Conjecture: $n>2$ is prime iff $sum_{k=1}^{n-1}left(3^k-2right)^{n-1} ;equiv; n cdot 2^{n-1}-1...
$begingroup$
This question is closely related to: Conjectured primality test
Can you provide a proof or a counterexample for the following claim :
Conjecture. Let $n$ be a natural number greater than $2$. Then $n$ is prime if and only if
$$sum_{k=1}^{n-1}left(3^k-2right)^{n-1} ;equiv;
n cdot 2^{n-1}-1 pmod{frac{3^n-1}{2}}$$
You can run this test here.
I was searching for a counterexample using the following two PARI/GP programs :
CE1(n1,n2)=
{
forcomposite(n=n1,n2,
s=sum(k=1,n-1,lift(Mod(3^k-2,(3^n-1)/2)^(n-1)));
if((Mod(s,(3^n-1)/2)==n*2^(n-1)-1),print("n="n)))
}
CE2(n1,n2)=
{
forprime(n=n1,n2,
s=sum(k=1,n-1,lift(Mod(3^k-2,(3^n-1)/2)^(n-1)));
if(!(Mod(s,(3^n-1)/2)==n*2^(n-1)-1),print("n="n)))
}
REMARK
More generally we can formulate the following criterion :
Let $b$ , $a$ and $n$ be a natural numbers, $b>ageq 1$, $n>2$ and $n notin {4,8,9}$. Then $n$ is prime if and only if
$$displaystylesum_{k=1}^{n}left(b^kpm aright)^{n-1} equiv n cdot a^{n-1} pmod{frac{b^n-1}{b-1}}$$
elementary-number-theory prime-numbers examples-counterexamples primality-test
$endgroup$
add a comment |
$begingroup$
This question is closely related to: Conjectured primality test
Can you provide a proof or a counterexample for the following claim :
Conjecture. Let $n$ be a natural number greater than $2$. Then $n$ is prime if and only if
$$sum_{k=1}^{n-1}left(3^k-2right)^{n-1} ;equiv;
n cdot 2^{n-1}-1 pmod{frac{3^n-1}{2}}$$
You can run this test here.
I was searching for a counterexample using the following two PARI/GP programs :
CE1(n1,n2)=
{
forcomposite(n=n1,n2,
s=sum(k=1,n-1,lift(Mod(3^k-2,(3^n-1)/2)^(n-1)));
if((Mod(s,(3^n-1)/2)==n*2^(n-1)-1),print("n="n)))
}
CE2(n1,n2)=
{
forprime(n=n1,n2,
s=sum(k=1,n-1,lift(Mod(3^k-2,(3^n-1)/2)^(n-1)));
if(!(Mod(s,(3^n-1)/2)==n*2^(n-1)-1),print("n="n)))
}
REMARK
More generally we can formulate the following criterion :
Let $b$ , $a$ and $n$ be a natural numbers, $b>ageq 1$, $n>2$ and $n notin {4,8,9}$. Then $n$ is prime if and only if
$$displaystylesum_{k=1}^{n}left(b^kpm aright)^{n-1} equiv n cdot a^{n-1} pmod{frac{b^n-1}{b-1}}$$
elementary-number-theory prime-numbers examples-counterexamples primality-test
$endgroup$
2
$begingroup$
To how high a value of $n$ is it known to hold?
$endgroup$
– Oscar Lanzi
Feb 2 at 10:50
1
$begingroup$
@OscarLanzi For all $n$ up to $5000$.
$endgroup$
– Peđa Terzić
Feb 2 at 12:54
add a comment |
$begingroup$
This question is closely related to: Conjectured primality test
Can you provide a proof or a counterexample for the following claim :
Conjecture. Let $n$ be a natural number greater than $2$. Then $n$ is prime if and only if
$$sum_{k=1}^{n-1}left(3^k-2right)^{n-1} ;equiv;
n cdot 2^{n-1}-1 pmod{frac{3^n-1}{2}}$$
You can run this test here.
I was searching for a counterexample using the following two PARI/GP programs :
CE1(n1,n2)=
{
forcomposite(n=n1,n2,
s=sum(k=1,n-1,lift(Mod(3^k-2,(3^n-1)/2)^(n-1)));
if((Mod(s,(3^n-1)/2)==n*2^(n-1)-1),print("n="n)))
}
CE2(n1,n2)=
{
forprime(n=n1,n2,
s=sum(k=1,n-1,lift(Mod(3^k-2,(3^n-1)/2)^(n-1)));
if(!(Mod(s,(3^n-1)/2)==n*2^(n-1)-1),print("n="n)))
}
REMARK
More generally we can formulate the following criterion :
Let $b$ , $a$ and $n$ be a natural numbers, $b>ageq 1$, $n>2$ and $n notin {4,8,9}$. Then $n$ is prime if and only if
$$displaystylesum_{k=1}^{n}left(b^kpm aright)^{n-1} equiv n cdot a^{n-1} pmod{frac{b^n-1}{b-1}}$$
elementary-number-theory prime-numbers examples-counterexamples primality-test
$endgroup$
This question is closely related to: Conjectured primality test
Can you provide a proof or a counterexample for the following claim :
Conjecture. Let $n$ be a natural number greater than $2$. Then $n$ is prime if and only if
$$sum_{k=1}^{n-1}left(3^k-2right)^{n-1} ;equiv;
n cdot 2^{n-1}-1 pmod{frac{3^n-1}{2}}$$
You can run this test here.
I was searching for a counterexample using the following two PARI/GP programs :
CE1(n1,n2)=
{
forcomposite(n=n1,n2,
s=sum(k=1,n-1,lift(Mod(3^k-2,(3^n-1)/2)^(n-1)));
if((Mod(s,(3^n-1)/2)==n*2^(n-1)-1),print("n="n)))
}
CE2(n1,n2)=
{
forprime(n=n1,n2,
s=sum(k=1,n-1,lift(Mod(3^k-2,(3^n-1)/2)^(n-1)));
if(!(Mod(s,(3^n-1)/2)==n*2^(n-1)-1),print("n="n)))
}
REMARK
More generally we can formulate the following criterion :
Let $b$ , $a$ and $n$ be a natural numbers, $b>ageq 1$, $n>2$ and $n notin {4,8,9}$. Then $n$ is prime if and only if
$$displaystylesum_{k=1}^{n}left(b^kpm aright)^{n-1} equiv n cdot a^{n-1} pmod{frac{b^n-1}{b-1}}$$
elementary-number-theory prime-numbers examples-counterexamples primality-test
elementary-number-theory prime-numbers examples-counterexamples primality-test
edited Feb 4 at 8:07
Peđa Terzić
asked Feb 2 at 9:43


Peđa TerzićPeđa Terzić
7,66922674
7,66922674
2
$begingroup$
To how high a value of $n$ is it known to hold?
$endgroup$
– Oscar Lanzi
Feb 2 at 10:50
1
$begingroup$
@OscarLanzi For all $n$ up to $5000$.
$endgroup$
– Peđa Terzić
Feb 2 at 12:54
add a comment |
2
$begingroup$
To how high a value of $n$ is it known to hold?
$endgroup$
– Oscar Lanzi
Feb 2 at 10:50
1
$begingroup$
@OscarLanzi For all $n$ up to $5000$.
$endgroup$
– Peđa Terzić
Feb 2 at 12:54
2
2
$begingroup$
To how high a value of $n$ is it known to hold?
$endgroup$
– Oscar Lanzi
Feb 2 at 10:50
$begingroup$
To how high a value of $n$ is it known to hold?
$endgroup$
– Oscar Lanzi
Feb 2 at 10:50
1
1
$begingroup$
@OscarLanzi For all $n$ up to $5000$.
$endgroup$
– Peđa Terzić
Feb 2 at 12:54
$begingroup$
@OscarLanzi For all $n$ up to $5000$.
$endgroup$
– Peđa Terzić
Feb 2 at 12:54
add a comment |
0
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2
$begingroup$
To how high a value of $n$ is it known to hold?
$endgroup$
– Oscar Lanzi
Feb 2 at 10:50
1
$begingroup$
@OscarLanzi For all $n$ up to $5000$.
$endgroup$
– Peđa Terzić
Feb 2 at 12:54