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Does $n = 2p$ where $p$ is prime, have fewer prime pairs than $n neq 2p?$
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Does an even integer $n = 2p$ where $p$ is prime, have relatively fewer prime pairs $(p_1, p_2)$ such that $p_1 + p_2 = n$, than an even integer $n neq 2p$?
prime-numbers
edited yesterday
Robert Soupe
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temp watts
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Does an even integer $n = 2p$ where $p$ is prime, have relatively fewer prime pairs $(p_1, p_2)$ such that $p_1 + p_2 = n$, than an even integer $n neq 2p$?
prime-numbers
edited yesterday
Robert Soupe
10.6k21948
asked yesterday
temp watts
62
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temp watts is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1
Presumably you are comparing with $n$ twice a composite number
– Henry
yesterday
It's more a matter of how large $n$ is. Suppose $p = 7$, then $n = 14$ and you just have 3 + 11 = 7 + 7. Suppose instead $n = 720$, which is twice 360, clearly not a prime. You have 709 + 11 = 701 + 19 = 691 + 29 = 683 + 37 = 677 + 43 = etc.
– Robert Soupe
yesterday
add a comment |
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up vote
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down vote
favorite
Does an even integer $n = 2p$ where $p$ is prime, have relatively fewer prime pairs $(p_1, p_2)$ such that $p_1 + p_2 = n$, than an even integer $n neq 2p$?
prime-numbers
edited yesterday
Robert Soupe
10.6k21948
asked yesterday
temp watts
62
New contributor
temp watts is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Does an even integer $n = 2p$ where $p$ is prime, have relatively fewer prime pairs $(p_1, p_2)$ such that $p_1 + p_2 = n$, than an even integer $n neq 2p$?
prime-numbers
prime-numbers
edited yesterday
Robert Soupe
10.6k21948
asked yesterday
temp watts
62
New contributor
temp watts is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited yesterday
Robert Soupe
10.6k21948
asked yesterday
temp watts
62
New contributor
temp watts is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Robert Soupe
10.6k21948
edited yesterday
Robert Soupe
10.6k21948
edited yesterday
Robert Soupe
10.6k21948
10.6k21948
asked yesterday
temp watts
62
New contributor
temp watts is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
temp watts
62
asked yesterday
temp watts
62
62
New contributor
temp watts is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
temp watts is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
temp watts is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Presumably you are comparing with $n$ twice a composite number
– Henry
yesterday
It's more a matter of how large $n$ is. Suppose $p = 7$, then $n = 14$ and you just have 3 + 11 = 7 + 7. Suppose instead $n = 720$, which is twice 360, clearly not a prime. You have 709 + 11 = 701 + 19 = 691 + 29 = 683 + 37 = 677 + 43 = etc.
– Robert Soupe
yesterday
add a comment |
1
Presumably you are comparing with $n$ twice a composite number
– Henry
yesterday
It's more a matter of how large $n$ is. Suppose $p = 7$, then $n = 14$ and you just have 3 + 11 = 7 + 7. Suppose instead $n = 720$, which is twice 360, clearly not a prime. You have 709 + 11 = 701 + 19 = 691 + 29 = 683 + 37 = 677 + 43 = etc.
– Robert Soupe
yesterday
1
1
Presumably you are comparing with $n$ twice a composite number
– Henry
yesterday
Presumably you are comparing with $n$ twice a composite number
– Henry
yesterday
It's more a matter of how large $n$ is. Suppose $p = 7$, then $n = 14$ and you just have 3 + 11 = 7 + 7. Suppose instead $n = 720$, which is twice 360, clearly not a prime. You have 709 + 11 = 701 + 19 = 691 + 29 = 683 + 37 = 677 + 43 = etc.
– Robert Soupe
yesterday
It's more a matter of how large $n$ is. Suppose $p = 7$, then $n = 14$ and you just have 3 + 11 = 7 + 7. Suppose instead $n = 720$, which is twice 360, clearly not a prime. You have 709 + 11 = 701 + 19 = 691 + 29 = 683 + 37 = 677 + 43 = etc.
– Robert Soupe
yesterday
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
This is just an attempt to understand the question which I think could be an interesting one but seems a little ill defined.
Define a set $S_m={ (p_1,p_2): p_1,p_2 in mathbb{P} text{ such that } 2m=p_1+p_2}$. Where $mathbb{P}$ is the set of primes.
Define a function $f(m)=|S_m|$.
So now we have some machinery in place. $f$ counts the number of ways we can write the double of a number as the sum of two primes. For example
$f(7)=3$ because we can write $14$ in $3$ different ways. $14=11+3=7+7=3+11$.
Note that you may decide not to count them in this way but if that's the case you should really specify how you want to count this in your question...
$f(6)=2$, because $12=5+7=7+5$.
$f(8)=4$, because $16= 5+11=11+5= 3+13=13+3$.
So we have $f(6)<f(7)<f(8)$.
Now after we build up some machinery... what's a good way to frame the question?
Something like this?
Maybe let $operatorname{avg}(n) = frac{1}{n}sum^n_{j=1} f(j)$
and let $operatorname{avg_p}(n) = frac{1}{pi(n)}sum^n_{p in mathbb{P}} f(p)$
Where $pi(n)$ is the number of primes less than $n$.
Then we can ask how these functions compare as $n$ grows?
Anyway... I think that this is one reasonable interpretation of this question but I am just trying to make the point that there are probably many possible interpretations you could ask. You might have to do some thinking about how to formulate this question so there is something to really investigate and not leave so much interpretation up to others...
answered yesterday
Mason
1,6501325
Thank you for your reply. Your interpretation of what I wanted to say is quite accurate. I apologize that my math and writing skills need work. Anyway, what I wanted to know is, if you graph f(x) where x is not prime, and you graph f(p) where p is prime, will the curve for f(x) always lie above f(p)?
– temp watts
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
This is just an attempt to understand the question which I think could be an interesting one but seems a little ill defined.
Define a set $S_m={ (p_1,p_2): p_1,p_2 in mathbb{P} text{ such that } 2m=p_1+p_2}$. Where $mathbb{P}$ is the set of primes.
Define a function $f(m)=|S_m|$.
So now we have some machinery in place. $f$ counts the number of ways we can write the double of a number as the sum of two primes. For example
$f(7)=3$ because we can write $14$ in $3$ different ways. $14=11+3=7+7=3+11$.
Note that you may decide not to count them in this way but if that's the case you should really specify how you want to count this in your question...
$f(6)=2$, because $12=5+7=7+5$.
$f(8)=4$, because $16= 5+11=11+5= 3+13=13+3$.
So we have $f(6)<f(7)<f(8)$.
Now after we build up some machinery... what's a good way to frame the question?
Something like this?
Maybe let $operatorname{avg}(n) = frac{1}{n}sum^n_{j=1} f(j)$
and let $operatorname{avg_p}(n) = frac{1}{pi(n)}sum^n_{p in mathbb{P}} f(p)$
Where $pi(n)$ is the number of primes less than $n$.
Then we can ask how these functions compare as $n$ grows?
Anyway... I think that this is one reasonable interpretation of this question but I am just trying to make the point that there are probably many possible interpretations you could ask. You might have to do some thinking about how to formulate this question so there is something to really investigate and not leave so much interpretation up to others...
answered yesterday
Mason
1,6501325
Thank you for your reply. Your interpretation of what I wanted to say is quite accurate. I apologize that my math and writing skills need work. Anyway, what I wanted to know is, if you graph f(x) where x is not prime, and you graph f(p) where p is prime, will the curve for f(x) always lie above f(p)?
– temp watts
yesterday
add a comment |
up vote
1
down vote
This is just an attempt to understand the question which I think could be an interesting one but seems a little ill defined.
Define a set $S_m={ (p_1,p_2): p_1,p_2 in mathbb{P} text{ such that } 2m=p_1+p_2}$. Where $mathbb{P}$ is the set of primes.
Define a function $f(m)=|S_m|$.
So now we have some machinery in place. $f$ counts the number of ways we can write the double of a number as the sum of two primes. For example
$f(7)=3$ because we can write $14$ in $3$ different ways. $14=11+3=7+7=3+11$.
Note that you may decide not to count them in this way but if that's the case you should really specify how you want to count this in your question...
$f(6)=2$, because $12=5+7=7+5$.
$f(8)=4$, because $16= 5+11=11+5= 3+13=13+3$.
So we have $f(6)<f(7)<f(8)$.
Now after we build up some machinery... what's a good way to frame the question?
Something like this?
Maybe let $operatorname{avg}(n) = frac{1}{n}sum^n_{j=1} f(j)$
and let $operatorname{avg_p}(n) = frac{1}{pi(n)}sum^n_{p in mathbb{P}} f(p)$
Where $pi(n)$ is the number of primes less than $n$.
Then we can ask how these functions compare as $n$ grows?
Anyway... I think that this is one reasonable interpretation of this question but I am just trying to make the point that there are probably many possible interpretations you could ask. You might have to do some thinking about how to formulate this question so there is something to really investigate and not leave so much interpretation up to others...
answered yesterday
Mason
1,6501325
Thank you for your reply. Your interpretation of what I wanted to say is quite accurate. I apologize that my math and writing skills need work. Anyway, what I wanted to know is, if you graph f(x) where x is not prime, and you graph f(p) where p is prime, will the curve for f(x) always lie above f(p)?
– temp watts
yesterday
add a comment |
up vote
1
down vote
up vote
1
down vote
This is just an attempt to understand the question which I think could be an interesting one but seems a little ill defined.
Define a set $S_m={ (p_1,p_2): p_1,p_2 in mathbb{P} text{ such that } 2m=p_1+p_2}$. Where $mathbb{P}$ is the set of primes.
Define a function $f(m)=|S_m|$.
So now we have some machinery in place. $f$ counts the number of ways we can write the double of a number as the sum of two primes. For example
$f(7)=3$ because we can write $14$ in $3$ different ways. $14=11+3=7+7=3+11$.
Note that you may decide not to count them in this way but if that's the case you should really specify how you want to count this in your question...
$f(6)=2$, because $12=5+7=7+5$.
$f(8)=4$, because $16= 5+11=11+5= 3+13=13+3$.
So we have $f(6)<f(7)<f(8)$.
Now after we build up some machinery... what's a good way to frame the question?
Something like this?
Maybe let $operatorname{avg}(n) = frac{1}{n}sum^n_{j=1} f(j)$
and let $operatorname{avg_p}(n) = frac{1}{pi(n)}sum^n_{p in mathbb{P}} f(p)$
Where $pi(n)$ is the number of primes less than $n$.
Then we can ask how these functions compare as $n$ grows?
Anyway... I think that this is one reasonable interpretation of this question but I am just trying to make the point that there are probably many possible interpretations you could ask. You might have to do some thinking about how to formulate this question so there is something to really investigate and not leave so much interpretation up to others...
answered yesterday
Mason
1,6501325
This is just an attempt to understand the question which I think could be an interesting one but seems a little ill defined.
Define a set $S_m={ (p_1,p_2): p_1,p_2 in mathbb{P} text{ such that } 2m=p_1+p_2}$. Where $mathbb{P}$ is the set of primes.
Define a function $f(m)=|S_m|$.
So now we have some machinery in place. $f$ counts the number of ways we can write the double of a number as the sum of two primes. For example
$f(7)=3$ because we can write $14$ in $3$ different ways. $14=11+3=7+7=3+11$.
Note that you may decide not to count them in this way but if that's the case you should really specify how you want to count this in your question...
$f(6)=2$, because $12=5+7=7+5$.
$f(8)=4$, because $16= 5+11=11+5= 3+13=13+3$.
So we have $f(6)<f(7)<f(8)$.
Now after we build up some machinery... what's a good way to frame the question?
Something like this?
Maybe let $operatorname{avg}(n) = frac{1}{n}sum^n_{j=1} f(j)$
and let $operatorname{avg_p}(n) = frac{1}{pi(n)}sum^n_{p in mathbb{P}} f(p)$
Where $pi(n)$ is the number of primes less than $n$.
Then we can ask how these functions compare as $n$ grows?
Anyway... I think that this is one reasonable interpretation of this question but I am just trying to make the point that there are probably many possible interpretations you could ask. You might have to do some thinking about how to formulate this question so there is something to really investigate and not leave so much interpretation up to others...
answered yesterday
Mason
1,6501325
edited yesterday
answered yesterday
Mason
1,6501325
answered yesterday
Mason
1,6501325
answered yesterday
Mason
1,6501325
1,6501325
Thank you for your reply. Your interpretation of what I wanted to say is quite accurate. I apologize that my math and writing skills need work. Anyway, what I wanted to know is, if you graph f(x) where x is not prime, and you graph f(p) where p is prime, will the curve for f(x) always lie above f(p)?
– temp watts
yesterday
add a comment |
Thank you for your reply. Your interpretation of what I wanted to say is quite accurate. I apologize that my math and writing skills need work. Anyway, what I wanted to know is, if you graph f(x) where x is not prime, and you graph f(p) where p is prime, will the curve for f(x) always lie above f(p)?
– temp watts
yesterday
Thank you for your reply. Your interpretation of what I wanted to say is quite accurate. I apologize that my math and writing skills need work. Anyway, what I wanted to know is, if you graph f(x) where x is not prime, and you graph f(p) where p is prime, will the curve for f(x) always lie above f(p)?
– temp watts
yesterday
Thank you for your reply. Your interpretation of what I wanted to say is quite accurate. I apologize that my math and writing skills need work. Anyway, what I wanted to know is, if you graph f(x) where x is not prime, and you graph f(p) where p is prime, will the curve for f(x) always lie above f(p)?
– temp watts
yesterday
add a comment |
temp watts is a new contributor. Be nice, and check out our Code of Conduct.
temp watts is a new contributor. Be nice, and check out our Code of Conduct.
temp watts is a new contributor. Be nice, and check out our Code of Conduct.
temp watts is a new contributor. Be nice, and check out our Code of Conduct.
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1
Presumably you are comparing with $n$ twice a composite number
– Henry
yesterday
It's more a matter of how large $n$ is. Suppose $p = 7$, then $n = 14$ and you just have 3 + 11 = 7 + 7. Suppose instead $n = 720$, which is twice 360, clearly not a prime. You have 709 + 11 = 701 + 19 = 691 + 29 = 683 + 37 = 677 + 43 = etc.
– Robert Soupe
yesterday