20140814, 05:45  #1 
Jul 2014
Montenegro
32_{8} Posts 
Conjectured Primality Test for Specific Class of k6^n1

20140814, 12:46  #2 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 

20140814, 14:04  #3 
Aug 2006
3^{2}·5·7·19 Posts 
How far have you tested this?

20140814, 14:59  #4 
Jul 2014
Montenegro
2×13 Posts 
maxima code to test this conjecture : Code:

20140814, 16:17  #5 
Nov 2003
2^{2}·5·373 Posts 

20140814, 17:13  #6 
Aug 2006
3^{2}·5·7·19 Posts 

20140814, 17:52  #7 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2×5×23×41 Posts 
P_{6}(x) = x^6  6x^4 + 9x^2  2
(not 1), because P_{6}(x) = P_{2}(P_{3}(x)), where P_{2}(x) = x^22, P_{3}(x) = x^33x P_{3k}(x) = P_{k}(P_{3}(x)), so S0 = P_{9k}(3) = P_{k}(P_{9}(3)) = P_{k}(5778) Anyway, it looks like a PRP test, but if one wanted to find violations of the test, then for large values it is best to implement P_{2}(x) and P_{3}(x) with FFT, and chain them. (P_{2}(x) is the same as in LL test, so this implementation already exists.) Last fiddled with by Batalov on 20140814 at 18:04 
20140814, 18:08  #8  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
Code:
(09:10) gp > (a+b)^6 %1 = a^6 + 6*b*a^5 + 15*b^2*a^4 + 20*b^3*a^3 + 15*b^4*a^2 + 6*b^5*a + b^6 (09:10) gp > (ab)^6 %2 = a^6  6*b*a^5 + 15*b^2*a^4  20*b^3*a^3 + 15*b^4*a^2  6*b^5*a + b^6 (09:10) gp > %1+%2 %3 = 2*a^6 + 30*b^2*a^4 + 30*b^4*a^2 + 2*b^6 Code:
(09:16) gp > (x^24)*(x^24) %5 = x^4  8*x^2 + 16 (09:16) gp > (x^24)*(x^24)*(x^24) %6 = x^6  12*x^4 + 48*x^2  64 multiplying this by 2^6 gives you: Code:
(09:35) gp > %9/64 %11 = x^6  6*x^4 + 9*x^2  1 Last fiddled with by science_man_88 on 20140814 at 18:10 

20140814, 18:15  #9  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
10010011010110_{2} Posts 
Quote:


20140814, 22:00  #10 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
22326_{8} Posts 
What is the reason for restriction k ≡ 3,9 (mod 10)?
It is easy to see that k ≡ 0,2,5,7 (mod 10) are not working (many counterexamples), but what about k ≡ 1,4,6,8 (mod 10)? None of them have counterexamples so far. Code:
# Pari/GP allocatemem(1800000000); KK=7000; v=vector(KK); v[1]=x; v[2]=x^22; for(i=3,KK,v[i]=x*v[i1]v[i2]) P(k,x)=eval(v[k]) t6(k,n)=s=Mod(P(k,5778),k*6^n1);for(k=1,n2,s=s*(s^23);s=s^22);s==0 #for example 1 (mod 10) forstep(k=1,KK,10,for(n=3,500,if(t6(k,n) && !ispseudoprime(k*6^n1),print(k" "n)))) 
20140814, 22:26  #11  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
Last fiddled with by science_man_88 on 20140814 at 22:29 Reason: added would be 

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