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 2 hours ago
Tito Piezas IIITito Piezas III
27.4k366174
$endgroup$
|
show 7 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 2 hours 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
2 hours ago
$begingroup$
@JohnOmielan: A typo. I meant all $n>0$. I will correct it.
$endgroup$
– Tito Piezas III
2 hours 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
1 hour ago
1
$begingroup$
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
1 hour ago
$begingroup$
@ClaudeLeibovici: Thanks for checking, Claude! However, that table would be too huge for MSE. :)
$endgroup$
– Tito Piezas III
1 hour ago
|
show 7 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 2 hours 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 2 hours ago
Tito Piezas IIITito Piezas III
27.4k366174
asked 2 hours ago
Tito Piezas IIITito Piezas III
27.4k366174
edited 1 hour ago
Tito Piezas III
asked 2 hours ago
Tito Piezas IIITito Piezas III
27.4k366174
asked 2 hours ago
Tito Piezas IIITito Piezas III
27.4k366174
asked 2 hours 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
2 hours ago
$begingroup$
@JohnOmielan: A typo. I meant all $n>0$. I will correct it.
$endgroup$
– Tito Piezas III
2 hours 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
1 hour ago
1
$begingroup$
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
1 hour ago
$begingroup$
@ClaudeLeibovici: Thanks for checking, Claude! However, that table would be too huge for MSE. :)
$endgroup$
– Tito Piezas III
1 hour ago
|
show 7 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
2 hours ago
$begingroup$
@JohnOmielan: A typo. I meant all $n>0$. I will correct it.
$endgroup$
– Tito Piezas III
2 hours 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
1 hour ago
1
$begingroup$
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
1 hour ago
$begingroup$
@ClaudeLeibovici: Thanks for checking, Claude! However, that table would be too huge for MSE. :)
$endgroup$
– Tito Piezas III
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
2 hours 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
2 hours ago
$begingroup$
@JohnOmielan: A typo. I meant all $n>0$. I will correct it.
$endgroup$
– Tito Piezas III
2 hours ago
$begingroup$
@JohnOmielan: A typo. I meant all $n>0$. I will correct it.
$endgroup$
– Tito Piezas III
2 hours 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
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
1 hour ago
1
1
$begingroup$
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
1 hour 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
1 hour ago
$begingroup$
@ClaudeLeibovici: Thanks for checking, Claude! However, that table would be too huge for MSE. :)
$endgroup$
– Tito Piezas III
1 hour ago
$begingroup$
@ClaudeLeibovici: Thanks for checking, Claude! However, that table would be too huge for MSE. :)
$endgroup$
– Tito Piezas III
1 hour ago
|
show 7 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 1 hour ago
Tito Piezas III
27.4k366174
answered 1 hour 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
1 hour ago
$begingroup$
Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
$endgroup$
– Tito Piezas III
1 hour ago
$begingroup$
This really surprises me since I thought the equation will be eventually false...
$endgroup$
– Seewoo Lee
58 mins ago
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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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 1 hour ago
Tito Piezas III
27.4k366174
answered 1 hour 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
1 hour ago
$begingroup$
Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
$endgroup$
– Tito Piezas III
1 hour ago
$begingroup$
This really surprises me since I thought the equation will be eventually false...
$endgroup$
– Seewoo Lee
58 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 1 hour ago
Tito Piezas III
27.4k366174
answered 1 hour 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
1 hour ago
$begingroup$
Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
$endgroup$
– Tito Piezas III
1 hour ago
$begingroup$
This really surprises me since I thought the equation will be eventually false...
$endgroup$
– Seewoo Lee
58 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 1 hour ago
Tito Piezas III
27.4k366174
answered 1 hour 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 1 hour ago
Tito Piezas III
27.4k366174
answered 1 hour ago
Greg MartinGreg Martin
35.5k23364
edited 1 hour ago
Tito Piezas III
27.4k366174
edited 1 hour ago
Tito Piezas III
27.4k366174
edited 1 hour ago
Tito Piezas III
27.4k366174
27.4k366174
answered 1 hour ago
Greg MartinGreg Martin
35.5k23364
answered 1 hour ago
Greg MartinGreg Martin
35.5k23364
answered 1 hour 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
1 hour ago
$begingroup$
Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
$endgroup$
– Tito Piezas III
1 hour ago
$begingroup$
This really surprises me since I thought the equation will be eventually false...
$endgroup$
– Seewoo Lee
58 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
1 hour ago
$begingroup$
Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
$endgroup$
– Tito Piezas III
1 hour ago
$begingroup$
This really surprises me since I thought the equation will be eventually false...
$endgroup$
– Seewoo Lee
58 mins ago
1
1
$begingroup$
MSE never ceases to amaze me how fast some people can figure out the answer.
$endgroup$
– Tito Piezas III
1 hour ago
$begingroup$
MSE never ceases to amaze me how fast some people can figure out the answer.
$endgroup$
– Tito Piezas III
1 hour ago
$begingroup$
Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
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– Tito Piezas III
1 hour ago
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Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
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– Tito Piezas III
1 hour ago
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This really surprises me since I thought the equation will be eventually false...
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– Seewoo Lee
58 mins ago
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This really surprises me since I thought the equation will be eventually false...
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– Seewoo Lee
58 mins ago
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Note your proposed equation doesn't hold for $n = 0$ as $I(4) = 0$, $I(5) = 1$ and $I(6) = 4$.
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– John Omielan
2 hours ago
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@JohnOmielan: A typo. I meant all $n>0$. I will correct it.
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– Tito Piezas III
2 hours ago
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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.
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– John Omielan
1 hour ago
1
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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.
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– Claude Leibovici
1 hour ago
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@ClaudeLeibovici: Thanks for checking, Claude! However, that table would be too huge for MSE. :)
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– Tito Piezas III
1 hour ago