Why do all primes $n>3$ satisfy $,309mid 20^n-13^n-7^n$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Number of divisors excluding set of primes SCalculating $e^3x pmod27$Help understanding number theory definitionWhat's wrong with this proof of the infinity of primes?Unclear on why Meissel's approach to counting primes worksIf $a^n+n^bmid c^n+n^d$ for every $n$ then $c=a^k$ and $d=kb$ .What does x equivalent to 2 mod 15 mean?Olympiad problem: Prove that there are infinitely many primes which divide the sequence $ S(k) _kin mathbbN$Primes of Atlantis: a definition and a problemWhat is an irreducible prime?

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Why do all primes $n>3$ satisfy $,309mid 20^n-13^n-7^n$



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Number of divisors excluding set of primes SCalculating $e^3x pmod27$Help understanding number theory definitionWhat's wrong with this proof of the infinity of primes?Unclear on why Meissel's approach to counting primes worksIf $a^n+n^bmid c^n+n^d$ for every $n$ then $c=a^k$ and $d=kb$ .What does x equivalent to 2 mod 15 mean?Olympiad problem: Prove that there are infinitely many primes which divide the sequence $ S(k) _kin mathbbN$Primes of Atlantis: a definition and a problemWhat is an irreducible prime?










5












$begingroup$


Solve the following... $309|(20^n-13^n-7^n)$ in $mathbbZ^+$.
I invested lotof time to it and finally went to WolframAlpha for help by typing...
Solve $309k=20^n-13^n-7^n$ over the integers. It returned the following...enter image description hereenter image description here Note that all the primes are generated here. Can someone please explain why? Thanks! EDIT: the formulas suggested in the answerare too complicated, but this is so simple!










share|cite|improve this question









New contributor




Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$











  • $begingroup$
    Is it false that it generates all the primes?
    $endgroup$
    – Shamim Akhtar
    4 hours ago










  • $begingroup$
    It is easy to picture why there might be a solution for all prime $n$. However, the interesting part of the pattern is that $n=5cdot 5$ and $n=5cdot 7$ both appear, yet neither $n=3cdot 3$ nor $n=5cdot 5$ appear. Which odd composites admit solutions?
    $endgroup$
    – Mark Fischler
    4 hours ago










  • $begingroup$
    There are no even numbers ..
    $endgroup$
    – Shamim Akhtar
    4 hours ago







  • 3




    $begingroup$
    It seems like $n equiv pm1 pmod 6$.
    $endgroup$
    – md2perpe
    4 hours ago










  • $begingroup$
    Yeah right but why do only specific composites arrive?
    $endgroup$
    – Shamim Akhtar
    4 hours ago















5












$begingroup$


Solve the following... $309|(20^n-13^n-7^n)$ in $mathbbZ^+$.
I invested lotof time to it and finally went to WolframAlpha for help by typing...
Solve $309k=20^n-13^n-7^n$ over the integers. It returned the following...enter image description hereenter image description here Note that all the primes are generated here. Can someone please explain why? Thanks! EDIT: the formulas suggested in the answerare too complicated, but this is so simple!










share|cite|improve this question









New contributor




Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$











  • $begingroup$
    Is it false that it generates all the primes?
    $endgroup$
    – Shamim Akhtar
    4 hours ago










  • $begingroup$
    It is easy to picture why there might be a solution for all prime $n$. However, the interesting part of the pattern is that $n=5cdot 5$ and $n=5cdot 7$ both appear, yet neither $n=3cdot 3$ nor $n=5cdot 5$ appear. Which odd composites admit solutions?
    $endgroup$
    – Mark Fischler
    4 hours ago










  • $begingroup$
    There are no even numbers ..
    $endgroup$
    – Shamim Akhtar
    4 hours ago







  • 3




    $begingroup$
    It seems like $n equiv pm1 pmod 6$.
    $endgroup$
    – md2perpe
    4 hours ago










  • $begingroup$
    Yeah right but why do only specific composites arrive?
    $endgroup$
    – Shamim Akhtar
    4 hours ago













5












5








5


4



$begingroup$


Solve the following... $309|(20^n-13^n-7^n)$ in $mathbbZ^+$.
I invested lotof time to it and finally went to WolframAlpha for help by typing...
Solve $309k=20^n-13^n-7^n$ over the integers. It returned the following...enter image description hereenter image description here Note that all the primes are generated here. Can someone please explain why? Thanks! EDIT: the formulas suggested in the answerare too complicated, but this is so simple!










share|cite|improve this question









New contributor




Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Solve the following... $309|(20^n-13^n-7^n)$ in $mathbbZ^+$.
I invested lotof time to it and finally went to WolframAlpha for help by typing...
Solve $309k=20^n-13^n-7^n$ over the integers. It returned the following...enter image description hereenter image description here Note that all the primes are generated here. Can someone please explain why? Thanks! EDIT: the formulas suggested in the answerare too complicated, but this is so simple!







elementary-number-theory






share|cite|improve this question









New contributor




Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









Bill Dubuque

214k29198660




214k29198660






New contributor




Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 4 hours ago









Shamim AkhtarShamim Akhtar

419




419




New contributor




Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











  • $begingroup$
    Is it false that it generates all the primes?
    $endgroup$
    – Shamim Akhtar
    4 hours ago










  • $begingroup$
    It is easy to picture why there might be a solution for all prime $n$. However, the interesting part of the pattern is that $n=5cdot 5$ and $n=5cdot 7$ both appear, yet neither $n=3cdot 3$ nor $n=5cdot 5$ appear. Which odd composites admit solutions?
    $endgroup$
    – Mark Fischler
    4 hours ago










  • $begingroup$
    There are no even numbers ..
    $endgroup$
    – Shamim Akhtar
    4 hours ago







  • 3




    $begingroup$
    It seems like $n equiv pm1 pmod 6$.
    $endgroup$
    – md2perpe
    4 hours ago










  • $begingroup$
    Yeah right but why do only specific composites arrive?
    $endgroup$
    – Shamim Akhtar
    4 hours ago
















  • $begingroup$
    Is it false that it generates all the primes?
    $endgroup$
    – Shamim Akhtar
    4 hours ago










  • $begingroup$
    It is easy to picture why there might be a solution for all prime $n$. However, the interesting part of the pattern is that $n=5cdot 5$ and $n=5cdot 7$ both appear, yet neither $n=3cdot 3$ nor $n=5cdot 5$ appear. Which odd composites admit solutions?
    $endgroup$
    – Mark Fischler
    4 hours ago










  • $begingroup$
    There are no even numbers ..
    $endgroup$
    – Shamim Akhtar
    4 hours ago







  • 3




    $begingroup$
    It seems like $n equiv pm1 pmod 6$.
    $endgroup$
    – md2perpe
    4 hours ago










  • $begingroup$
    Yeah right but why do only specific composites arrive?
    $endgroup$
    – Shamim Akhtar
    4 hours ago















$begingroup$
Is it false that it generates all the primes?
$endgroup$
– Shamim Akhtar
4 hours ago




$begingroup$
Is it false that it generates all the primes?
$endgroup$
– Shamim Akhtar
4 hours ago












$begingroup$
It is easy to picture why there might be a solution for all prime $n$. However, the interesting part of the pattern is that $n=5cdot 5$ and $n=5cdot 7$ both appear, yet neither $n=3cdot 3$ nor $n=5cdot 5$ appear. Which odd composites admit solutions?
$endgroup$
– Mark Fischler
4 hours ago




$begingroup$
It is easy to picture why there might be a solution for all prime $n$. However, the interesting part of the pattern is that $n=5cdot 5$ and $n=5cdot 7$ both appear, yet neither $n=3cdot 3$ nor $n=5cdot 5$ appear. Which odd composites admit solutions?
$endgroup$
– Mark Fischler
4 hours ago












$begingroup$
There are no even numbers ..
$endgroup$
– Shamim Akhtar
4 hours ago





$begingroup$
There are no even numbers ..
$endgroup$
– Shamim Akhtar
4 hours ago





3




3




$begingroup$
It seems like $n equiv pm1 pmod 6$.
$endgroup$
– md2perpe
4 hours ago




$begingroup$
It seems like $n equiv pm1 pmod 6$.
$endgroup$
– md2perpe
4 hours ago












$begingroup$
Yeah right but why do only specific composites arrive?
$endgroup$
– Shamim Akhtar
4 hours ago




$begingroup$
Yeah right but why do only specific composites arrive?
$endgroup$
– Shamim Akhtar
4 hours ago










3 Answers
3






active

oldest

votes


















11












$begingroup$

$20^6 equiv 13^6 equiv 7^6 equiv 229 bmod 309$, so $20^n equiv 13^n + 7^n mod 309$ if and only if $20^n+6 equiv 13^n+6 + 7^n+6 bmod 309$. Since it does work for $n=1$ and $n=5$, but not for $0$, $2$, $3$ or $4$, we find that $20^n equiv 13^n + 7^n bmod 309$ if and only if $n equiv 1$ or $5 bmod 6$.



All primes except $2$ and $3$ are congruent to $1$ or $5 bmod 6$. $2$ or $3$ can't divide a number congruent to $1$ or $5 bmod 6$, so it's not easy for a small number of this form to be composite. The first few composites are $25 = 5^2$, $35 = 5 cdot 7$, $49 = 7 cdot 7$, $55 = 5 cdot 11$, ...






share|cite|improve this answer











$endgroup$












  • $begingroup$
    +1 really cool.
    $endgroup$
    – Shamim Akhtar
    3 hours ago






  • 1




    $begingroup$
    @ShamimAkhtar If you had entered your sequence $,5,7,11,13,ldots, 65$ into OEIS you'd obtain a unique match: A007310 = Numbers congruent to $1$ or $5$ mod $6$. Once you know that the rest is easy. That's the first method one should try on problems like this (when the pattern is not already clear from prior knowledge).
    $endgroup$
    – Bill Dubuque
    1 hour ago



















1












$begingroup$

So far you have only shown evidence that some of the primes are generated here, not all of them. It's interesting how many there are, though. There are some polynomials that generate an exorbitant number of primes. Here are some examples. You will likely have to do a lot more work than just exploring the idea numerically to prove that all primes are generated. Even if you do, these functions aren't super interesting because they produce non-primes as well. You won't know whether a number that is generated is prime or not until you test it for primality, so it's not like it can be used to generate the $n^th$ prime without having to do lots more computation. Quite cool, though.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Can you explain why this works?
    $endgroup$
    – Shamim Akhtar
    4 hours ago










  • $begingroup$
    I tapped the show more but wolframalpha doesnt show more
    $endgroup$
    – Shamim Akhtar
    4 hours ago






  • 1




    $begingroup$
    I'm afraid I canot
    $endgroup$
    – Michael Stachowsky
    4 hours ago










  • $begingroup$
    Is this a standard question in mse? I too am afraid if any moderator marks this as off topic
    $endgroup$
    – Shamim Akhtar
    4 hours ago






  • 1




    $begingroup$
    It's a perfectly acceptable question, although it's probably much more difficult to answer than we think. As a result, you're probably either not going to get an answer or get a very strange one. I could be wrong, number theory isn't my thing.
    $endgroup$
    – Michael Stachowsky
    4 hours ago


















1












$begingroup$

A trivial observation is that it does not generate all the odd primes: $3$ is missing.



However, I can verify that all primes up to $59359$ (other than $2$ and $3$) are represented.



This has a lot do to with the fact that $(a^2-ab+^2)$ is a apparently factor of $a^n-b^n-(a-b)^n$ for all prime $n>3$. Here, $a=20, b=13$.



But that fact is at least as interesting as your observation, and I don't know how to prove it.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    How can you verify?
    $endgroup$
    – Shamim Akhtar
    3 hours ago






  • 1




    $begingroup$
    Mathematica is very fast in determining residues mod 309.
    $endgroup$
    – Mark Fischler
    3 hours ago











Your Answer








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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









11












$begingroup$

$20^6 equiv 13^6 equiv 7^6 equiv 229 bmod 309$, so $20^n equiv 13^n + 7^n mod 309$ if and only if $20^n+6 equiv 13^n+6 + 7^n+6 bmod 309$. Since it does work for $n=1$ and $n=5$, but not for $0$, $2$, $3$ or $4$, we find that $20^n equiv 13^n + 7^n bmod 309$ if and only if $n equiv 1$ or $5 bmod 6$.



All primes except $2$ and $3$ are congruent to $1$ or $5 bmod 6$. $2$ or $3$ can't divide a number congruent to $1$ or $5 bmod 6$, so it's not easy for a small number of this form to be composite. The first few composites are $25 = 5^2$, $35 = 5 cdot 7$, $49 = 7 cdot 7$, $55 = 5 cdot 11$, ...






share|cite|improve this answer











$endgroup$












  • $begingroup$
    +1 really cool.
    $endgroup$
    – Shamim Akhtar
    3 hours ago






  • 1




    $begingroup$
    @ShamimAkhtar If you had entered your sequence $,5,7,11,13,ldots, 65$ into OEIS you'd obtain a unique match: A007310 = Numbers congruent to $1$ or $5$ mod $6$. Once you know that the rest is easy. That's the first method one should try on problems like this (when the pattern is not already clear from prior knowledge).
    $endgroup$
    – Bill Dubuque
    1 hour ago
















11












$begingroup$

$20^6 equiv 13^6 equiv 7^6 equiv 229 bmod 309$, so $20^n equiv 13^n + 7^n mod 309$ if and only if $20^n+6 equiv 13^n+6 + 7^n+6 bmod 309$. Since it does work for $n=1$ and $n=5$, but not for $0$, $2$, $3$ or $4$, we find that $20^n equiv 13^n + 7^n bmod 309$ if and only if $n equiv 1$ or $5 bmod 6$.



All primes except $2$ and $3$ are congruent to $1$ or $5 bmod 6$. $2$ or $3$ can't divide a number congruent to $1$ or $5 bmod 6$, so it's not easy for a small number of this form to be composite. The first few composites are $25 = 5^2$, $35 = 5 cdot 7$, $49 = 7 cdot 7$, $55 = 5 cdot 11$, ...






share|cite|improve this answer











$endgroup$












  • $begingroup$
    +1 really cool.
    $endgroup$
    – Shamim Akhtar
    3 hours ago






  • 1




    $begingroup$
    @ShamimAkhtar If you had entered your sequence $,5,7,11,13,ldots, 65$ into OEIS you'd obtain a unique match: A007310 = Numbers congruent to $1$ or $5$ mod $6$. Once you know that the rest is easy. That's the first method one should try on problems like this (when the pattern is not already clear from prior knowledge).
    $endgroup$
    – Bill Dubuque
    1 hour ago














11












11








11





$begingroup$

$20^6 equiv 13^6 equiv 7^6 equiv 229 bmod 309$, so $20^n equiv 13^n + 7^n mod 309$ if and only if $20^n+6 equiv 13^n+6 + 7^n+6 bmod 309$. Since it does work for $n=1$ and $n=5$, but not for $0$, $2$, $3$ or $4$, we find that $20^n equiv 13^n + 7^n bmod 309$ if and only if $n equiv 1$ or $5 bmod 6$.



All primes except $2$ and $3$ are congruent to $1$ or $5 bmod 6$. $2$ or $3$ can't divide a number congruent to $1$ or $5 bmod 6$, so it's not easy for a small number of this form to be composite. The first few composites are $25 = 5^2$, $35 = 5 cdot 7$, $49 = 7 cdot 7$, $55 = 5 cdot 11$, ...






share|cite|improve this answer











$endgroup$



$20^6 equiv 13^6 equiv 7^6 equiv 229 bmod 309$, so $20^n equiv 13^n + 7^n mod 309$ if and only if $20^n+6 equiv 13^n+6 + 7^n+6 bmod 309$. Since it does work for $n=1$ and $n=5$, but not for $0$, $2$, $3$ or $4$, we find that $20^n equiv 13^n + 7^n bmod 309$ if and only if $n equiv 1$ or $5 bmod 6$.



All primes except $2$ and $3$ are congruent to $1$ or $5 bmod 6$. $2$ or $3$ can't divide a number congruent to $1$ or $5 bmod 6$, so it's not easy for a small number of this form to be composite. The first few composites are $25 = 5^2$, $35 = 5 cdot 7$, $49 = 7 cdot 7$, $55 = 5 cdot 11$, ...







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 3 hours ago

























answered 3 hours ago









Robert IsraelRobert Israel

333k23223484




333k23223484











  • $begingroup$
    +1 really cool.
    $endgroup$
    – Shamim Akhtar
    3 hours ago






  • 1




    $begingroup$
    @ShamimAkhtar If you had entered your sequence $,5,7,11,13,ldots, 65$ into OEIS you'd obtain a unique match: A007310 = Numbers congruent to $1$ or $5$ mod $6$. Once you know that the rest is easy. That's the first method one should try on problems like this (when the pattern is not already clear from prior knowledge).
    $endgroup$
    – Bill Dubuque
    1 hour ago

















  • $begingroup$
    +1 really cool.
    $endgroup$
    – Shamim Akhtar
    3 hours ago






  • 1




    $begingroup$
    @ShamimAkhtar If you had entered your sequence $,5,7,11,13,ldots, 65$ into OEIS you'd obtain a unique match: A007310 = Numbers congruent to $1$ or $5$ mod $6$. Once you know that the rest is easy. That's the first method one should try on problems like this (when the pattern is not already clear from prior knowledge).
    $endgroup$
    – Bill Dubuque
    1 hour ago
















$begingroup$
+1 really cool.
$endgroup$
– Shamim Akhtar
3 hours ago




$begingroup$
+1 really cool.
$endgroup$
– Shamim Akhtar
3 hours ago




1




1




$begingroup$
@ShamimAkhtar If you had entered your sequence $,5,7,11,13,ldots, 65$ into OEIS you'd obtain a unique match: A007310 = Numbers congruent to $1$ or $5$ mod $6$. Once you know that the rest is easy. That's the first method one should try on problems like this (when the pattern is not already clear from prior knowledge).
$endgroup$
– Bill Dubuque
1 hour ago





$begingroup$
@ShamimAkhtar If you had entered your sequence $,5,7,11,13,ldots, 65$ into OEIS you'd obtain a unique match: A007310 = Numbers congruent to $1$ or $5$ mod $6$. Once you know that the rest is easy. That's the first method one should try on problems like this (when the pattern is not already clear from prior knowledge).
$endgroup$
– Bill Dubuque
1 hour ago












1












$begingroup$

So far you have only shown evidence that some of the primes are generated here, not all of them. It's interesting how many there are, though. There are some polynomials that generate an exorbitant number of primes. Here are some examples. You will likely have to do a lot more work than just exploring the idea numerically to prove that all primes are generated. Even if you do, these functions aren't super interesting because they produce non-primes as well. You won't know whether a number that is generated is prime or not until you test it for primality, so it's not like it can be used to generate the $n^th$ prime without having to do lots more computation. Quite cool, though.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Can you explain why this works?
    $endgroup$
    – Shamim Akhtar
    4 hours ago










  • $begingroup$
    I tapped the show more but wolframalpha doesnt show more
    $endgroup$
    – Shamim Akhtar
    4 hours ago






  • 1




    $begingroup$
    I'm afraid I canot
    $endgroup$
    – Michael Stachowsky
    4 hours ago










  • $begingroup$
    Is this a standard question in mse? I too am afraid if any moderator marks this as off topic
    $endgroup$
    – Shamim Akhtar
    4 hours ago






  • 1




    $begingroup$
    It's a perfectly acceptable question, although it's probably much more difficult to answer than we think. As a result, you're probably either not going to get an answer or get a very strange one. I could be wrong, number theory isn't my thing.
    $endgroup$
    – Michael Stachowsky
    4 hours ago















1












$begingroup$

So far you have only shown evidence that some of the primes are generated here, not all of them. It's interesting how many there are, though. There are some polynomials that generate an exorbitant number of primes. Here are some examples. You will likely have to do a lot more work than just exploring the idea numerically to prove that all primes are generated. Even if you do, these functions aren't super interesting because they produce non-primes as well. You won't know whether a number that is generated is prime or not until you test it for primality, so it's not like it can be used to generate the $n^th$ prime without having to do lots more computation. Quite cool, though.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Can you explain why this works?
    $endgroup$
    – Shamim Akhtar
    4 hours ago










  • $begingroup$
    I tapped the show more but wolframalpha doesnt show more
    $endgroup$
    – Shamim Akhtar
    4 hours ago






  • 1




    $begingroup$
    I'm afraid I canot
    $endgroup$
    – Michael Stachowsky
    4 hours ago










  • $begingroup$
    Is this a standard question in mse? I too am afraid if any moderator marks this as off topic
    $endgroup$
    – Shamim Akhtar
    4 hours ago






  • 1




    $begingroup$
    It's a perfectly acceptable question, although it's probably much more difficult to answer than we think. As a result, you're probably either not going to get an answer or get a very strange one. I could be wrong, number theory isn't my thing.
    $endgroup$
    – Michael Stachowsky
    4 hours ago













1












1








1





$begingroup$

So far you have only shown evidence that some of the primes are generated here, not all of them. It's interesting how many there are, though. There are some polynomials that generate an exorbitant number of primes. Here are some examples. You will likely have to do a lot more work than just exploring the idea numerically to prove that all primes are generated. Even if you do, these functions aren't super interesting because they produce non-primes as well. You won't know whether a number that is generated is prime or not until you test it for primality, so it's not like it can be used to generate the $n^th$ prime without having to do lots more computation. Quite cool, though.






share|cite|improve this answer









$endgroup$



So far you have only shown evidence that some of the primes are generated here, not all of them. It's interesting how many there are, though. There are some polynomials that generate an exorbitant number of primes. Here are some examples. You will likely have to do a lot more work than just exploring the idea numerically to prove that all primes are generated. Even if you do, these functions aren't super interesting because they produce non-primes as well. You won't know whether a number that is generated is prime or not until you test it for primality, so it's not like it can be used to generate the $n^th$ prime without having to do lots more computation. Quite cool, though.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 4 hours ago









Michael StachowskyMichael Stachowsky

1,409418




1,409418











  • $begingroup$
    Can you explain why this works?
    $endgroup$
    – Shamim Akhtar
    4 hours ago










  • $begingroup$
    I tapped the show more but wolframalpha doesnt show more
    $endgroup$
    – Shamim Akhtar
    4 hours ago






  • 1




    $begingroup$
    I'm afraid I canot
    $endgroup$
    – Michael Stachowsky
    4 hours ago










  • $begingroup$
    Is this a standard question in mse? I too am afraid if any moderator marks this as off topic
    $endgroup$
    – Shamim Akhtar
    4 hours ago






  • 1




    $begingroup$
    It's a perfectly acceptable question, although it's probably much more difficult to answer than we think. As a result, you're probably either not going to get an answer or get a very strange one. I could be wrong, number theory isn't my thing.
    $endgroup$
    – Michael Stachowsky
    4 hours ago
















  • $begingroup$
    Can you explain why this works?
    $endgroup$
    – Shamim Akhtar
    4 hours ago










  • $begingroup$
    I tapped the show more but wolframalpha doesnt show more
    $endgroup$
    – Shamim Akhtar
    4 hours ago






  • 1




    $begingroup$
    I'm afraid I canot
    $endgroup$
    – Michael Stachowsky
    4 hours ago










  • $begingroup$
    Is this a standard question in mse? I too am afraid if any moderator marks this as off topic
    $endgroup$
    – Shamim Akhtar
    4 hours ago






  • 1




    $begingroup$
    It's a perfectly acceptable question, although it's probably much more difficult to answer than we think. As a result, you're probably either not going to get an answer or get a very strange one. I could be wrong, number theory isn't my thing.
    $endgroup$
    – Michael Stachowsky
    4 hours ago















$begingroup$
Can you explain why this works?
$endgroup$
– Shamim Akhtar
4 hours ago




$begingroup$
Can you explain why this works?
$endgroup$
– Shamim Akhtar
4 hours ago












$begingroup$
I tapped the show more but wolframalpha doesnt show more
$endgroup$
– Shamim Akhtar
4 hours ago




$begingroup$
I tapped the show more but wolframalpha doesnt show more
$endgroup$
– Shamim Akhtar
4 hours ago




1




1




$begingroup$
I'm afraid I canot
$endgroup$
– Michael Stachowsky
4 hours ago




$begingroup$
I'm afraid I canot
$endgroup$
– Michael Stachowsky
4 hours ago












$begingroup$
Is this a standard question in mse? I too am afraid if any moderator marks this as off topic
$endgroup$
– Shamim Akhtar
4 hours ago




$begingroup$
Is this a standard question in mse? I too am afraid if any moderator marks this as off topic
$endgroup$
– Shamim Akhtar
4 hours ago




1




1




$begingroup$
It's a perfectly acceptable question, although it's probably much more difficult to answer than we think. As a result, you're probably either not going to get an answer or get a very strange one. I could be wrong, number theory isn't my thing.
$endgroup$
– Michael Stachowsky
4 hours ago




$begingroup$
It's a perfectly acceptable question, although it's probably much more difficult to answer than we think. As a result, you're probably either not going to get an answer or get a very strange one. I could be wrong, number theory isn't my thing.
$endgroup$
– Michael Stachowsky
4 hours ago











1












$begingroup$

A trivial observation is that it does not generate all the odd primes: $3$ is missing.



However, I can verify that all primes up to $59359$ (other than $2$ and $3$) are represented.



This has a lot do to with the fact that $(a^2-ab+^2)$ is a apparently factor of $a^n-b^n-(a-b)^n$ for all prime $n>3$. Here, $a=20, b=13$.



But that fact is at least as interesting as your observation, and I don't know how to prove it.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    How can you verify?
    $endgroup$
    – Shamim Akhtar
    3 hours ago






  • 1




    $begingroup$
    Mathematica is very fast in determining residues mod 309.
    $endgroup$
    – Mark Fischler
    3 hours ago















1












$begingroup$

A trivial observation is that it does not generate all the odd primes: $3$ is missing.



However, I can verify that all primes up to $59359$ (other than $2$ and $3$) are represented.



This has a lot do to with the fact that $(a^2-ab+^2)$ is a apparently factor of $a^n-b^n-(a-b)^n$ for all prime $n>3$. Here, $a=20, b=13$.



But that fact is at least as interesting as your observation, and I don't know how to prove it.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    How can you verify?
    $endgroup$
    – Shamim Akhtar
    3 hours ago






  • 1




    $begingroup$
    Mathematica is very fast in determining residues mod 309.
    $endgroup$
    – Mark Fischler
    3 hours ago













1












1








1





$begingroup$

A trivial observation is that it does not generate all the odd primes: $3$ is missing.



However, I can verify that all primes up to $59359$ (other than $2$ and $3$) are represented.



This has a lot do to with the fact that $(a^2-ab+^2)$ is a apparently factor of $a^n-b^n-(a-b)^n$ for all prime $n>3$. Here, $a=20, b=13$.



But that fact is at least as interesting as your observation, and I don't know how to prove it.






share|cite|improve this answer











$endgroup$



A trivial observation is that it does not generate all the odd primes: $3$ is missing.



However, I can verify that all primes up to $59359$ (other than $2$ and $3$) are represented.



This has a lot do to with the fact that $(a^2-ab+^2)$ is a apparently factor of $a^n-b^n-(a-b)^n$ for all prime $n>3$. Here, $a=20, b=13$.



But that fact is at least as interesting as your observation, and I don't know how to prove it.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 3 hours ago

























answered 3 hours ago









Mark FischlerMark Fischler

34.6k12552




34.6k12552











  • $begingroup$
    How can you verify?
    $endgroup$
    – Shamim Akhtar
    3 hours ago






  • 1




    $begingroup$
    Mathematica is very fast in determining residues mod 309.
    $endgroup$
    – Mark Fischler
    3 hours ago
















  • $begingroup$
    How can you verify?
    $endgroup$
    – Shamim Akhtar
    3 hours ago






  • 1




    $begingroup$
    Mathematica is very fast in determining residues mod 309.
    $endgroup$
    – Mark Fischler
    3 hours ago















$begingroup$
How can you verify?
$endgroup$
– Shamim Akhtar
3 hours ago




$begingroup$
How can you verify?
$endgroup$
– Shamim Akhtar
3 hours ago




1




1




$begingroup$
Mathematica is very fast in determining residues mod 309.
$endgroup$
– Mark Fischler
3 hours ago




$begingroup$
Mathematica is very fast in determining residues mod 309.
$endgroup$
– Mark Fischler
3 hours ago










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