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 $n neq 2p$?










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    Presumably you are comparing with $n$ twice a composite number
    – Henry
    Nov 19 at 17:51










  • 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
    Nov 20 at 3:30
















1














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 $n neq 2p$?










share|cite|improve this question




















  • 1




    Presumably you are comparing with $n$ twice a composite number
    – Henry
    Nov 19 at 17:51










  • 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
    Nov 20 at 3:30














1












1








1







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 $n neq 2p$?










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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






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edited Nov 20 at 3:26









Robert Soupe

10.9k21949




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asked Nov 19 at 17:48









temp watts

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  • 1




    Presumably you are comparing with $n$ twice a composite number
    – Henry
    Nov 19 at 17:51










  • 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
    Nov 20 at 3:30














  • 1




    Presumably you are comparing with $n$ twice a composite number
    – Henry
    Nov 19 at 17:51










  • 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
    Nov 20 at 3:30








1




1




Presumably you are comparing with $n$ twice a composite number
– Henry
Nov 19 at 17:51




Presumably you are comparing with $n$ twice a composite number
– Henry
Nov 19 at 17:51












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
Nov 20 at 3:30




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
Nov 20 at 3:30










1 Answer
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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...






share|cite|improve this answer























  • 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
    Nov 19 at 19:52













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1 Answer
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1 Answer
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active

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active

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1














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...






share|cite|improve this answer























  • 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
    Nov 19 at 19:52


















1














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...






share|cite|improve this answer























  • 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
    Nov 19 at 19:52
















1












1








1






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...






share|cite|improve this answer














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...







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 19 at 18:35

























answered Nov 19 at 18:24









Mason

1,9341530




1,9341530












  • 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
    Nov 19 at 19:52




















  • 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
    Nov 19 at 19:52


















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
Nov 19 at 19:52






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
Nov 19 at 19:52




















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