One real root of $f(t)$ seems to be the number,
$$ gamma = -0.62923 cdots $$
(OEIS: http://oeis.org/A078756 )
Since everything related to primes has something to do with the Riemann Zeta function, I wonder, what is the relation to the stuff above to the Riemann Zeta function?
If someone knows of any reference or has any idea, that would be great.
Is there for example a way to compute the real root $gamma$?
What is the relation of $gamma$ to the other complex roots?
What other properties do the numbers $b_n$ have?
etc.
Thanks for your help.
Edit:
I found a conjectural way to compute $gamma$ and using the Euler Product a link to Riemann Zeta function:
$$zeta(s) = prod_p frac11-(2+sum_q<p b_p-1-q)^-s$$
and for $gamma$ numerical coincidences suggest that:
$$lim_n rightarrow infty fracb_nb_n+1 = gamma = -0.629233cdots$$
which could be one way to define $gamma$. Then one has to show, that this limit exists and that $f(gamma) = 0$.
If someone has an idea in this direction,that would be nice.
Second edit:
Here is the computation for the recurence relation of $a_n,k$:
For $kge 1$ we have on the one hand:
$$log(f(t)^k)' = frack f(t)^k-1 f'(t)f(t)^k = k fracf'(t)f(t) = sum_n=-1^infty k b_n t^n$$
On the other hand it is
$$f(t)^k = sum_n=0^infty a_n,kt^n$$
which means that
$$log(f(t)^k)' = fracsum_n=0^infty n a_n,k t^n-1sum_n=0^infty a_n,k t^n$$
Hence it follows, that (by multiplying with the denominator):
$$sum_n=0^infty n a_n,k t^n-1 = k (sum_n=0^infty a_n,k t^n) (sum_n=0 b_n-1t^n) frac1t$$
After multiplying with $t$ and using the Cauchy product formula we get:
$$sum_n=0^infty n a_n,k t^n = sum_n=0^infty k( sum_v=0^n a_v,kb_n-1-v)t^n$$
and comparing coefficients we find that:
$$n a_n,k = k ( sum_v=0^n a_v,k b_n-1-v)$$
and with $b_-1 = 2$ it follows after solving this equation for $a_n,k$ that:
$$ a_n,k = frackn-2k sum_v=0^n-1 a_v,k b_n-1-v$$
Especially for $k=1$ and $n=p$ prime we find that:
$$ p = 2+ sum_q<p b_p-1-q$$
number-theory prime-numbers riemann-zeta
$endgroup$
add a comment |
$begingroup$
Let $H(t) = sum_n=1 ^infty pi(n)t^n$ where $pi(n)$ is the prime counting function. This is the Hilbert series of some $mathbbQ$-vector space. By the prime number theorem, the radius of convergence is $1$. Observing $pi(n) = pi(n-1)+1$ if $n$ is prime and $pi(n) = pi(n-1)$ if $n$ is composite, we might rewrite this as $H(t) = f(t)/(1-t)$ where $f(t) =sum_p text prime t^p$. Define the sequence $b_n$ for $n=-1,0,1,2,cdots$ as $ f'(t)/f(t) = sum_n=-1 b_n t^n$. Then we can recover the primes from this sequence:
$$ p = 2 + sum_q <p, q text prime b_p-1-q$$
For example:
For example the first coefficients are given by the series:
$$2*t^(-1) + 1 + (-1)*t + 4*t^2 + (-5)*t^3 + 11*t^4 + (-16)*t^5 + 22*t^6 + (-37)*t^7 + 67*t^8 + (-101)*t^9 + 166*t^10 + (-260)*t^11 + 404*t^12 + (-652)*t^13 + cdots $$
so $b_-1 = 2, b_0 = 1, b_1 = 1 , b_2 = 4$ etc.
We have for example:
$$3 = 2+b_0 = 2+1$$
$$5 = 2+b_2+b_1 = 2+4-1$$
$$7 = 2+b_4+b_3+b_1 = 2+11-5-1$$
Let $a_n,k$ denote the number of ordered ways of writing $n$ as a sum of $k$ primes.
Then after some calculation, one finds that:
$$a_n,k = frackn-2k sum_v=0^n-1 a_v,k b_n-1-v$$
which is a recurence relation.
Furthermore if $alpha_n$ $n=0,1,2,3,cdots$ are all roots of $f(t)$ not equal to zero, then for $nge 0$
$$ b_n = - sum_k=0 ^ infty frac1alpha_k^n+1$$
The numbers $b_n$ might be computed inductively using:
$$ n a_n,1 = sum_v=0^n b_v-1 a_n-v,1$$
from which one sees, that $b_n in mathbbZ$.
One real root of $f(t)$ seems to be the number,
$$ gamma = -0.62923 cdots $$
(OEIS: http://oeis.org/A078756 )
Since everything related to primes has something to do with the Riemann Zeta function, I wonder, what is the relation to the stuff above to the Riemann Zeta function?
If someone knows of any reference or has any idea, that would be great.
Is there for example a way to compute the real root $gamma$?
What is the relation of $gamma$ to the other complex roots?
What other properties do the numbers $b_n$ have?
etc.
Thanks for your help.
Edit:
I found a conjectural way to compute $gamma$ and using the Euler Product a link to Riemann Zeta function:
$$zeta(s) = prod_p frac11-(2+sum_q<p b_p-1-q)^-s$$
and for $gamma$ numerical coincidences suggest that:
$$lim_n rightarrow infty fracb_nb_n+1 = gamma = -0.629233cdots$$
which could be one way to define $gamma$. Then one has to show, that this limit exists and that $f(gamma) = 0$.
If someone has an idea in this direction,that would be nice.
Second edit:
Here is the computation for the recurence relation of $a_n,k$:
For $kge 1$ we have on the one hand:
$$log(f(t)^k)' = frack f(t)^k-1 f'(t)f(t)^k = k fracf'(t)f(t) = sum_n=-1^infty k b_n t^n$$
On the other hand it is
$$f(t)^k = sum_n=0^infty a_n,kt^n$$
which means that
$$log(f(t)^k)' = fracsum_n=0^infty n a_n,k t^n-1sum_n=0^infty a_n,k t^n$$
Hence it follows, that (by multiplying with the denominator):
$$sum_n=0^infty n a_n,k t^n-1 = k (sum_n=0^infty a_n,k t^n) (sum_n=0 b_n-1t^n) frac1t$$
After multiplying with $t$ and using the Cauchy product formula we get:
$$sum_n=0^infty n a_n,k t^n = sum_n=0^infty k( sum_v=0^n a_v,kb_n-1-v)t^n$$
and comparing coefficients we find that:
$$n a_n,k = k ( sum_v=0^n a_v,k b_n-1-v)$$
and with $b_-1 = 2$ it follows after solving this equation for $a_n,k$ that:
$$ a_n,k = frackn-2k sum_v=0^n-1 a_v,k b_n-1-v$$
Especially for $k=1$ and $n=p$ prime we find that:
$$ p = 2+ sum_q<p b_p-1-q$$
number-theory prime-numbers riemann-zeta
$endgroup$
add a comment |
$begingroup$
Let $H(t) = sum_n=1 ^infty pi(n)t^n$ where $pi(n)$ is the prime counting function. This is the Hilbert series of some $mathbbQ$-vector space. By the prime number theorem, the radius of convergence is $1$. Observing $pi(n) = pi(n-1)+1$ if $n$ is prime and $pi(n) = pi(n-1)$ if $n$ is composite, we might rewrite this as $H(t) = f(t)/(1-t)$ where $f(t) =sum_p text prime t^p$. Define the sequence $b_n$ for $n=-1,0,1,2,cdots$ as $ f'(t)/f(t) = sum_n=-1 b_n t^n$. Then we can recover the primes from this sequence:
$$ p = 2 + sum_q <p, q text prime b_p-1-q$$
For example:
For example the first coefficients are given by the series:
$$2*t^(-1) + 1 + (-1)*t + 4*t^2 + (-5)*t^3 + 11*t^4 + (-16)*t^5 + 22*t^6 + (-37)*t^7 + 67*t^8 + (-101)*t^9 + 166*t^10 + (-260)*t^11 + 404*t^12 + (-652)*t^13 + cdots $$
so $b_-1 = 2, b_0 = 1, b_1 = 1 , b_2 = 4$ etc.
We have for example:
$$3 = 2+b_0 = 2+1$$
$$5 = 2+b_2+b_1 = 2+4-1$$
$$7 = 2+b_4+b_3+b_1 = 2+11-5-1$$
Let $a_n,k$ denote the number of ordered ways of writing $n$ as a sum of $k$ primes.
Then after some calculation, one finds that:
$$a_n,k = frackn-2k sum_v=0^n-1 a_v,k b_n-1-v$$
which is a recurence relation.
Furthermore if $alpha_n$ $n=0,1,2,3,cdots$ are all roots of $f(t)$ not equal to zero, then for $nge 0$
$$ b_n = - sum_k=0 ^ infty frac1alpha_k^n+1$$
The numbers $b_n$ might be computed inductively using:
$$ n a_n,1 = sum_v=0^n b_v-1 a_n-v,1$$
from which one sees, that $b_n in mathbbZ$.
One real root of $f(t)$ seems to be the number,
$$ gamma = -0.62923 cdots $$
(OEIS: http://oeis.org/A078756 )
Since everything related to primes has something to do with the Riemann Zeta function, I wonder, what is the relation to the stuff above to the Riemann Zeta function?
If someone knows of any reference or has any idea, that would be great.
Is there for example a way to compute the real root $gamma$?
What is the relation of $gamma$ to the other complex roots?
What other properties do the numbers $b_n$ have?
etc.
Thanks for your help.
Edit:
I found a conjectural way to compute $gamma$ and using the Euler Product a link to Riemann Zeta function:
$$zeta(s) = prod_p frac11-(2+sum_q<p b_p-1-q)^-s$$
and for $gamma$ numerical coincidences suggest that:
$$lim_n rightarrow infty fracb_nb_n+1 = gamma = -0.629233cdots$$
which could be one way to define $gamma$. Then one has to show, that this limit exists and that $f(gamma) = 0$.
If someone has an idea in this direction,that would be nice.
Second edit:
Here is the computation for the recurence relation of $a_n,k$:
For $kge 1$ we have on the one hand:
$$log(f(t)^k)' = frack f(t)^k-1 f'(t)f(t)^k = k fracf'(t)f(t) = sum_n=-1^infty k b_n t^n$$
On the other hand it is
$$f(t)^k = sum_n=0^infty a_n,kt^n$$
which means that
$$log(f(t)^k)' = fracsum_n=0^infty n a_n,k t^n-1sum_n=0^infty a_n,k t^n$$
Hence it follows, that (by multiplying with the denominator):
$$sum_n=0^infty n a_n,k t^n-1 = k (sum_n=0^infty a_n,k t^n) (sum_n=0 b_n-1t^n) frac1t$$
After multiplying with $t$ and using the Cauchy product formula we get:
$$sum_n=0^infty n a_n,k t^n = sum_n=0^infty k( sum_v=0^n a_v,kb_n-1-v)t^n$$
and comparing coefficients we find that:
$$n a_n,k = k ( sum_v=0^n a_v,k b_n-1-v)$$
and with $b_-1 = 2$ it follows after solving this equation for $a_n,k$ that:
$$ a_n,k = frackn-2k sum_v=0^n-1 a_v,k b_n-1-v$$
Especially for $k=1$ and $n=p$ prime we find that:
$$ p = 2+ sum_q<p b_p-1-q$$
number-theory prime-numbers riemann-zeta
$endgroup$
Let $H(t) = sum_n=1 ^infty pi(n)t^n$ where $pi(n)$ is the prime counting function. This is the Hilbert series of some $mathbbQ$-vector space. By the prime number theorem, the radius of convergence is $1$. Observing $pi(n) = pi(n-1)+1$ if $n$ is prime and $pi(n) = pi(n-1)$ if $n$ is composite, we might rewrite this as $H(t) = f(t)/(1-t)$ where $f(t) =sum_p text prime t^p$. Define the sequence $b_n$ for $n=-1,0,1,2,cdots$ as $ f'(t)/f(t) = sum_n=-1 b_n t^n$. Then we can recover the primes from this sequence:
$$ p = 2 + sum_q <p, q text prime b_p-1-q$$
For example:
For example the first coefficients are given by the series:
$$2*t^(-1) + 1 + (-1)*t + 4*t^2 + (-5)*t^3 + 11*t^4 + (-16)*t^5 + 22*t^6 + (-37)*t^7 + 67*t^8 + (-101)*t^9 + 166*t^10 + (-260)*t^11 + 404*t^12 + (-652)*t^13 + cdots $$
so $b_-1 = 2, b_0 = 1, b_1 = 1 , b_2 = 4$ etc.
We have for example:
$$3 = 2+b_0 = 2+1$$
$$5 = 2+b_2+b_1 = 2+4-1$$
$$7 = 2+b_4+b_3+b_1 = 2+11-5-1$$
Let $a_n,k$ denote the number of ordered ways of writing $n$ as a sum of $k$ primes.
Then after some calculation, one finds that:
$$a_n,k = frackn-2k sum_v=0^n-1 a_v,k b_n-1-v$$
which is a recurence relation.
Furthermore if $alpha_n$ $n=0,1,2,3,cdots$ are all roots of $f(t)$ not equal to zero, then for $nge 0$
$$ b_n = - sum_k=0 ^ infty frac1alpha_k^n+1$$
The numbers $b_n$ might be computed inductively using:
$$ n a_n,1 = sum_v=0^n b_v-1 a_n-v,1$$
from which one sees, that $b_n in mathbbZ$.
One real root of $f(t)$ seems to be the number,
$$ gamma = -0.62923 cdots $$
(OEIS: http://oeis.org/A078756 )
Since everything related to primes has something to do with the Riemann Zeta function, I wonder, what is the relation to the stuff above to the Riemann Zeta function?
If someone knows of any reference or has any idea, that would be great.
Is there for example a way to compute the real root $gamma$?
What is the relation of $gamma$ to the other complex roots?
What other properties do the numbers $b_n$ have?
etc.
Thanks for your help.
Edit:
I found a conjectural way to compute $gamma$ and using the Euler Product a link to Riemann Zeta function:
$$zeta(s) = prod_p frac11-(2+sum_q<p b_p-1-q)^-s$$
and for $gamma$ numerical coincidences suggest that:
$$lim_n rightarrow infty fracb_nb_n+1 = gamma = -0.629233cdots$$
which could be one way to define $gamma$. Then one has to show, that this limit exists and that $f(gamma) = 0$.
If someone has an idea in this direction,that would be nice.
Second edit:
Here is the computation for the recurence relation of $a_n,k$:
For $kge 1$ we have on the one hand:
$$log(f(t)^k)' = frack f(t)^k-1 f'(t)f(t)^k = k fracf'(t)f(t) = sum_n=-1^infty k b_n t^n$$
On the other hand it is
$$f(t)^k = sum_n=0^infty a_n,kt^n$$
which means that
$$log(f(t)^k)' = fracsum_n=0^infty n a_n,k t^n-1sum_n=0^infty a_n,k t^n$$
Hence it follows, that (by multiplying with the denominator):
$$sum_n=0^infty n a_n,k t^n-1 = k (sum_n=0^infty a_n,k t^n) (sum_n=0 b_n-1t^n) frac1t$$
After multiplying with $t$ and using the Cauchy product formula we get:
$$sum_n=0^infty n a_n,k t^n = sum_n=0^infty k( sum_v=0^n a_v,kb_n-1-v)t^n$$
and comparing coefficients we find that:
$$n a_n,k = k ( sum_v=0^n a_v,k b_n-1-v)$$
and with $b_-1 = 2$ it follows after solving this equation for $a_n,k$ that:
$$ a_n,k = frackn-2k sum_v=0^n-1 a_v,k b_n-1-v$$
Especially for $k=1$ and $n=p$ prime we find that:
$$ p = 2+ sum_q<p b_p-1-q$$
number-theory prime-numbers riemann-zeta
number-theory prime-numbers riemann-zeta
edited Mar 28 at 14:09
stackExchangeUser
asked Mar 27 at 8:12
stackExchangeUserstackExchangeUser
1,237513
1,237513
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First of all $f(t) = sum_p t^p$ means $f'(t)/f(t) = sum_n b_n t^n$ where $b_n=0$ for $n < -1$ and $sum_p b_n-p = casesn if n+1 is prime\ 0 otherwise$.
$f(e^-x)$ is the inverse Mellin transform of $Gamma(s)sum_p p^-s$ while things like $1/f(t), log f(t), f'(t)/f(t)$ are quite inaccessible, in the same way that $zeta(s)$ is accessible from only the integers while $1/zeta(s),log zeta(s),zeta'(s)/zeta(s)$ need the primes.
The number of ordered ways to write $n$ as a sum of $m$ primes are the coefficients of $f(t)^m$ and the number of ordered ways to write $n$ as a sum of primes are the coefficients of $frac11-f(t)-1 = sum_m=1^infty f(t)^m$.
$frac1f(t)=frac1t^2(1-(1-fracf(t)t^2))= t^-2sum_m=0^infty (-1)^m(fracf(t)t^2-1)^m$ where the coefficients of $(fracf(t)t^2-1)^m$ are the number of ordered ways to write $n+2m$ as a sum of $m$ primes $ge 3$. $frac1t^2-f(t)=t^-2sum_m=0^infty (fracf(t)t^2-1)^m$.
Let $g(x) = sum_p^k e^-p^k x log p$ the inverse Mellin transform of $Gamma(s) frac-zeta'(s)zeta(s)$. We have the explicit formula $g(x) = sum Res(Gamma(s) frac-zeta'(s)zeta(s) x^-s) = x^-1- sum_rho Gamma(rho) x^-rho-sum_k=0^infty (a_k+b_k log(x))x^k$. A corresponding explicit formula exists for $f(t)$ but it will be messy because $Gamma(s)sum_p p^-s$ has many branch points and a naturaly boundary $Re(s)=0$.
We don't know anything on the zeros and particular values of $f(t)$ and since it is analytic only for $|t|<1$ then $f'(t)/f(t)$ won't be equal to a sum over $f$'s zeros. The most of what we know is the asymptotic of $f(t)$ as $t to 1$, the error terms depends on the RH. It will give the asymptotic of $log f(t)$ and possibly for $f'(t)/f(t)$, as $t to 1$
If $f(t)$ has a zero on $|t|< 1$, let $z_0$ be one with minimal absolute value, assume there is no other zero on $|z_0|$, then $fracf'(t)f(t)-frac1t-z_0$ is analytic for $|t|le |z_0|+epsilon$ so that $b_n = z_0^-n+O(|z_0|+epsilon)^-n$ and $lim_n to infty b_n/b_n+1 = z_0$. By numerical approximation you can show such a $z_0$ exists in which case $z_0 in (-1,0)$. So your claim $z_0= -gamma$ is that $f'(t)/f(t)$ is analytic for $|t| < gamma$. It is plausible there are some heuristics for such a thing given $fracf(t)1-t = sum_n ge 2 pi(n) t^n approxsum_n ge 2 fracnlog n t^n$
$endgroup$
$begingroup$
Thanks for the detailed answer. One could not hope for more. I will need some time to go through this and maybe I have some question. Thanks again.
$endgroup$
– stackExchangeUser
Mar 28 at 13:01
$begingroup$
I don't understand the part with the number of ordered ways of writing $n$ as a sum of $k$ primes. I edited the question with the computation of the recurence relation. Maybe you can have a look?
$endgroup$
– stackExchangeUser
Mar 28 at 14:13
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3164216%2fa-series-related-to-prime-numbers%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First of all $f(t) = sum_p t^p$ means $f'(t)/f(t) = sum_n b_n t^n$ where $b_n=0$ for $n < -1$ and $sum_p b_n-p = casesn if n+1 is prime\ 0 otherwise$.
$f(e^-x)$ is the inverse Mellin transform of $Gamma(s)sum_p p^-s$ while things like $1/f(t), log f(t), f'(t)/f(t)$ are quite inaccessible, in the same way that $zeta(s)$ is accessible from only the integers while $1/zeta(s),log zeta(s),zeta'(s)/zeta(s)$ need the primes.
The number of ordered ways to write $n$ as a sum of $m$ primes are the coefficients of $f(t)^m$ and the number of ordered ways to write $n$ as a sum of primes are the coefficients of $frac11-f(t)-1 = sum_m=1^infty f(t)^m$.
$frac1f(t)=frac1t^2(1-(1-fracf(t)t^2))= t^-2sum_m=0^infty (-1)^m(fracf(t)t^2-1)^m$ where the coefficients of $(fracf(t)t^2-1)^m$ are the number of ordered ways to write $n+2m$ as a sum of $m$ primes $ge 3$. $frac1t^2-f(t)=t^-2sum_m=0^infty (fracf(t)t^2-1)^m$.
Let $g(x) = sum_p^k e^-p^k x log p$ the inverse Mellin transform of $Gamma(s) frac-zeta'(s)zeta(s)$. We have the explicit formula $g(x) = sum Res(Gamma(s) frac-zeta'(s)zeta(s) x^-s) = x^-1- sum_rho Gamma(rho) x^-rho-sum_k=0^infty (a_k+b_k log(x))x^k$. A corresponding explicit formula exists for $f(t)$ but it will be messy because $Gamma(s)sum_p p^-s$ has many branch points and a naturaly boundary $Re(s)=0$.
We don't know anything on the zeros and particular values of $f(t)$ and since it is analytic only for $|t|<1$ then $f'(t)/f(t)$ won't be equal to a sum over $f$'s zeros. The most of what we know is the asymptotic of $f(t)$ as $t to 1$, the error terms depends on the RH. It will give the asymptotic of $log f(t)$ and possibly for $f'(t)/f(t)$, as $t to 1$
If $f(t)$ has a zero on $|t|< 1$, let $z_0$ be one with minimal absolute value, assume there is no other zero on $|z_0|$, then $fracf'(t)f(t)-frac1t-z_0$ is analytic for $|t|le |z_0|+epsilon$ so that $b_n = z_0^-n+O(|z_0|+epsilon)^-n$ and $lim_n to infty b_n/b_n+1 = z_0$. By numerical approximation you can show such a $z_0$ exists in which case $z_0 in (-1,0)$. So your claim $z_0= -gamma$ is that $f'(t)/f(t)$ is analytic for $|t| < gamma$. It is plausible there are some heuristics for such a thing given $fracf(t)1-t = sum_n ge 2 pi(n) t^n approxsum_n ge 2 fracnlog n t^n$
$endgroup$
$begingroup$
Thanks for the detailed answer. One could not hope for more. I will need some time to go through this and maybe I have some question. Thanks again.
$endgroup$
– stackExchangeUser
Mar 28 at 13:01
$begingroup$
I don't understand the part with the number of ordered ways of writing $n$ as a sum of $k$ primes. I edited the question with the computation of the recurence relation. Maybe you can have a look?
$endgroup$
– stackExchangeUser
Mar 28 at 14:13
add a comment |
$begingroup$
First of all $f(t) = sum_p t^p$ means $f'(t)/f(t) = sum_n b_n t^n$ where $b_n=0$ for $n < -1$ and $sum_p b_n-p = casesn if n+1 is prime\ 0 otherwise$.
$f(e^-x)$ is the inverse Mellin transform of $Gamma(s)sum_p p^-s$ while things like $1/f(t), log f(t), f'(t)/f(t)$ are quite inaccessible, in the same way that $zeta(s)$ is accessible from only the integers while $1/zeta(s),log zeta(s),zeta'(s)/zeta(s)$ need the primes.
The number of ordered ways to write $n$ as a sum of $m$ primes are the coefficients of $f(t)^m$ and the number of ordered ways to write $n$ as a sum of primes are the coefficients of $frac11-f(t)-1 = sum_m=1^infty f(t)^m$.
$frac1f(t)=frac1t^2(1-(1-fracf(t)t^2))= t^-2sum_m=0^infty (-1)^m(fracf(t)t^2-1)^m$ where the coefficients of $(fracf(t)t^2-1)^m$ are the number of ordered ways to write $n+2m$ as a sum of $m$ primes $ge 3$. $frac1t^2-f(t)=t^-2sum_m=0^infty (fracf(t)t^2-1)^m$.
Let $g(x) = sum_p^k e^-p^k x log p$ the inverse Mellin transform of $Gamma(s) frac-zeta'(s)zeta(s)$. We have the explicit formula $g(x) = sum Res(Gamma(s) frac-zeta'(s)zeta(s) x^-s) = x^-1- sum_rho Gamma(rho) x^-rho-sum_k=0^infty (a_k+b_k log(x))x^k$. A corresponding explicit formula exists for $f(t)$ but it will be messy because $Gamma(s)sum_p p^-s$ has many branch points and a naturaly boundary $Re(s)=0$.
We don't know anything on the zeros and particular values of $f(t)$ and since it is analytic only for $|t|<1$ then $f'(t)/f(t)$ won't be equal to a sum over $f$'s zeros. The most of what we know is the asymptotic of $f(t)$ as $t to 1$, the error terms depends on the RH. It will give the asymptotic of $log f(t)$ and possibly for $f'(t)/f(t)$, as $t to 1$
If $f(t)$ has a zero on $|t|< 1$, let $z_0$ be one with minimal absolute value, assume there is no other zero on $|z_0|$, then $fracf'(t)f(t)-frac1t-z_0$ is analytic for $|t|le |z_0|+epsilon$ so that $b_n = z_0^-n+O(|z_0|+epsilon)^-n$ and $lim_n to infty b_n/b_n+1 = z_0$. By numerical approximation you can show such a $z_0$ exists in which case $z_0 in (-1,0)$. So your claim $z_0= -gamma$ is that $f'(t)/f(t)$ is analytic for $|t| < gamma$. It is plausible there are some heuristics for such a thing given $fracf(t)1-t = sum_n ge 2 pi(n) t^n approxsum_n ge 2 fracnlog n t^n$
$endgroup$
$begingroup$
Thanks for the detailed answer. One could not hope for more. I will need some time to go through this and maybe I have some question. Thanks again.
$endgroup$
– stackExchangeUser
Mar 28 at 13:01
$begingroup$
I don't understand the part with the number of ordered ways of writing $n$ as a sum of $k$ primes. I edited the question with the computation of the recurence relation. Maybe you can have a look?
$endgroup$
– stackExchangeUser
Mar 28 at 14:13
add a comment |
$begingroup$
First of all $f(t) = sum_p t^p$ means $f'(t)/f(t) = sum_n b_n t^n$ where $b_n=0$ for $n < -1$ and $sum_p b_n-p = casesn if n+1 is prime\ 0 otherwise$.
$f(e^-x)$ is the inverse Mellin transform of $Gamma(s)sum_p p^-s$ while things like $1/f(t), log f(t), f'(t)/f(t)$ are quite inaccessible, in the same way that $zeta(s)$ is accessible from only the integers while $1/zeta(s),log zeta(s),zeta'(s)/zeta(s)$ need the primes.
The number of ordered ways to write $n$ as a sum of $m$ primes are the coefficients of $f(t)^m$ and the number of ordered ways to write $n$ as a sum of primes are the coefficients of $frac11-f(t)-1 = sum_m=1^infty f(t)^m$.
$frac1f(t)=frac1t^2(1-(1-fracf(t)t^2))= t^-2sum_m=0^infty (-1)^m(fracf(t)t^2-1)^m$ where the coefficients of $(fracf(t)t^2-1)^m$ are the number of ordered ways to write $n+2m$ as a sum of $m$ primes $ge 3$. $frac1t^2-f(t)=t^-2sum_m=0^infty (fracf(t)t^2-1)^m$.
Let $g(x) = sum_p^k e^-p^k x log p$ the inverse Mellin transform of $Gamma(s) frac-zeta'(s)zeta(s)$. We have the explicit formula $g(x) = sum Res(Gamma(s) frac-zeta'(s)zeta(s) x^-s) = x^-1- sum_rho Gamma(rho) x^-rho-sum_k=0^infty (a_k+b_k log(x))x^k$. A corresponding explicit formula exists for $f(t)$ but it will be messy because $Gamma(s)sum_p p^-s$ has many branch points and a naturaly boundary $Re(s)=0$.
We don't know anything on the zeros and particular values of $f(t)$ and since it is analytic only for $|t|<1$ then $f'(t)/f(t)$ won't be equal to a sum over $f$'s zeros. The most of what we know is the asymptotic of $f(t)$ as $t to 1$, the error terms depends on the RH. It will give the asymptotic of $log f(t)$ and possibly for $f'(t)/f(t)$, as $t to 1$
If $f(t)$ has a zero on $|t|< 1$, let $z_0$ be one with minimal absolute value, assume there is no other zero on $|z_0|$, then $fracf'(t)f(t)-frac1t-z_0$ is analytic for $|t|le |z_0|+epsilon$ so that $b_n = z_0^-n+O(|z_0|+epsilon)^-n$ and $lim_n to infty b_n/b_n+1 = z_0$. By numerical approximation you can show such a $z_0$ exists in which case $z_0 in (-1,0)$. So your claim $z_0= -gamma$ is that $f'(t)/f(t)$ is analytic for $|t| < gamma$. It is plausible there are some heuristics for such a thing given $fracf(t)1-t = sum_n ge 2 pi(n) t^n approxsum_n ge 2 fracnlog n t^n$
$endgroup$
First of all $f(t) = sum_p t^p$ means $f'(t)/f(t) = sum_n b_n t^n$ where $b_n=0$ for $n < -1$ and $sum_p b_n-p = casesn if n+1 is prime\ 0 otherwise$.
$f(e^-x)$ is the inverse Mellin transform of $Gamma(s)sum_p p^-s$ while things like $1/f(t), log f(t), f'(t)/f(t)$ are quite inaccessible, in the same way that $zeta(s)$ is accessible from only the integers while $1/zeta(s),log zeta(s),zeta'(s)/zeta(s)$ need the primes.
The number of ordered ways to write $n$ as a sum of $m$ primes are the coefficients of $f(t)^m$ and the number of ordered ways to write $n$ as a sum of primes are the coefficients of $frac11-f(t)-1 = sum_m=1^infty f(t)^m$.
$frac1f(t)=frac1t^2(1-(1-fracf(t)t^2))= t^-2sum_m=0^infty (-1)^m(fracf(t)t^2-1)^m$ where the coefficients of $(fracf(t)t^2-1)^m$ are the number of ordered ways to write $n+2m$ as a sum of $m$ primes $ge 3$. $frac1t^2-f(t)=t^-2sum_m=0^infty (fracf(t)t^2-1)^m$.
Let $g(x) = sum_p^k e^-p^k x log p$ the inverse Mellin transform of $Gamma(s) frac-zeta'(s)zeta(s)$. We have the explicit formula $g(x) = sum Res(Gamma(s) frac-zeta'(s)zeta(s) x^-s) = x^-1- sum_rho Gamma(rho) x^-rho-sum_k=0^infty (a_k+b_k log(x))x^k$. A corresponding explicit formula exists for $f(t)$ but it will be messy because $Gamma(s)sum_p p^-s$ has many branch points and a naturaly boundary $Re(s)=0$.
We don't know anything on the zeros and particular values of $f(t)$ and since it is analytic only for $|t|<1$ then $f'(t)/f(t)$ won't be equal to a sum over $f$'s zeros. The most of what we know is the asymptotic of $f(t)$ as $t to 1$, the error terms depends on the RH. It will give the asymptotic of $log f(t)$ and possibly for $f'(t)/f(t)$, as $t to 1$
If $f(t)$ has a zero on $|t|< 1$, let $z_0$ be one with minimal absolute value, assume there is no other zero on $|z_0|$, then $fracf'(t)f(t)-frac1t-z_0$ is analytic for $|t|le |z_0|+epsilon$ so that $b_n = z_0^-n+O(|z_0|+epsilon)^-n$ and $lim_n to infty b_n/b_n+1 = z_0$. By numerical approximation you can show such a $z_0$ exists in which case $z_0 in (-1,0)$. So your claim $z_0= -gamma$ is that $f'(t)/f(t)$ is analytic for $|t| < gamma$. It is plausible there are some heuristics for such a thing given $fracf(t)1-t = sum_n ge 2 pi(n) t^n approxsum_n ge 2 fracnlog n t^n$
edited Mar 28 at 13:02
answered Mar 28 at 12:55
reunsreuns
20.9k21354
20.9k21354
$begingroup$
Thanks for the detailed answer. One could not hope for more. I will need some time to go through this and maybe I have some question. Thanks again.
$endgroup$
– stackExchangeUser
Mar 28 at 13:01
$begingroup$
I don't understand the part with the number of ordered ways of writing $n$ as a sum of $k$ primes. I edited the question with the computation of the recurence relation. Maybe you can have a look?
$endgroup$
– stackExchangeUser
Mar 28 at 14:13
add a comment |
$begingroup$
Thanks for the detailed answer. One could not hope for more. I will need some time to go through this and maybe I have some question. Thanks again.
$endgroup$
– stackExchangeUser
Mar 28 at 13:01
$begingroup$
I don't understand the part with the number of ordered ways of writing $n$ as a sum of $k$ primes. I edited the question with the computation of the recurence relation. Maybe you can have a look?
$endgroup$
– stackExchangeUser
Mar 28 at 14:13
$begingroup$
Thanks for the detailed answer. One could not hope for more. I will need some time to go through this and maybe I have some question. Thanks again.
$endgroup$
– stackExchangeUser
Mar 28 at 13:01
$begingroup$
Thanks for the detailed answer. One could not hope for more. I will need some time to go through this and maybe I have some question. Thanks again.
$endgroup$
– stackExchangeUser
Mar 28 at 13:01
$begingroup$
I don't understand the part with the number of ordered ways of writing $n$ as a sum of $k$ primes. I edited the question with the computation of the recurence relation. Maybe you can have a look?
$endgroup$
– stackExchangeUser
Mar 28 at 14:13
$begingroup$
I don't understand the part with the number of ordered ways of writing $n$ as a sum of $k$ primes. I edited the question with the computation of the recurence relation. Maybe you can have a look?
$endgroup$
– stackExchangeUser
Mar 28 at 14:13
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3164216%2fa-series-related-to-prime-numbers%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown