Arithmetic Progression of Primes of length 7
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Determine the least possible value of the largest term in an
arithmetic progression of seven distinct primes.
I don't have a great understanding of how to start the problem. I have seen approaches of searching for sequences with a difference between terms of $(2cdot3cdot5cdot7)$ and then seeing what the largest number is.
How would you go about starting this question? Is it possible to bound the largest term?
Edit - commentators mention that we can search for multiples that differ by 30 instead of 210. Why is this the case?
algebra-precalculus elementary-number-theory prime-numbers contest-math
edited Nov 21 at 12:45
Klangen
1,36511131
asked May 4 at 16:51
Abe
540115
|
show 2 more comments
up vote
2
down vote
favorite
Determine the least possible value of the largest term in an
arithmetic progression of seven distinct primes.
I don't have a great understanding of how to start the problem. I have seen approaches of searching for sequences with a difference between terms of $(2cdot3cdot5cdot7)$ and then seeing what the largest number is.
How would you go about starting this question? Is it possible to bound the largest term?
Edit - commentators mention that we can search for multiples that differ by 30 instead of 210. Why is this the case?
algebra-precalculus elementary-number-theory prime-numbers contest-math
edited Nov 21 at 12:45
Klangen
1,36511131
asked May 4 at 16:51
Abe
540115
6
Have you seen this already?
– B. Mehta
May 4 at 16:58
2
The primes in question would be a sequence like $p,p+k,p+2k,dots,p+6k$. By the pigeonhole principle, one of these must be divisible by $7$, or else $k$ must be divisible by $7$ (same with $2,3,5$). If we start with $p=7,k=30$ we almost make it, except for $187=11cdot 17$. This means that we have to use $k=210=2cdot3cdot5cdot7$...
– abiessu
May 4 at 17:04
@abiessu, We can try multiples of $30$ below $210$ as well, and there is in fact a lower possible value than the one you suggest.
– B. Mehta
May 4 at 17:18
2
You could start with $-113$ or lower and go up in steps of $30$… somewhere you need a positivity criterion.
– Macavity
May 5 at 4:38
1
By the way, here is an example I found: $58n^2+1$. It is not a prime arithmetic progression, really, because it includes an exponent greater than $1$, but it does have length $7$ from $n=1$. How I found it, just trial and error. Another more complex one with length $7$ is $$811+10sum_{k=0}^n(5k+1)$$ from $n=1$. Believe it or not, they are not too difficult to find. Go here to factorize integers and see whether or not they are prime $longrightarrow$ alpertron.com.ar/ECM.HTM
– user477343
May 11 at 2:08
|
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Determine the least possible value of the largest term in an
arithmetic progression of seven distinct primes.
I don't have a great understanding of how to start the problem. I have seen approaches of searching for sequences with a difference between terms of $(2cdot3cdot5cdot7)$ and then seeing what the largest number is.
How would you go about starting this question? Is it possible to bound the largest term?
Edit - commentators mention that we can search for multiples that differ by 30 instead of 210. Why is this the case?
algebra-precalculus elementary-number-theory prime-numbers contest-math
edited Nov 21 at 12:45
Klangen
1,36511131
asked May 4 at 16:51
Abe
540115
Determine the least possible value of the largest term in an
arithmetic progression of seven distinct primes.
I don't have a great understanding of how to start the problem. I have seen approaches of searching for sequences with a difference between terms of $(2cdot3cdot5cdot7)$ and then seeing what the largest number is.
How would you go about starting this question? Is it possible to bound the largest term?
Edit - commentators mention that we can search for multiples that differ by 30 instead of 210. Why is this the case?
algebra-precalculus elementary-number-theory prime-numbers contest-math
algebra-precalculus elementary-number-theory prime-numbers contest-math
edited Nov 21 at 12:45
Klangen
1,36511131
asked May 4 at 16:51
Abe
540115
edited Nov 21 at 12:45
Klangen
1,36511131
asked May 4 at 16:51
Abe
540115
edited Nov 21 at 12:45
Klangen
1,36511131
edited Nov 21 at 12:45
Klangen
1,36511131
edited Nov 21 at 12:45
Klangen
1,36511131
1,36511131
asked May 4 at 16:51
Abe
540115
asked May 4 at 16:51
Abe
540115
asked May 4 at 16:51
Abe
540115
540115
6
Have you seen this already?
– B. Mehta
May 4 at 16:58
2
The primes in question would be a sequence like $p,p+k,p+2k,dots,p+6k$. By the pigeonhole principle, one of these must be divisible by $7$, or else $k$ must be divisible by $7$ (same with $2,3,5$). If we start with $p=7,k=30$ we almost make it, except for $187=11cdot 17$. This means that we have to use $k=210=2cdot3cdot5cdot7$...
– abiessu
May 4 at 17:04
@abiessu, We can try multiples of $30$ below $210$ as well, and there is in fact a lower possible value than the one you suggest.
– B. Mehta
May 4 at 17:18
2
You could start with $-113$ or lower and go up in steps of $30$… somewhere you need a positivity criterion.
– Macavity
May 5 at 4:38
1
By the way, here is an example I found: $58n^2+1$. It is not a prime arithmetic progression, really, because it includes an exponent greater than $1$, but it does have length $7$ from $n=1$. How I found it, just trial and error. Another more complex one with length $7$ is $$811+10sum_{k=0}^n(5k+1)$$ from $n=1$. Believe it or not, they are not too difficult to find. Go here to factorize integers and see whether or not they are prime $longrightarrow$ alpertron.com.ar/ECM.HTM
– user477343
May 11 at 2:08
|
show 2 more comments
6
Have you seen this already?
– B. Mehta
May 4 at 16:58
2
The primes in question would be a sequence like $p,p+k,p+2k,dots,p+6k$. By the pigeonhole principle, one of these must be divisible by $7$, or else $k$ must be divisible by $7$ (same with $2,3,5$). If we start with $p=7,k=30$ we almost make it, except for $187=11cdot 17$. This means that we have to use $k=210=2cdot3cdot5cdot7$...
– abiessu
May 4 at 17:04
@abiessu, We can try multiples of $30$ below $210$ as well, and there is in fact a lower possible value than the one you suggest.
– B. Mehta
May 4 at 17:18
2
You could start with $-113$ or lower and go up in steps of $30$… somewhere you need a positivity criterion.
– Macavity
May 5 at 4:38
1
By the way, here is an example I found: $58n^2+1$. It is not a prime arithmetic progression, really, because it includes an exponent greater than $1$, but it does have length $7$ from $n=1$. How I found it, just trial and error. Another more complex one with length $7$ is $$811+10sum_{k=0}^n(5k+1)$$ from $n=1$. Believe it or not, they are not too difficult to find. Go here to factorize integers and see whether or not they are prime $longrightarrow$ alpertron.com.ar/ECM.HTM
– user477343
May 11 at 2:08
6
6
Have you seen this already?
– B. Mehta
May 4 at 16:58
Have you seen this already?
– B. Mehta
May 4 at 16:58
2
2
The primes in question would be a sequence like $p,p+k,p+2k,dots,p+6k$. By the pigeonhole principle, one of these must be divisible by $7$, or else $k$ must be divisible by $7$ (same with $2,3,5$). If we start with $p=7,k=30$ we almost make it, except for $187=11cdot 17$. This means that we have to use $k=210=2cdot3cdot5cdot7$...
– abiessu
May 4 at 17:04
The primes in question would be a sequence like $p,p+k,p+2k,dots,p+6k$. By the pigeonhole principle, one of these must be divisible by $7$, or else $k$ must be divisible by $7$ (same with $2,3,5$). If we start with $p=7,k=30$ we almost make it, except for $187=11cdot 17$. This means that we have to use $k=210=2cdot3cdot5cdot7$...
– abiessu
May 4 at 17:04
@abiessu, We can try multiples of $30$ below $210$ as well, and there is in fact a lower possible value than the one you suggest.
– B. Mehta
May 4 at 17:18
@abiessu, We can try multiples of $30$ below $210$ as well, and there is in fact a lower possible value than the one you suggest.
– B. Mehta
May 4 at 17:18
2
2
You could start with $-113$ or lower and go up in steps of $30$… somewhere you need a positivity criterion.
– Macavity
May 5 at 4:38
You could start with $-113$ or lower and go up in steps of $30$… somewhere you need a positivity criterion.
– Macavity
May 5 at 4:38
1
1
By the way, here is an example I found: $58n^2+1$. It is not a prime arithmetic progression, really, because it includes an exponent greater than $1$, but it does have length $7$ from $n=1$. How I found it, just trial and error. Another more complex one with length $7$ is $$811+10sum_{k=0}^n(5k+1)$$ from $n=1$. Believe it or not, they are not too difficult to find. Go here to factorize integers and see whether or not they are prime $longrightarrow$ alpertron.com.ar/ECM.HTM
– user477343
May 11 at 2:08
By the way, here is an example I found: $58n^2+1$. It is not a prime arithmetic progression, really, because it includes an exponent greater than $1$, but it does have length $7$ from $n=1$. How I found it, just trial and error. Another more complex one with length $7$ is $$811+10sum_{k=0}^n(5k+1)$$ from $n=1$. Believe it or not, they are not too difficult to find. Go here to factorize integers and see whether or not they are prime $longrightarrow$ alpertron.com.ar/ECM.HTM
– user477343
May 11 at 2:08
|
show 2 more comments
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6
Have you seen this already?
– B. Mehta
May 4 at 16:58
2
The primes in question would be a sequence like $p,p+k,p+2k,dots,p+6k$. By the pigeonhole principle, one of these must be divisible by $7$, or else $k$ must be divisible by $7$ (same with $2,3,5$). If we start with $p=7,k=30$ we almost make it, except for $187=11cdot 17$. This means that we have to use $k=210=2cdot3cdot5cdot7$...
– abiessu
May 4 at 17:04
@abiessu, We can try multiples of $30$ below $210$ as well, and there is in fact a lower possible value than the one you suggest.
– B. Mehta
May 4 at 17:18
2
You could start with $-113$ or lower and go up in steps of $30$… somewhere you need a positivity criterion.
– Macavity
May 5 at 4:38
1
By the way, here is an example I found: $58n^2+1$. It is not a prime arithmetic progression, really, because it includes an exponent greater than $1$, but it does have length $7$ from $n=1$. How I found it, just trial and error. Another more complex one with length $7$ is $$811+10sum_{k=0}^n(5k+1)$$ from $n=1$. Believe it or not, they are not too difficult to find. Go here to factorize integers and see whether or not they are prime $longrightarrow$ alpertron.com.ar/ECM.HTM
– user477343
May 11 at 2:08