Are prime numbers really random?












11












$begingroup$


While practicing to code for my college course I stumbled upon this and would like to know if this is something new or significant as I haven't found anything resembling it on the internet.



$ p_1, p_2, ..., p_n = text{consecutive prime numbers starting from 2} \$



$ q_1 = p_2p_3...p_n \
q_2 = p_1p_3...p_n \
... \
q_n = p_1p_2...p_{n-1} \$



$ r_1 in {1, ..., p_1-1} \
... \
r_n in {1, ..., p_n-1} \$



$ s= p_1p_2...p_n \$



$ x equiv q_1r_1+...+q_nr_n :text{mod}: s \$



$ x_2 = text{the second smallest congruence} \
text{All congruences less than } x_2^2 text{ are also every prime number between } p_n text{ and } x_2^2 \$



$ text{example:} \$



$ p_1, p_2, ..., p_n = 2, 3, 5 \$



$ q_1 = 3cdot5 = 15 \
q_2 = 2cdot5 = 10 \
q_3 = 2cdot3 = 6 \$



$ r_1 in {1} \
r_2 in {1, 2} \
r_3 in {1, 2, 3, 4} \$



$ s = 2cdot3cdot5=30 \$



$ x_2 equiv 7 equiv 15cdot1 +10cdot1 +6cdot2 :text{mod}: 30 \
7^2 = 49 \$



$ text{The full sequence of primes is clearly not random.} \$



$ 0 \
color{blue}{1 equiv 31 equiv 15cdot1 +10cdot1 +6cdot1 :text{mod}: 30} \
1 \
2 \
3 \
4 \
color{red}{5} \$



$ 6 \
color{blue}{7 equiv 37 equiv 15cdot1 +10cdot1 +6cdot2 :text{mod}: 30} \
8 \
9 \
10 \
color{red}{11 equiv 41 equiv 15cdot1 +10cdot2 +6cdot1 :text{mod}: 30} \$



$ 12 \
color{blue}{13 equiv 43 equiv 15cdot1 +10cdot1 +6cdot3 :text{mod}: 30} \
14 \
15 \
16 \
color{red}{17 equiv 47 equiv 15cdot1 +10cdot2 +6cdot2 :text{mod}: 30} \$



$ 18 \
color{blue}{19 equiv 15cdot1 +10cdot1 +6cdot4 :text{mod}: 30} \
20 \
21 \
22 \
color{red}{23 equiv 15cdot1 +10cdot2 +6cdot3 :text{mod}: 30} \$



$ 24 \
color{blue}{25} \
26 \
27 \
28 \
color{red}{29 equiv 15cdot1 +10cdot2 +6cdot4 :text{mod}: 30} \$






prime-numbers modular-arithmetic







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  • 1




    $begingroup$
    Just one question, what do you mean with second smallest congruence?
    $endgroup$
    – Stan Tendijck
    1 hour ago










  • $begingroup$
    @StanTendijck um the second smallest x
    $endgroup$
    – user644904
    1 hour ago








  • 1




    $begingroup$
    @user644904 Pretty interesting observation in my opinion.
    $endgroup$
    – Larry
    1 hour ago








  • 1




    $begingroup$
    I am going to code it up. I'll get back to you in approximately 15 minutes.
    $endgroup$
    – Stan Tendijck
    1 hour ago






  • 1




    $begingroup$
    I believe that this can be setup a bit more cleanly: Let $p_i$ denote the $i$-th prime. For a particular $n$, define $s:=p_1p_2cdots p_n$ and $q_i:=s/p_i$ for $i=1,2,ldots,n$. Let $x_2$ be the second-smallest member of the set $$S = left{sum_{i=1}^n q_i r_i bmod s;middle|; 1 leq r_i < p_i right}$$ From there, you seem to be conjecturing that the members of $S$ in some range are exactly the primes in that range, but I'm not entirely clear on the formulation.
    $endgroup$
    – Blue
    9 mins ago


















11












$begingroup$


While practicing to code for my college course I stumbled upon this and would like to know if this is something new or significant as I haven't found anything resembling it on the internet.



$ p_1, p_2, ..., p_n = text{consecutive prime numbers starting from 2} \$



$ q_1 = p_2p_3...p_n \
q_2 = p_1p_3...p_n \
... \
q_n = p_1p_2...p_{n-1} \$



$ r_1 in {1, ..., p_1-1} \
... \
r_n in {1, ..., p_n-1} \$



$ s= p_1p_2...p_n \$



$ x equiv q_1r_1+...+q_nr_n :text{mod}: s \$



$ x_2 = text{the second smallest congruence} \
text{All congruences less than } x_2^2 text{ are also every prime number between } p_n text{ and } x_2^2 \$



$ text{example:} \$



$ p_1, p_2, ..., p_n = 2, 3, 5 \$



$ q_1 = 3cdot5 = 15 \
q_2 = 2cdot5 = 10 \
q_3 = 2cdot3 = 6 \$



$ r_1 in {1} \
r_2 in {1, 2} \
r_3 in {1, 2, 3, 4} \$



$ s = 2cdot3cdot5=30 \$



$ x_2 equiv 7 equiv 15cdot1 +10cdot1 +6cdot2 :text{mod}: 30 \
7^2 = 49 \$



$ text{The full sequence of primes is clearly not random.} \$



$ 0 \
color{blue}{1 equiv 31 equiv 15cdot1 +10cdot1 +6cdot1 :text{mod}: 30} \
1 \
2 \
3 \
4 \
color{red}{5} \$



$ 6 \
color{blue}{7 equiv 37 equiv 15cdot1 +10cdot1 +6cdot2 :text{mod}: 30} \
8 \
9 \
10 \
color{red}{11 equiv 41 equiv 15cdot1 +10cdot2 +6cdot1 :text{mod}: 30} \$



$ 12 \
color{blue}{13 equiv 43 equiv 15cdot1 +10cdot1 +6cdot3 :text{mod}: 30} \
14 \
15 \
16 \
color{red}{17 equiv 47 equiv 15cdot1 +10cdot2 +6cdot2 :text{mod}: 30} \$



$ 18 \
color{blue}{19 equiv 15cdot1 +10cdot1 +6cdot4 :text{mod}: 30} \
20 \
21 \
22 \
color{red}{23 equiv 15cdot1 +10cdot2 +6cdot3 :text{mod}: 30} \$



$ 24 \
color{blue}{25} \
26 \
27 \
28 \
color{red}{29 equiv 15cdot1 +10cdot2 +6cdot4 :text{mod}: 30} \$






prime-numbers modular-arithmetic







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user644904 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 1




    $begingroup$
    Just one question, what do you mean with second smallest congruence?
    $endgroup$
    – Stan Tendijck
    1 hour ago










  • $begingroup$
    @StanTendijck um the second smallest x
    $endgroup$
    – user644904
    1 hour ago








  • 1




    $begingroup$
    @user644904 Pretty interesting observation in my opinion.
    $endgroup$
    – Larry
    1 hour ago








  • 1




    $begingroup$
    I am going to code it up. I'll get back to you in approximately 15 minutes.
    $endgroup$
    – Stan Tendijck
    1 hour ago






  • 1




    $begingroup$
    I believe that this can be setup a bit more cleanly: Let $p_i$ denote the $i$-th prime. For a particular $n$, define $s:=p_1p_2cdots p_n$ and $q_i:=s/p_i$ for $i=1,2,ldots,n$. Let $x_2$ be the second-smallest member of the set $$S = left{sum_{i=1}^n q_i r_i bmod s;middle|; 1 leq r_i < p_i right}$$ From there, you seem to be conjecturing that the members of $S$ in some range are exactly the primes in that range, but I'm not entirely clear on the formulation.
    $endgroup$
    – Blue
    9 mins ago
















11












11








11


11



$begingroup$


While practicing to code for my college course I stumbled upon this and would like to know if this is something new or significant as I haven't found anything resembling it on the internet.



$ p_1, p_2, ..., p_n = text{consecutive prime numbers starting from 2} \$



$ q_1 = p_2p_3...p_n \
q_2 = p_1p_3...p_n \
... \
q_n = p_1p_2...p_{n-1} \$



$ r_1 in {1, ..., p_1-1} \
... \
r_n in {1, ..., p_n-1} \$



$ s= p_1p_2...p_n \$



$ x equiv q_1r_1+...+q_nr_n :text{mod}: s \$



$ x_2 = text{the second smallest congruence} \
text{All congruences less than } x_2^2 text{ are also every prime number between } p_n text{ and } x_2^2 \$



$ text{example:} \$



$ p_1, p_2, ..., p_n = 2, 3, 5 \$



$ q_1 = 3cdot5 = 15 \
q_2 = 2cdot5 = 10 \
q_3 = 2cdot3 = 6 \$



$ r_1 in {1} \
r_2 in {1, 2} \
r_3 in {1, 2, 3, 4} \$



$ s = 2cdot3cdot5=30 \$



$ x_2 equiv 7 equiv 15cdot1 +10cdot1 +6cdot2 :text{mod}: 30 \
7^2 = 49 \$



$ text{The full sequence of primes is clearly not random.} \$



$ 0 \
color{blue}{1 equiv 31 equiv 15cdot1 +10cdot1 +6cdot1 :text{mod}: 30} \
1 \
2 \
3 \
4 \
color{red}{5} \$



$ 6 \
color{blue}{7 equiv 37 equiv 15cdot1 +10cdot1 +6cdot2 :text{mod}: 30} \
8 \
9 \
10 \
color{red}{11 equiv 41 equiv 15cdot1 +10cdot2 +6cdot1 :text{mod}: 30} \$



$ 12 \
color{blue}{13 equiv 43 equiv 15cdot1 +10cdot1 +6cdot3 :text{mod}: 30} \
14 \
15 \
16 \
color{red}{17 equiv 47 equiv 15cdot1 +10cdot2 +6cdot2 :text{mod}: 30} \$



$ 18 \
color{blue}{19 equiv 15cdot1 +10cdot1 +6cdot4 :text{mod}: 30} \
20 \
21 \
22 \
color{red}{23 equiv 15cdot1 +10cdot2 +6cdot3 :text{mod}: 30} \$



$ 24 \
color{blue}{25} \
26 \
27 \
28 \
color{red}{29 equiv 15cdot1 +10cdot2 +6cdot4 :text{mod}: 30} \$






prime-numbers modular-arithmetic







share|cite|improve this question









New contributor




user644904 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$




While practicing to code for my college course I stumbled upon this and would like to know if this is something new or significant as I haven't found anything resembling it on the internet.



$ p_1, p_2, ..., p_n = text{consecutive prime numbers starting from 2} \$



$ q_1 = p_2p_3...p_n \
q_2 = p_1p_3...p_n \
... \
q_n = p_1p_2...p_{n-1} \$



$ r_1 in {1, ..., p_1-1} \
... \
r_n in {1, ..., p_n-1} \$



$ s= p_1p_2...p_n \$



$ x equiv q_1r_1+...+q_nr_n :text{mod}: s \$



$ x_2 = text{the second smallest congruence} \
text{All congruences less than } x_2^2 text{ are also every prime number between } p_n text{ and } x_2^2 \$



$ text{example:} \$



$ p_1, p_2, ..., p_n = 2, 3, 5 \$



$ q_1 = 3cdot5 = 15 \
q_2 = 2cdot5 = 10 \
q_3 = 2cdot3 = 6 \$



$ r_1 in {1} \
r_2 in {1, 2} \
r_3 in {1, 2, 3, 4} \$



$ s = 2cdot3cdot5=30 \$



$ x_2 equiv 7 equiv 15cdot1 +10cdot1 +6cdot2 :text{mod}: 30 \
7^2 = 49 \$



$ text{The full sequence of primes is clearly not random.} \$



$ 0 \
color{blue}{1 equiv 31 equiv 15cdot1 +10cdot1 +6cdot1 :text{mod}: 30} \
1 \
2 \
3 \
4 \
color{red}{5} \$



$ 6 \
color{blue}{7 equiv 37 equiv 15cdot1 +10cdot1 +6cdot2 :text{mod}: 30} \
8 \
9 \
10 \
color{red}{11 equiv 41 equiv 15cdot1 +10cdot2 +6cdot1 :text{mod}: 30} \$



$ 12 \
color{blue}{13 equiv 43 equiv 15cdot1 +10cdot1 +6cdot3 :text{mod}: 30} \
14 \
15 \
16 \
color{red}{17 equiv 47 equiv 15cdot1 +10cdot2 +6cdot2 :text{mod}: 30} \$



$ 18 \
color{blue}{19 equiv 15cdot1 +10cdot1 +6cdot4 :text{mod}: 30} \
20 \
21 \
22 \
color{red}{23 equiv 15cdot1 +10cdot2 +6cdot3 :text{mod}: 30} \$



$ 24 \
color{blue}{25} \
26 \
27 \
28 \
color{red}{29 equiv 15cdot1 +10cdot2 +6cdot4 :text{mod}: 30} \$






prime-numbers modular-arithmetic




prime-numbers modular-arithmetic






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share|cite|improve this question









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share|cite|improve this question




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edited 2 mins ago







user644904













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asked 1 hour ago









user644904user644904

564




564




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New contributor





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Check out our Code of Conduct.








  • 1




    $begingroup$
    Just one question, what do you mean with second smallest congruence?
    $endgroup$
    – Stan Tendijck
    1 hour ago










  • $begingroup$
    @StanTendijck um the second smallest x
    $endgroup$
    – user644904
    1 hour ago








  • 1




    $begingroup$
    @user644904 Pretty interesting observation in my opinion.
    $endgroup$
    – Larry
    1 hour ago








  • 1




    $begingroup$
    I am going to code it up. I'll get back to you in approximately 15 minutes.
    $endgroup$
    – Stan Tendijck
    1 hour ago






  • 1




    $begingroup$
    I believe that this can be setup a bit more cleanly: Let $p_i$ denote the $i$-th prime. For a particular $n$, define $s:=p_1p_2cdots p_n$ and $q_i:=s/p_i$ for $i=1,2,ldots,n$. Let $x_2$ be the second-smallest member of the set $$S = left{sum_{i=1}^n q_i r_i bmod s;middle|; 1 leq r_i < p_i right}$$ From there, you seem to be conjecturing that the members of $S$ in some range are exactly the primes in that range, but I'm not entirely clear on the formulation.
    $endgroup$
    – Blue
    9 mins ago
















  • 1




    $begingroup$
    Just one question, what do you mean with second smallest congruence?
    $endgroup$
    – Stan Tendijck
    1 hour ago










  • $begingroup$
    @StanTendijck um the second smallest x
    $endgroup$
    – user644904
    1 hour ago








  • 1




    $begingroup$
    @user644904 Pretty interesting observation in my opinion.
    $endgroup$
    – Larry
    1 hour ago








  • 1




    $begingroup$
    I am going to code it up. I'll get back to you in approximately 15 minutes.
    $endgroup$
    – Stan Tendijck
    1 hour ago






  • 1




    $begingroup$
    I believe that this can be setup a bit more cleanly: Let $p_i$ denote the $i$-th prime. For a particular $n$, define $s:=p_1p_2cdots p_n$ and $q_i:=s/p_i$ for $i=1,2,ldots,n$. Let $x_2$ be the second-smallest member of the set $$S = left{sum_{i=1}^n q_i r_i bmod s;middle|; 1 leq r_i < p_i right}$$ From there, you seem to be conjecturing that the members of $S$ in some range are exactly the primes in that range, but I'm not entirely clear on the formulation.
    $endgroup$
    – Blue
    9 mins ago










1




1




$begingroup$
Just one question, what do you mean with second smallest congruence?
$endgroup$
– Stan Tendijck
1 hour ago




$begingroup$
Just one question, what do you mean with second smallest congruence?
$endgroup$
– Stan Tendijck
1 hour ago












$begingroup$
@StanTendijck um the second smallest x
$endgroup$
– user644904
1 hour ago






$begingroup$
@StanTendijck um the second smallest x
$endgroup$
– user644904
1 hour ago






1




1




$begingroup$
@user644904 Pretty interesting observation in my opinion.
$endgroup$
– Larry
1 hour ago






$begingroup$
@user644904 Pretty interesting observation in my opinion.
$endgroup$
– Larry
1 hour ago






1




1




$begingroup$
I am going to code it up. I'll get back to you in approximately 15 minutes.
$endgroup$
– Stan Tendijck
1 hour ago




$begingroup$
I am going to code it up. I'll get back to you in approximately 15 minutes.
$endgroup$
– Stan Tendijck
1 hour ago




1




1




$begingroup$
I believe that this can be setup a bit more cleanly: Let $p_i$ denote the $i$-th prime. For a particular $n$, define $s:=p_1p_2cdots p_n$ and $q_i:=s/p_i$ for $i=1,2,ldots,n$. Let $x_2$ be the second-smallest member of the set $$S = left{sum_{i=1}^n q_i r_i bmod s;middle|; 1 leq r_i < p_i right}$$ From there, you seem to be conjecturing that the members of $S$ in some range are exactly the primes in that range, but I'm not entirely clear on the formulation.
$endgroup$
– Blue
9 mins ago






$begingroup$
I believe that this can be setup a bit more cleanly: Let $p_i$ denote the $i$-th prime. For a particular $n$, define $s:=p_1p_2cdots p_n$ and $q_i:=s/p_i$ for $i=1,2,ldots,n$. Let $x_2$ be the second-smallest member of the set $$S = left{sum_{i=1}^n q_i r_i bmod s;middle|; 1 leq r_i < p_i right}$$ From there, you seem to be conjecturing that the members of $S$ in some range are exactly the primes in that range, but I'm not entirely clear on the formulation.
$endgroup$
– Blue
9 mins ago












2 Answers
2






active

oldest

votes


















1












$begingroup$

I had to say, I was suspicious when I read the question but I coded up the problem and up to at least $n=8$ it definitely works. For $n=9$, it becomes a bit problematic for my computer/programming skills to handle since we need to work with $prod_{i=1}^{9}(p_i-1)$ possibilities in working with the $r_i$ but as far as I could run it at the moment, it also seems to work for $n=9$. I'll update my answer if I found something else.



Maybe something that is worth noting because it definitely increased my interest in this question. I believe that for $n=9$, $x_2=29$ and almost all primes smaller than $29^2$ are generated which means it might not only be true but it even generates a lot of primes!



Edit: $n=9$ is confirmed to work.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the verification
    $endgroup$
    – user644904
    34 mins ago



















1












$begingroup$

Interesting that you marked 25, which is not a prime, similar to 6x+1 and 6x-1, which means it's not a primality test, but still quite interesting.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yeah, but I can't comment (need 50 reputation), so I posted it here.
    $endgroup$
    – James
    49 mins ago










  • $begingroup$
    Its not a primality test
    $endgroup$
    – user644904
    35 mins ago










  • $begingroup$
    I didn't say you meant it to be, I'm just pointing out it's not unique to only prime numbers.
    $endgroup$
    – James
    32 mins ago












  • $begingroup$
    Yes you're correct
    $endgroup$
    – user644904
    32 mins ago






  • 1




    $begingroup$
    If it was unique to prime numbers, then this would be ground-breaking.
    $endgroup$
    – James
    23 mins ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

I had to say, I was suspicious when I read the question but I coded up the problem and up to at least $n=8$ it definitely works. For $n=9$, it becomes a bit problematic for my computer/programming skills to handle since we need to work with $prod_{i=1}^{9}(p_i-1)$ possibilities in working with the $r_i$ but as far as I could run it at the moment, it also seems to work for $n=9$. I'll update my answer if I found something else.



Maybe something that is worth noting because it definitely increased my interest in this question. I believe that for $n=9$, $x_2=29$ and almost all primes smaller than $29^2$ are generated which means it might not only be true but it even generates a lot of primes!



Edit: $n=9$ is confirmed to work.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the verification
    $endgroup$
    – user644904
    34 mins ago
















1












$begingroup$

I had to say, I was suspicious when I read the question but I coded up the problem and up to at least $n=8$ it definitely works. For $n=9$, it becomes a bit problematic for my computer/programming skills to handle since we need to work with $prod_{i=1}^{9}(p_i-1)$ possibilities in working with the $r_i$ but as far as I could run it at the moment, it also seems to work for $n=9$. I'll update my answer if I found something else.



Maybe something that is worth noting because it definitely increased my interest in this question. I believe that for $n=9$, $x_2=29$ and almost all primes smaller than $29^2$ are generated which means it might not only be true but it even generates a lot of primes!



Edit: $n=9$ is confirmed to work.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the verification
    $endgroup$
    – user644904
    34 mins ago














1












1








1





$begingroup$

I had to say, I was suspicious when I read the question but I coded up the problem and up to at least $n=8$ it definitely works. For $n=9$, it becomes a bit problematic for my computer/programming skills to handle since we need to work with $prod_{i=1}^{9}(p_i-1)$ possibilities in working with the $r_i$ but as far as I could run it at the moment, it also seems to work for $n=9$. I'll update my answer if I found something else.



Maybe something that is worth noting because it definitely increased my interest in this question. I believe that for $n=9$, $x_2=29$ and almost all primes smaller than $29^2$ are generated which means it might not only be true but it even generates a lot of primes!



Edit: $n=9$ is confirmed to work.






share|cite|improve this answer











$endgroup$



I had to say, I was suspicious when I read the question but I coded up the problem and up to at least $n=8$ it definitely works. For $n=9$, it becomes a bit problematic for my computer/programming skills to handle since we need to work with $prod_{i=1}^{9}(p_i-1)$ possibilities in working with the $r_i$ but as far as I could run it at the moment, it also seems to work for $n=9$. I'll update my answer if I found something else.



Maybe something that is worth noting because it definitely increased my interest in this question. I believe that for $n=9$, $x_2=29$ and almost all primes smaller than $29^2$ are generated which means it might not only be true but it even generates a lot of primes!



Edit: $n=9$ is confirmed to work.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 43 mins ago

























answered 51 mins ago









Stan TendijckStan Tendijck

1,541310




1,541310












  • $begingroup$
    Thanks for the verification
    $endgroup$
    – user644904
    34 mins ago


















  • $begingroup$
    Thanks for the verification
    $endgroup$
    – user644904
    34 mins ago
















$begingroup$
Thanks for the verification
$endgroup$
– user644904
34 mins ago




$begingroup$
Thanks for the verification
$endgroup$
– user644904
34 mins ago











1












$begingroup$

Interesting that you marked 25, which is not a prime, similar to 6x+1 and 6x-1, which means it's not a primality test, but still quite interesting.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yeah, but I can't comment (need 50 reputation), so I posted it here.
    $endgroup$
    – James
    49 mins ago










  • $begingroup$
    Its not a primality test
    $endgroup$
    – user644904
    35 mins ago










  • $begingroup$
    I didn't say you meant it to be, I'm just pointing out it's not unique to only prime numbers.
    $endgroup$
    – James
    32 mins ago












  • $begingroup$
    Yes you're correct
    $endgroup$
    – user644904
    32 mins ago






  • 1




    $begingroup$
    If it was unique to prime numbers, then this would be ground-breaking.
    $endgroup$
    – James
    23 mins ago
















1












$begingroup$

Interesting that you marked 25, which is not a prime, similar to 6x+1 and 6x-1, which means it's not a primality test, but still quite interesting.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yeah, but I can't comment (need 50 reputation), so I posted it here.
    $endgroup$
    – James
    49 mins ago










  • $begingroup$
    Its not a primality test
    $endgroup$
    – user644904
    35 mins ago










  • $begingroup$
    I didn't say you meant it to be, I'm just pointing out it's not unique to only prime numbers.
    $endgroup$
    – James
    32 mins ago












  • $begingroup$
    Yes you're correct
    $endgroup$
    – user644904
    32 mins ago






  • 1




    $begingroup$
    If it was unique to prime numbers, then this would be ground-breaking.
    $endgroup$
    – James
    23 mins ago














1












1








1





$begingroup$

Interesting that you marked 25, which is not a prime, similar to 6x+1 and 6x-1, which means it's not a primality test, but still quite interesting.






share|cite|improve this answer











$endgroup$



Interesting that you marked 25, which is not a prime, similar to 6x+1 and 6x-1, which means it's not a primality test, but still quite interesting.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 38 mins ago

























answered 54 mins ago









JamesJames

218




218












  • $begingroup$
    Yeah, but I can't comment (need 50 reputation), so I posted it here.
    $endgroup$
    – James
    49 mins ago










  • $begingroup$
    Its not a primality test
    $endgroup$
    – user644904
    35 mins ago










  • $begingroup$
    I didn't say you meant it to be, I'm just pointing out it's not unique to only prime numbers.
    $endgroup$
    – James
    32 mins ago












  • $begingroup$
    Yes you're correct
    $endgroup$
    – user644904
    32 mins ago






  • 1




    $begingroup$
    If it was unique to prime numbers, then this would be ground-breaking.
    $endgroup$
    – James
    23 mins ago


















  • $begingroup$
    Yeah, but I can't comment (need 50 reputation), so I posted it here.
    $endgroup$
    – James
    49 mins ago










  • $begingroup$
    Its not a primality test
    $endgroup$
    – user644904
    35 mins ago










  • $begingroup$
    I didn't say you meant it to be, I'm just pointing out it's not unique to only prime numbers.
    $endgroup$
    – James
    32 mins ago












  • $begingroup$
    Yes you're correct
    $endgroup$
    – user644904
    32 mins ago






  • 1




    $begingroup$
    If it was unique to prime numbers, then this would be ground-breaking.
    $endgroup$
    – James
    23 mins ago
















$begingroup$
Yeah, but I can't comment (need 50 reputation), so I posted it here.
$endgroup$
– James
49 mins ago




$begingroup$
Yeah, but I can't comment (need 50 reputation), so I posted it here.
$endgroup$
– James
49 mins ago












$begingroup$
Its not a primality test
$endgroup$
– user644904
35 mins ago




$begingroup$
Its not a primality test
$endgroup$
– user644904
35 mins ago












$begingroup$
I didn't say you meant it to be, I'm just pointing out it's not unique to only prime numbers.
$endgroup$
– James
32 mins ago






$begingroup$
I didn't say you meant it to be, I'm just pointing out it's not unique to only prime numbers.
$endgroup$
– James
32 mins ago














$begingroup$
Yes you're correct
$endgroup$
– user644904
32 mins ago




$begingroup$
Yes you're correct
$endgroup$
– user644904
32 mins ago




1




1




$begingroup$
If it was unique to prime numbers, then this would be ground-breaking.
$endgroup$
– James
23 mins ago




$begingroup$
If it was unique to prime numbers, then this would be ground-breaking.
$endgroup$
– James
23 mins ago










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