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Primality test for $N=2cdot 3^n-1$

2

3

$begingroup$

Can you provide proof or counterexample for the following claim:

Let $P_m(x)=2^-mcdot left(left(x-sqrtx^2-4right)^m+left(x+sqrtx^2-4right)^mright)$ . Let $N=2cdot 3^n-1$ such that $n>2$ . Let $S_i=P_3(S_i-1)$ with $S_0$ equal to the modular $P_6(P_3(4)) phantom5 textmod phantom5 N$ . Then , $N$ is prime iff $S_n-2 equiv 2 pmodN$

You can run this test here.

OEIS sequence of exponents $n$ for wich $N$ is prime

I was searching for counterexample using the following PARI/GP code :

CE231(lb,ub)=

for(n=lb,ub,
N=2*3^n-1;
S=Mod(2*polchebyshev(6,1,polchebyshev(3,1,2)),N);
ctr=1;
while(ctr<=n-2,
S=Mod(2*polchebyshev(3,1,S/2),N);
ctr+=1);
if(S==2 && !ispseudoprime(N),print(n)))

I have verified this claim for all $n$ up to $7000$ .

elementary-number-theory prime-numbers examples-counterexamples primality-test

share|cite|improve this question

edited Apr 8 at 7:11

Peđa Terzić

asked Mar 20 '15 at 7:59

Peđa TerzićPeđa Terzić

7,66422674

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  FYI, $P_3(x) equiv x^3 - 3x$.
  $endgroup$
  – kennytm
  Mar 22 '15 at 9:41
  

  
  
  
  

add a comment | 

2

3

$begingroup$

Can you provide proof or counterexample for the following claim:

Let $P_m(x)=2^-mcdot left(left(x-sqrtx^2-4right)^m+left(x+sqrtx^2-4right)^mright)$ . Let $N=2cdot 3^n-1$ such that $n>2$ . Let $S_i=P_3(S_i-1)$ with $S_0$ equal to the modular $P_6(P_3(4)) phantom5 textmod phantom5 N$ . Then , $N$ is prime iff $S_n-2 equiv 2 pmodN$

You can run this test here.

OEIS sequence of exponents $n$ for wich $N$ is prime

I was searching for counterexample using the following PARI/GP code :

CE231(lb,ub)=

for(n=lb,ub,
N=2*3^n-1;
S=Mod(2*polchebyshev(6,1,polchebyshev(3,1,2)),N);
ctr=1;
while(ctr<=n-2,
S=Mod(2*polchebyshev(3,1,S/2),N);
ctr+=1);
if(S==2 && !ispseudoprime(N),print(n)))

I have verified this claim for all $n$ up to $7000$ .

elementary-number-theory prime-numbers examples-counterexamples primality-test

share|cite|improve this question

edited Apr 8 at 7:11

Peđa Terzić

asked Mar 20 '15 at 7:59

Peđa TerzićPeđa Terzić

7,66422674

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  FYI, $P_3(x) equiv x^3 - 3x$.
  $endgroup$
  – kennytm
  Mar 22 '15 at 9:41
  

  
  
  
  

add a comment | 

$begingroup$

Can you provide proof or counterexample for the following claim:

Let $P_m(x)=2^-mcdot left(left(x-sqrtx^2-4right)^m+left(x+sqrtx^2-4right)^mright)$ . Let $N=2cdot 3^n-1$ such that $n>2$ . Let $S_i=P_3(S_i-1)$ with $S_0$ equal to the modular $P_6(P_3(4)) phantom5 textmod phantom5 N$ . Then , $N$ is prime iff $S_n-2 equiv 2 pmodN$

You can run this test here.

OEIS sequence of exponents $n$ for wich $N$ is prime

I was searching for counterexample using the following PARI/GP code :

CE231(lb,ub)=

for(n=lb,ub,
N=2*3^n-1;
S=Mod(2*polchebyshev(6,1,polchebyshev(3,1,2)),N);
ctr=1;
while(ctr<=n-2,
S=Mod(2*polchebyshev(3,1,S/2),N);
ctr+=1);
if(S==2 && !ispseudoprime(N),print(n)))

I have verified this claim for all $n$ up to $7000$ .

elementary-number-theory prime-numbers examples-counterexamples primality-test

share|cite|improve this question

edited Apr 8 at 7:11

Peđa Terzić

asked Mar 20 '15 at 7:59

Peđa TerzićPeđa Terzić

7,66422674

$endgroup$

CE231(lb,ub)=

for(n=lb,ub,
N=2*3^n-1;
S=Mod(2*polchebyshev(6,1,polchebyshev(3,1,2)),N);
ctr=1;
while(ctr<=n-2,
S=Mod(2*polchebyshev(3,1,S/2),N);
ctr+=1);
if(S==2 && !ispseudoprime(N),print(n)))

share|cite|improve this question

edited Apr 8 at 7:11

Peđa Terzić

asked Mar 20 '15 at 7:59

Peđa TerzićPeđa Terzić

7,66422674

share|cite|improve this question

edited Apr 8 at 7:11

Peđa Terzić

asked Mar 20 '15 at 7:59

Peđa TerzićPeđa Terzić

7,66422674

asked Mar 20 '15 at 7:59

Peđa TerzićPeđa Terzić

7,66422674

asked Mar 20 '15 at 7:59

Peđa TerzićPeđa Terzić

7,66422674