Does $n = 2p$ where $p$ is prime, have fewer prime pairs than $n neq 2p?$
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
add a comment |
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
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
add a comment |
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
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 Nov 20 at 3:26
Robert Soupe
10.9k21949
10.9k21949
asked Nov 19 at 17:48
temp watts
212
212
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
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...
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
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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%2f3005249%2fdoes-n-2p-where-p-is-prime-have-fewer-prime-pairs-than-n-neq-2p%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
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...
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
add a comment |
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...
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
add a comment |
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...
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...
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
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
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
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3005249%2fdoes-n-2p-where-p-is-prime-have-fewer-prime-pairs-than-n-neq-2p%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
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