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 Feb 27 at 3:52
Tito Piezas IIITito Piezas III
27.7k367176
$endgroup$
|
show 12 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 Feb 27 at 3:52
Tito Piezas IIITito Piezas III
27.7k367176
$endgroup$
$begingroup$
Note your proposed equation doesn't hold for $n = 0$ as $I(4) = 0$, $I(5) = 1$ and $I(6) = 4$.
$endgroup$
– John Omielan
Feb 27 at 4:04
$begingroup$
@JohnOmielan: A typo. I meant all $n>0$. I will correct it.
$endgroup$
– Tito Piezas III
Feb 27 at 4:09
1
$begingroup$
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
Feb 27 at 4:47
5
$begingroup$
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
Feb 27 at 5:06
1
$begingroup$
Up to $k=540$, $I(k)$ is a prime for this list $${13,57,119,167,171,173,175,341,395,397,427,431,473,515,519}$$
$endgroup$
– Claude Leibovici
Feb 27 at 8:13
|
show 12 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 Feb 27 at 3:52
Tito Piezas IIITito Piezas III
27.7k367176
$endgroup$
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 Feb 27 at 3:52
Tito Piezas IIITito Piezas III
27.7k367176
asked Feb 27 at 3:52
Tito Piezas IIITito Piezas III
27.7k367176
edited Feb 27 at 5:22
Tito Piezas III
asked Feb 27 at 3:52
Tito Piezas IIITito Piezas III
27.7k367176
asked Feb 27 at 3:52
Tito Piezas IIITito Piezas III
27.7k367176
asked Feb 27 at 3:52
Tito Piezas IIITito Piezas III
27.7k367176
27.7k367176
$begingroup$
Note your proposed equation doesn't hold for $n = 0$ as $I(4) = 0$, $I(5) = 1$ and $I(6) = 4$.
$endgroup$
– John Omielan
Feb 27 at 4:04
$begingroup$
@JohnOmielan: A typo. I meant all $n>0$. I will correct it.
$endgroup$
– Tito Piezas III
Feb 27 at 4:09
1
$begingroup$
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
Feb 27 at 4:47
5
$begingroup$
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
Feb 27 at 5:06
1
$begingroup$
Up to $k=540$, $I(k)$ is a prime for this list $${13,57,119,167,171,173,175,341,395,397,427,431,473,515,519}$$
$endgroup$
– Claude Leibovici
Feb 27 at 8:13
|
show 12 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
Feb 27 at 4:04
$begingroup$
@JohnOmielan: A typo. I meant all $n>0$. I will correct it.
$endgroup$
– Tito Piezas III
Feb 27 at 4:09
1
$begingroup$
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
Feb 27 at 4:47
5
$begingroup$
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
Feb 27 at 5:06
1
$begingroup$
Up to $k=540$, $I(k)$ is a prime for this list $${13,57,119,167,171,173,175,341,395,397,427,431,473,515,519}$$
$endgroup$
– Claude Leibovici
Feb 27 at 8:13
$begingroup$
Note your proposed equation doesn't hold for $n = 0$ as $I(4) = 0$, $I(5) = 1$ and $I(6) = 4$.
$endgroup$
– John Omielan
Feb 27 at 4:04
$begingroup$
Note your proposed equation doesn't hold for $n = 0$ as $I(4) = 0$, $I(5) = 1$ and $I(6) = 4$.
$endgroup$
– John Omielan
Feb 27 at 4:04
$begingroup$
@JohnOmielan: A typo. I meant all $n>0$. I will correct it.
$endgroup$
– Tito Piezas III
Feb 27 at 4:09
$begingroup$
@JohnOmielan: A typo. I meant all $n>0$. I will correct it.
$endgroup$
– Tito Piezas III
Feb 27 at 4:09
1
1
$begingroup$
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
Feb 27 at 4:47
$begingroup$
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
Feb 27 at 4:47
5
5
$begingroup$
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
Feb 27 at 5:06
$begingroup$
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
Feb 27 at 5:06
1
1
$begingroup$
Up to $k=540$, $I(k)$ is a prime for this list $${13,57,119,167,171,173,175,341,395,397,427,431,473,515,519}$$
$endgroup$
– Claude Leibovici
Feb 27 at 8:13
$begingroup$
Up to $k=540$, $I(k)$ is a prime for this list $${13,57,119,167,171,173,175,341,395,397,427,431,473,515,519}$$
$endgroup$
– Claude Leibovici
Feb 27 at 8:13
|
show 12 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
begin{align}
I(6n+6) &{}-2I(6n+5)+I(6n+4) \
&= sum_{m le 6n+4} r(m)big((6n+6-m)-2(6n+5-m) +(6m+4-m)big) + r(6n+5) \&= 0 + r(6n+5);
end{align}
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 $n=6$.
answered Feb 27 at 5:07
Greg MartinGreg Martin
36k23565
$endgroup$
14
$begingroup$
MSE never ceases to amaze me how fast some people can figure out the answer.
$endgroup$
– Tito Piezas III
Feb 27 at 5:21
2
$begingroup$
Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
$endgroup$
– Tito Piezas III
Feb 27 at 5:31
3
$begingroup$
This really surprises me since I thought the equation will be eventually false...
$endgroup$
– Seewoo Lee
Feb 27 at 5:36
$begingroup$
@stressedout it's arguably worse due to the HNQ list
$endgroup$
– qwr
Feb 28 at 17:39
add a comment |
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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
begin{align}
I(6n+6) &{}-2I(6n+5)+I(6n+4) \
&= sum_{m le 6n+4} r(m)big((6n+6-m)-2(6n+5-m) +(6m+4-m)big) + r(6n+5) \&= 0 + r(6n+5);
end{align}
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 $n=6$.
answered Feb 27 at 5:07
Greg MartinGreg Martin
36k23565
$endgroup$
14
$begingroup$
MSE never ceases to amaze me how fast some people can figure out the answer.
$endgroup$
– Tito Piezas III
Feb 27 at 5:21
2
$begingroup$
Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
$endgroup$
– Tito Piezas III
Feb 27 at 5:31
3
$begingroup$
This really surprises me since I thought the equation will be eventually false...
$endgroup$
– Seewoo Lee
Feb 27 at 5:36
$begingroup$
@stressedout it's arguably worse due to the HNQ list
$endgroup$
– qwr
Feb 28 at 17:39
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
begin{align}
I(6n+6) &{}-2I(6n+5)+I(6n+4) \
&= sum_{m le 6n+4} r(m)big((6n+6-m)-2(6n+5-m) +(6m+4-m)big) + r(6n+5) \&= 0 + r(6n+5);
end{align}
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 $n=6$.
answered Feb 27 at 5:07
Greg MartinGreg Martin
36k23565
$endgroup$
14
$begingroup$
MSE never ceases to amaze me how fast some people can figure out the answer.
$endgroup$
– Tito Piezas III
Feb 27 at 5:21
2
$begingroup$
Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
$endgroup$
– Tito Piezas III
Feb 27 at 5:31
3
$begingroup$
This really surprises me since I thought the equation will be eventually false...
$endgroup$
– Seewoo Lee
Feb 27 at 5:36
$begingroup$
@stressedout it's arguably worse due to the HNQ list
$endgroup$
– qwr
Feb 28 at 17:39
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
begin{align}
I(6n+6) &{}-2I(6n+5)+I(6n+4) \
&= sum_{m le 6n+4} r(m)big((6n+6-m)-2(6n+5-m) +(6m+4-m)big) + r(6n+5) \&= 0 + r(6n+5);
end{align}
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 $n=6$.
answered Feb 27 at 5:07
Greg MartinGreg Martin
36k23565
$endgroup$
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
begin{align}
I(6n+6) &{}-2I(6n+5)+I(6n+4) \
&= sum_{m le 6n+4} r(m)big((6n+6-m)-2(6n+5-m) +(6m+4-m)big) + r(6n+5) \&= 0 + r(6n+5);
end{align}
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 $n=6$.
answered Feb 27 at 5:07
Greg MartinGreg Martin
36k23565
edited Feb 27 at 20:23
answered Feb 27 at 5:07
Greg MartinGreg Martin
36k23565
answered Feb 27 at 5:07
Greg MartinGreg Martin
36k23565
answered Feb 27 at 5:07
Greg MartinGreg Martin
36k23565
36k23565
14
$begingroup$
MSE never ceases to amaze me how fast some people can figure out the answer.
$endgroup$
– Tito Piezas III
Feb 27 at 5:21
2
$begingroup$
Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
$endgroup$
– Tito Piezas III
Feb 27 at 5:31
3
$begingroup$
This really surprises me since I thought the equation will be eventually false...
$endgroup$
– Seewoo Lee
Feb 27 at 5:36
$begingroup$
@stressedout it's arguably worse due to the HNQ list
$endgroup$
– qwr
Feb 28 at 17:39
add a comment |
14
$begingroup$
MSE never ceases to amaze me how fast some people can figure out the answer.
$endgroup$
– Tito Piezas III
Feb 27 at 5:21
2
$begingroup$
Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
$endgroup$
– Tito Piezas III
Feb 27 at 5:31
3
$begingroup$
This really surprises me since I thought the equation will be eventually false...
$endgroup$
– Seewoo Lee
Feb 27 at 5:36
$begingroup$
@stressedout it's arguably worse due to the HNQ list
$endgroup$
– qwr
Feb 28 at 17:39
14
14
$begingroup$
MSE never ceases to amaze me how fast some people can figure out the answer.
$endgroup$
– Tito Piezas III
Feb 27 at 5:21
$begingroup$
MSE never ceases to amaze me how fast some people can figure out the answer.
$endgroup$
– Tito Piezas III
Feb 27 at 5:21
2
2
$begingroup$
Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
$endgroup$
– Tito Piezas III
Feb 27 at 5:31
$begingroup$
Greg, do you know how to address Ultradark's question regarding when $I(k)$ is prime?
$endgroup$
– Tito Piezas III
Feb 27 at 5:31
3
3
$begingroup$
This really surprises me since I thought the equation will be eventually false...
$endgroup$
– Seewoo Lee
Feb 27 at 5:36
$begingroup$
This really surprises me since I thought the equation will be eventually false...
$endgroup$
– Seewoo Lee
Feb 27 at 5:36
$begingroup$
@stressedout it's arguably worse due to the HNQ list
$endgroup$
– qwr
Feb 28 at 17:39
$begingroup$
@stressedout it's arguably worse due to the HNQ list
$endgroup$
– qwr
Feb 28 at 17:39
add a comment |
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$begingroup$
Note your proposed equation doesn't hold for $n = 0$ as $I(4) = 0$, $I(5) = 1$ and $I(6) = 4$.
$endgroup$
– John Omielan
Feb 27 at 4:04
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@JohnOmielan: A typo. I meant all $n>0$. I will correct it.
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– Tito Piezas III
Feb 27 at 4:09
1
$begingroup$
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
Feb 27 at 4:47
5
$begingroup$
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
Feb 27 at 5:06
1
$begingroup$
Up to $k=540$, $I(k)$ is a prime for this list $${13,57,119,167,171,173,175,341,395,397,427,431,473,515,519}$$
$endgroup$
– Claude Leibovici
Feb 27 at 8:13