A curious equality of integrals involving the prime counting function?
$begingroup$
This post discusses the integral,
$$I(k)=int_0^kpi(x)pi(k-x)dx$$
where $pi(x)$ is the prime-counting function. For example,
$$I(13)=int_0^{13}pi(x)pi(13-x)dx = 73$$
Using WolframAlpha, the first 50 values for $k=1,2,3,dots$ are,
$$I(k) = 0, 0, 0, 0, 1, 4, 8, 14, 22, 32, 45, 58, 73, 90, 110, 132, 158, 184, 214, 246, 282, 320, 363, 406, 455, 506, 562, 618, 678, 738, 804, 872, 944, 1018, 1099, 1180, 1269, 1358, 1450, 1544, 1644, 1744, 1852, 1962, 2078, 2196, 2321, 2446, 2581, 2718,dots$$
While trying to find if the above sequence obeyed a pattern, I noticed a rather unexpected relationship:
Q: For all $n>0$, is it true,
$$I(6n+4) - 2,I(6n+5) + I(6n+6) overset{color{red}?}= 0$$
Example, for $n=1,2$, then
$$I(10)-2I(11)+I(12)=32-2*45+58 = 0$$
$$I(16)-2I(17)+I(18)=132-2*158+184= 0$$
and so on.
integration definite-integrals prime-numbers
asked 1 hour ago
Tito Piezas IIITito Piezas III
27.4k366174
$endgroup$
|
show 3 more comments
$begingroup$
This post discusses the integral,
$$I(k)=int_0^kpi(x)pi(k-x)dx$$
where $pi(x)$ is the prime-counting function. For example,
$$I(13)=int_0^{13}pi(x)pi(13-x)dx = 73$$
Using WolframAlpha, the first 50 values for $k=1,2,3,dots$ are,
$$I(k) = 0, 0, 0, 0, 1, 4, 8, 14, 22, 32, 45, 58, 73, 90, 110, 132, 158, 184, 214, 246, 282, 320, 363, 406, 455, 506, 562, 618, 678, 738, 804, 872, 944, 1018, 1099, 1180, 1269, 1358, 1450, 1544, 1644, 1744, 1852, 1962, 2078, 2196, 2321, 2446, 2581, 2718,dots$$
While trying to find if the above sequence obeyed a pattern, I noticed a rather unexpected relationship:
Q: For all $n>0$, is it true,
$$I(6n+4) - 2,I(6n+5) + I(6n+6) overset{color{red}?}= 0$$
Example, for $n=1,2$, then
$$I(10)-2I(11)+I(12)=32-2*45+58 = 0$$
$$I(16)-2I(17)+I(18)=132-2*158+184= 0$$
and so on.
integration definite-integrals prime-numbers
asked 1 hour ago
Tito Piezas IIITito Piezas III
27.4k366174
$endgroup$
$begingroup$
Note your proposed equation doesn't hold for $n = 0$ as $I(4) = 0$, $I(5) = 1$ and $I(6) = 4$.
$endgroup$
– John Omielan
1 hour ago
$begingroup$
@JohnOmielan: A typo. I meant all $n>0$. I will correct it.
$endgroup$
– Tito Piezas III
1 hour ago
$begingroup$
I have checked to confirm what you're asking is true for $n$ up to $18$. However, I have my doubts it'll always work, partially because it doesn't work for $n = 0$. Also, a similar type condition is that $I(6n) - 2I(6n + 1) + I(6n + 2) = 2$, which holds for $1 le n le 5$, but at $n = 6$, the LHS becomes $0$ instead. If I get a chance, I will investigate your equation to see if I can figure out why it's true for at least the first $18$ values and, more importantly, will it always stay true. Regardless, though, it's an excellent observation you've made, even if it doesn't always hold.
$endgroup$
– John Omielan
52 mins ago
$begingroup$
I checked your result up to $n=533$ (for $n geq 534$, I have problems. Would you be interested by a huge table of $I(k)$ (I was able to generate it up to $k=540$). This is a very interesting problem.
$endgroup$
– Claude Leibovici
34 mins ago
$begingroup$
@ClaudeLeibovici: Thanks for checking, Claude! However, that table would be too huge for MSE. :)
$endgroup$
– Tito Piezas III
31 mins ago
|
show 3 more comments
$begingroup$
This post discusses the integral,
$$I(k)=int_0^kpi(x)pi(k-x)dx$$
where $pi(x)$ is the prime-counting function. For example,
$$I(13)=int_0^{13}pi(x)pi(13-x)dx = 73$$
Using WolframAlpha, the first 50 values for $k=1,2,3,dots$ are,
$$I(k) = 0, 0, 0, 0, 1, 4, 8, 14, 22, 32, 45, 58, 73, 90, 110, 132, 158, 184, 214, 246, 282, 320, 363, 406, 455, 506, 562, 618, 678, 738, 804, 872, 944, 1018, 1099, 1180, 1269, 1358, 1450, 1544, 1644, 1744, 1852, 1962, 2078, 2196, 2321, 2446, 2581, 2718,dots$$
While trying to find if the above sequence obeyed a pattern, I noticed a rather unexpected relationship:
Q: For all $n>0$, is it true,
$$I(6n+4) - 2,I(6n+5) + I(6n+6) overset{color{red}?}= 0$$
Example, for $n=1,2$, then
$$I(10)-2I(11)+I(12)=32-2*45+58 = 0$$
$$I(16)-2I(17)+I(18)=132-2*158+184= 0$$
and so on.
integration definite-integrals prime-numbers
asked 1 hour ago
Tito Piezas IIITito Piezas III
27.4k366174
$endgroup$
This post discusses the integral,
$$I(k)=int_0^kpi(x)pi(k-x)dx$$
where $pi(x)$ is the prime-counting function. For example,
$$I(13)=int_0^{13}pi(x)pi(13-x)dx = 73$$
Using WolframAlpha, the first 50 values for $k=1,2,3,dots$ are,
$$I(k) = 0, 0, 0, 0, 1, 4, 8, 14, 22, 32, 45, 58, 73, 90, 110, 132, 158, 184, 214, 246, 282, 320, 363, 406, 455, 506, 562, 618, 678, 738, 804, 872, 944, 1018, 1099, 1180, 1269, 1358, 1450, 1544, 1644, 1744, 1852, 1962, 2078, 2196, 2321, 2446, 2581, 2718,dots$$
While trying to find if the above sequence obeyed a pattern, I noticed a rather unexpected relationship:
Q: For all $n>0$, is it true,
$$I(6n+4) - 2,I(6n+5) + I(6n+6) overset{color{red}?}= 0$$
Example, for $n=1,2$, then
$$I(10)-2I(11)+I(12)=32-2*45+58 = 0$$
$$I(16)-2I(17)+I(18)=132-2*158+184= 0$$
and so on.
integration definite-integrals prime-numbers
integration definite-integrals prime-numbers
asked 1 hour ago
Tito Piezas IIITito Piezas III
27.4k366174
asked 1 hour ago
Tito Piezas IIITito Piezas III
27.4k366174
edited 17 mins ago
Tito Piezas III
asked 1 hour ago
Tito Piezas IIITito Piezas III
27.4k366174
asked 1 hour ago
Tito Piezas IIITito Piezas III
27.4k366174
asked 1 hour ago
Tito Piezas IIITito Piezas III
27.4k366174
27.4k366174
$begingroup$
Note your proposed equation doesn't hold for $n = 0$ as $I(4) = 0$, $I(5) = 1$ and $I(6) = 4$.
$endgroup$
– John Omielan
1 hour ago
$begingroup$
@JohnOmielan: A typo. I meant all $n>0$. I will correct it.
$endgroup$
– Tito Piezas III
1 hour ago
$begingroup$
I have checked to confirm what you're asking is true for $n$ up to $18$. However, I have my doubts it'll always work, partially because it doesn't work for $n = 0$. Also, a similar type condition is that $I(6n) - 2I(6n + 1) + I(6n + 2) = 2$, which holds for $1 le n le 5$, but at $n = 6$, the LHS becomes $0$ instead. If I get a chance, I will investigate your equation to see if I can figure out why it's true for at least the first $18$ values and, more importantly, will it always stay true. Regardless, though, it's an excellent observation you've made, even if it doesn't always hold.
$endgroup$
– John Omielan
52 mins ago
$begingroup$
I checked your result up to $n=533$ (for $n geq 534$, I have problems. Would you be interested by a huge table of $I(k)$ (I was able to generate it up to $k=540$). This is a very interesting problem.
$endgroup$
– Claude Leibovici
34 mins ago
$begingroup$
@ClaudeLeibovici: Thanks for checking, Claude! However, that table would be too huge for MSE. :)
$endgroup$
– Tito Piezas III
31 mins ago
|
show 3 more comments
$begingroup$
Note your proposed equation doesn't hold for $n = 0$ as $I(4) = 0$, $I(5) = 1$ and $I(6) = 4$.
$endgroup$
– John Omielan
1 hour ago
$begingroup$
@JohnOmielan: A typo. I meant all $n>0$. I will correct it.
$endgroup$
– Tito Piezas III
1 hour ago
$begingroup$
I have checked to confirm what you're asking is true for $n$ up to $18$. However, I have my doubts it'll always work, partially because it doesn't work for $n = 0$. Also, a similar type condition is that $I(6n) - 2I(6n + 1) + I(6n + 2) = 2$, which holds for $1 le n le 5$, but at $n = 6$, the LHS becomes $0$ instead. If I get a chance, I will investigate your equation to see if I can figure out why it's true for at least the first $18$ values and, more importantly, will it always stay true. Regardless, though, it's an excellent observation you've made, even if it doesn't always hold.
$endgroup$
– John Omielan
52 mins ago
$begingroup$
I checked your result up to $n=533$ (for $n geq 534$, I have problems. Would you be interested by a huge table of $I(k)$ (I was able to generate it up to $k=540$). This is a very interesting problem.
$endgroup$
– Claude Leibovici
34 mins ago
$begingroup$
@ClaudeLeibovici: Thanks for checking, Claude! However, that table would be too huge for MSE. :)
$endgroup$
– Tito Piezas III
31 mins ago
$begingroup$
Note your proposed equation doesn't hold for $n = 0$ as $I(4) = 0$, $I(5) = 1$ and $I(6) = 4$.
$endgroup$
– John Omielan
1 hour ago
$begingroup$
Note your proposed equation doesn't hold for $n = 0$ as $I(4) = 0$, $I(5) = 1$ and $I(6) = 4$.
$endgroup$
– John Omielan
1 hour ago
$begingroup$
@JohnOmielan: A typo. I meant all $n>0$. I will correct it.
$endgroup$
– Tito Piezas III
1 hour ago
$begingroup$
@JohnOmielan: A typo. I meant all $n>0$. I will correct it.
$endgroup$
– Tito Piezas III
1 hour ago
$begingroup$
I have checked to confirm what you're asking is true for $n$ up to $18$. However, I have my doubts it'll always work, partially because it doesn't work for $n = 0$. Also, a similar type condition is that $I(6n) - 2I(6n + 1) + I(6n + 2) = 2$, which holds for $1 le n le 5$, but at $n = 6$, the LHS becomes $0$ instead. If I get a chance, I will investigate your equation to see if I can figure out why it's true for at least the first $18$ values and, more importantly, will it always stay true. Regardless, though, it's an excellent observation you've made, even if it doesn't always hold.
$endgroup$
– John Omielan
52 mins ago
$begingroup$
I have checked to confirm what you're asking is true for $n$ up to $18$. However, I have my doubts it'll always work, partially because it doesn't work for $n = 0$. Also, a similar type condition is that $I(6n) - 2I(6n + 1) + I(6n + 2) = 2$, which holds for $1 le n le 5$, but at $n = 6$, the LHS becomes $0$ instead. If I get a chance, I will investigate your equation to see if I can figure out why it's true for at least the first $18$ values and, more importantly, will it always stay true. Regardless, though, it's an excellent observation you've made, even if it doesn't always hold.
$endgroup$
– John Omielan
52 mins ago
$begingroup$
I checked your result up to $n=533$ (for $n geq 534$, I have problems. Would you be interested by a huge table of $I(k)$ (I was able to generate it up to $k=540$). This is a very interesting problem.
$endgroup$
– Claude Leibovici
34 mins ago
$begingroup$
I checked your result up to $n=533$ (for $n geq 534$, I have problems. Would you be interested by a huge table of $I(k)$ (I was able to generate it up to $k=540$). This is a very interesting problem.
$endgroup$
– Claude Leibovici
34 mins ago
$begingroup$
@ClaudeLeibovici: Thanks for checking, Claude! However, that table would be too huge for MSE. :)
$endgroup$
– Tito Piezas III
31 mins ago
$begingroup$
@ClaudeLeibovici: Thanks for checking, Claude! However, that table would be too huge for MSE. :)
$endgroup$
– Tito Piezas III
31 mins ago
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The answer is yes. Sketch of solution:
$$
I(k) = int_0^k sum_{ple x} sum_{qle k-x} 1 ,dx = sum_p sum_{qle k-p} int_p^{k-q} dx = sum_p sum_{qle k-p} (k-(p+q)) = sum_{mle k} r(m)(k-m),
$$
where $r(m)$ is the number of ways of writing $m$ as the sum of two primes. Then
$$
I(6n+6)-2I(6n+5)+I(6n+4) = sum_{mle 6n+4} r(m)big( (6n+6-m)-2(6n+5-m)+(6m+4-m) big) + r(6n+5) = \0 + r(6n+5);
$$
and $r(6n+5)=0$ for every $nge1$, since the only way the odd integer $6n+5$ can be the sum of two primes is $6n+5=2+(6n+3)$, but $6n+3=3(2n+1)$ is always composite when $nge1$.
The same argument gives $I(6n+2)-2I(6n+1)+I(6n) = r(6n+1)$, which is $2$ if $6n-1$ is prime and $0$ otherwise; this is why (as observed by John Omielan) it equals $2$ for $1le nle 5$ but $0$ for $n=6$.
edited 20 mins ago
Tito Piezas III
27.4k366174
answered 33 mins ago
Greg MartinGreg Martin
35.5k23364
$endgroup$
1
$begingroup$
MSE never ceases to amaze me how fast some people can figure out the answer.
$endgroup$
– Tito Piezas III
19 mins ago
$begingroup$
Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
$endgroup$
– Tito Piezas III
9 mins ago
$begingroup$
This really surprises me since I thought the equation will be eventually false...
$endgroup$
– Seewoo Lee
4 mins ago
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%2f3128367%2fa-curious-equality-of-integrals-involving-the-prime-counting-function%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
$begingroup$
The answer is yes. Sketch of solution:
$$
I(k) = int_0^k sum_{ple x} sum_{qle k-x} 1 ,dx = sum_p sum_{qle k-p} int_p^{k-q} dx = sum_p sum_{qle k-p} (k-(p+q)) = sum_{mle k} r(m)(k-m),
$$
where $r(m)$ is the number of ways of writing $m$ as the sum of two primes. Then
$$
I(6n+6)-2I(6n+5)+I(6n+4) = sum_{mle 6n+4} r(m)big( (6n+6-m)-2(6n+5-m)+(6m+4-m) big) + r(6n+5) = \0 + r(6n+5);
$$
and $r(6n+5)=0$ for every $nge1$, since the only way the odd integer $6n+5$ can be the sum of two primes is $6n+5=2+(6n+3)$, but $6n+3=3(2n+1)$ is always composite when $nge1$.
The same argument gives $I(6n+2)-2I(6n+1)+I(6n) = r(6n+1)$, which is $2$ if $6n-1$ is prime and $0$ otherwise; this is why (as observed by John Omielan) it equals $2$ for $1le nle 5$ but $0$ for $n=6$.
edited 20 mins ago
Tito Piezas III
27.4k366174
answered 33 mins ago
Greg MartinGreg Martin
35.5k23364
$endgroup$
1
$begingroup$
MSE never ceases to amaze me how fast some people can figure out the answer.
$endgroup$
– Tito Piezas III
19 mins ago
$begingroup$
Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
$endgroup$
– Tito Piezas III
9 mins ago
$begingroup$
This really surprises me since I thought the equation will be eventually false...
$endgroup$
– Seewoo Lee
4 mins ago
add a comment |
$begingroup$
The answer is yes. Sketch of solution:
$$
I(k) = int_0^k sum_{ple x} sum_{qle k-x} 1 ,dx = sum_p sum_{qle k-p} int_p^{k-q} dx = sum_p sum_{qle k-p} (k-(p+q)) = sum_{mle k} r(m)(k-m),
$$
where $r(m)$ is the number of ways of writing $m$ as the sum of two primes. Then
$$
I(6n+6)-2I(6n+5)+I(6n+4) = sum_{mle 6n+4} r(m)big( (6n+6-m)-2(6n+5-m)+(6m+4-m) big) + r(6n+5) = \0 + r(6n+5);
$$
and $r(6n+5)=0$ for every $nge1$, since the only way the odd integer $6n+5$ can be the sum of two primes is $6n+5=2+(6n+3)$, but $6n+3=3(2n+1)$ is always composite when $nge1$.
The same argument gives $I(6n+2)-2I(6n+1)+I(6n) = r(6n+1)$, which is $2$ if $6n-1$ is prime and $0$ otherwise; this is why (as observed by John Omielan) it equals $2$ for $1le nle 5$ but $0$ for $n=6$.
edited 20 mins ago
Tito Piezas III
27.4k366174
answered 33 mins ago
Greg MartinGreg Martin
35.5k23364
$endgroup$
1
$begingroup$
MSE never ceases to amaze me how fast some people can figure out the answer.
$endgroup$
– Tito Piezas III
19 mins ago
$begingroup$
Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
$endgroup$
– Tito Piezas III
9 mins ago
$begingroup$
This really surprises me since I thought the equation will be eventually false...
$endgroup$
– Seewoo Lee
4 mins ago
add a comment |
$begingroup$
The answer is yes. Sketch of solution:
$$
I(k) = int_0^k sum_{ple x} sum_{qle k-x} 1 ,dx = sum_p sum_{qle k-p} int_p^{k-q} dx = sum_p sum_{qle k-p} (k-(p+q)) = sum_{mle k} r(m)(k-m),
$$
where $r(m)$ is the number of ways of writing $m$ as the sum of two primes. Then
$$
I(6n+6)-2I(6n+5)+I(6n+4) = sum_{mle 6n+4} r(m)big( (6n+6-m)-2(6n+5-m)+(6m+4-m) big) + r(6n+5) = \0 + r(6n+5);
$$
and $r(6n+5)=0$ for every $nge1$, since the only way the odd integer $6n+5$ can be the sum of two primes is $6n+5=2+(6n+3)$, but $6n+3=3(2n+1)$ is always composite when $nge1$.
The same argument gives $I(6n+2)-2I(6n+1)+I(6n) = r(6n+1)$, which is $2$ if $6n-1$ is prime and $0$ otherwise; this is why (as observed by John Omielan) it equals $2$ for $1le nle 5$ but $0$ for $n=6$.
edited 20 mins ago
Tito Piezas III
27.4k366174
answered 33 mins ago
Greg MartinGreg Martin
35.5k23364
$endgroup$
The answer is yes. Sketch of solution:
$$
I(k) = int_0^k sum_{ple x} sum_{qle k-x} 1 ,dx = sum_p sum_{qle k-p} int_p^{k-q} dx = sum_p sum_{qle k-p} (k-(p+q)) = sum_{mle k} r(m)(k-m),
$$
where $r(m)$ is the number of ways of writing $m$ as the sum of two primes. Then
$$
I(6n+6)-2I(6n+5)+I(6n+4) = sum_{mle 6n+4} r(m)big( (6n+6-m)-2(6n+5-m)+(6m+4-m) big) + r(6n+5) = \0 + r(6n+5);
$$
and $r(6n+5)=0$ for every $nge1$, since the only way the odd integer $6n+5$ can be the sum of two primes is $6n+5=2+(6n+3)$, but $6n+3=3(2n+1)$ is always composite when $nge1$.
The same argument gives $I(6n+2)-2I(6n+1)+I(6n) = r(6n+1)$, which is $2$ if $6n-1$ is prime and $0$ otherwise; this is why (as observed by John Omielan) it equals $2$ for $1le nle 5$ but $0$ for $n=6$.
edited 20 mins ago
Tito Piezas III
27.4k366174
answered 33 mins ago
Greg MartinGreg Martin
35.5k23364
edited 20 mins ago
Tito Piezas III
27.4k366174
edited 20 mins ago
Tito Piezas III
27.4k366174
edited 20 mins ago
Tito Piezas III
27.4k366174
27.4k366174
answered 33 mins ago
Greg MartinGreg Martin
35.5k23364
answered 33 mins ago
Greg MartinGreg Martin
35.5k23364
answered 33 mins ago
Greg MartinGreg Martin
35.5k23364
35.5k23364
1
$begingroup$
MSE never ceases to amaze me how fast some people can figure out the answer.
$endgroup$
– Tito Piezas III
19 mins ago
$begingroup$
Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
$endgroup$
– Tito Piezas III
9 mins ago
$begingroup$
This really surprises me since I thought the equation will be eventually false...
$endgroup$
– Seewoo Lee
4 mins ago
add a comment |
1
$begingroup$
MSE never ceases to amaze me how fast some people can figure out the answer.
$endgroup$
– Tito Piezas III
19 mins ago
$begingroup$
Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
$endgroup$
– Tito Piezas III
9 mins ago
$begingroup$
This really surprises me since I thought the equation will be eventually false...
$endgroup$
– Seewoo Lee
4 mins ago
1
1
$begingroup$
MSE never ceases to amaze me how fast some people can figure out the answer.
$endgroup$
– Tito Piezas III
19 mins ago
$begingroup$
MSE never ceases to amaze me how fast some people can figure out the answer.
$endgroup$
– Tito Piezas III
19 mins ago
$begingroup$
Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
$endgroup$
– Tito Piezas III
9 mins ago
$begingroup$
Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
$endgroup$
– Tito Piezas III
9 mins ago
$begingroup$
This really surprises me since I thought the equation will be eventually false...
$endgroup$
– Seewoo Lee
4 mins ago
$begingroup$
This really surprises me since I thought the equation will be eventually false...
$endgroup$
– Seewoo Lee
4 mins ago
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%2f3128367%2fa-curious-equality-of-integrals-involving-the-prime-counting-function%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
$begingroup$
Note your proposed equation doesn't hold for $n = 0$ as $I(4) = 0$, $I(5) = 1$ and $I(6) = 4$.
$endgroup$
– John Omielan
1 hour ago
$begingroup$
@JohnOmielan: A typo. I meant all $n>0$. I will correct it.
$endgroup$
– Tito Piezas III
1 hour ago
$begingroup$
I have checked to confirm what you're asking is true for $n$ up to $18$. However, I have my doubts it'll always work, partially because it doesn't work for $n = 0$. Also, a similar type condition is that $I(6n) - 2I(6n + 1) + I(6n + 2) = 2$, which holds for $1 le n le 5$, but at $n = 6$, the LHS becomes $0$ instead. If I get a chance, I will investigate your equation to see if I can figure out why it's true for at least the first $18$ values and, more importantly, will it always stay true. Regardless, though, it's an excellent observation you've made, even if it doesn't always hold.
$endgroup$
– John Omielan
52 mins ago
$begingroup$
I checked your result up to $n=533$ (for $n geq 534$, I have problems. Would you be interested by a huge table of $I(k)$ (I was able to generate it up to $k=540$). This is a very interesting problem.
$endgroup$
– Claude Leibovici
34 mins ago
$begingroup$
@ClaudeLeibovici: Thanks for checking, Claude! However, that table would be too huge for MSE. :)
$endgroup$
– Tito Piezas III
31 mins ago