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A curious equality of integrals involving the prime counting function?
How many primes does this sequence find?Tight bounds on the prime counting functionDirichlet prime counting function?closed form for integrals involving error functiona practical prime counting functionPrime counting functionRestricted equality involving prime numbersPrime counting function formulasProof for a prime number formula involving the prime counting functionProperty of Prime Counting FunctionApproximating 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
49 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
30 mins ago
$begingroup$
@ClaudeLeibovici: Thanks for checking, Claude! However, that table would be too huge for MSE. :)
$endgroup$
– Tito Piezas III
27 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 14 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
49 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
30 mins ago
$begingroup$
@ClaudeLeibovici: Thanks for checking, Claude! However, that table would be too huge for MSE. :)
$endgroup$
– Tito Piezas III
27 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
49 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
30 mins ago
$begingroup$
@ClaudeLeibovici: Thanks for checking, Claude! However, that table would be too huge for MSE. :)
$endgroup$
– Tito Piezas III
27 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
49 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
49 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
30 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
30 mins ago
$begingroup$
@ClaudeLeibovici: Thanks for checking, Claude! However, that table would be too huge for MSE. :)
$endgroup$
– Tito Piezas III
27 mins ago
$begingroup$
@ClaudeLeibovici: Thanks for checking, Claude! However, that table would be too huge for MSE. :)
$endgroup$
– Tito Piezas III
27 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 16 mins ago
Tito Piezas III
27.4k366174
answered 29 mins ago
Greg MartinGreg Martin
35.5k23263
$endgroup$
1
$begingroup$
MSE never ceases to amaze me how fast some people can figure out the answer.
$endgroup$
– Tito Piezas III
15 mins ago
$begingroup$
Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
$endgroup$
– Tito Piezas III
6 mins ago
$begingroup$
This really surprises me since I thought the equation will be eventually false...
$endgroup$
– Seewoo Lee
33 secs ago
add a comment |
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1 Answer
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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 16 mins ago
Tito Piezas III
27.4k366174
answered 29 mins ago
Greg MartinGreg Martin
35.5k23263
$endgroup$
1
$begingroup$
MSE never ceases to amaze me how fast some people can figure out the answer.
$endgroup$
– Tito Piezas III
15 mins ago
$begingroup$
Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
$endgroup$
– Tito Piezas III
6 mins ago
$begingroup$
This really surprises me since I thought the equation will be eventually false...
$endgroup$
– Seewoo Lee
33 secs 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 16 mins ago
Tito Piezas III
27.4k366174
answered 29 mins ago
Greg MartinGreg Martin
35.5k23263
$endgroup$
1
$begingroup$
MSE never ceases to amaze me how fast some people can figure out the answer.
$endgroup$
– Tito Piezas III
15 mins ago
$begingroup$
Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
$endgroup$
– Tito Piezas III
6 mins ago
$begingroup$
This really surprises me since I thought the equation will be eventually false...
$endgroup$
– Seewoo Lee
33 secs 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 16 mins ago
Tito Piezas III
27.4k366174
answered 29 mins ago
Greg MartinGreg Martin
35.5k23263
$endgroup$
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 16 mins ago
Tito Piezas III
27.4k366174
answered 29 mins ago
Greg MartinGreg Martin
35.5k23263
edited 16 mins ago
Tito Piezas III
27.4k366174
edited 16 mins ago
Tito Piezas III
27.4k366174
edited 16 mins ago
Tito Piezas III
27.4k366174
27.4k366174
answered 29 mins ago
Greg MartinGreg Martin
35.5k23263
answered 29 mins ago
Greg MartinGreg Martin
35.5k23263
answered 29 mins ago
Greg MartinGreg Martin
35.5k23263
35.5k23263
1
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MSE never ceases to amaze me how fast some people can figure out the answer.
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– Tito Piezas III
15 mins 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
6 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
33 secs 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
15 mins ago
$begingroup$
Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
$endgroup$
– Tito Piezas III
6 mins ago
$begingroup$
This really surprises me since I thought the equation will be eventually false...
$endgroup$
– Seewoo Lee
33 secs ago
1
1
$begingroup$
MSE never ceases to amaze me how fast some people can figure out the answer.
$endgroup$
– Tito Piezas III
15 mins ago
$begingroup$
MSE never ceases to amaze me how fast some people can figure out the answer.
$endgroup$
– Tito Piezas III
15 mins ago
$begingroup$
Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
$endgroup$
– Tito Piezas III
6 mins ago
$begingroup$
Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
$endgroup$
– Tito Piezas III
6 mins ago
$begingroup$
This really surprises me since I thought the equation will be eventually false...
$endgroup$
– Seewoo Lee
33 secs ago
$begingroup$
This really surprises me since I thought the equation will be eventually false...
$endgroup$
– Seewoo Lee
33 secs ago
add a comment |
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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
1 hour ago
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@JohnOmielan: A typo. I meant all $n>0$. I will correct it.
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– Tito Piezas III
1 hour 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
49 mins ago
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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
30 mins 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
27 mins ago