20150830, 10:41  #23 
Einyen
Dec 2003
Denmark
C6A_{16} Posts 
I did up to n=15k, I'm not doing it to 500k. There is no proof of either necessity or sufficiency?
The algorithm/conjecture would be no good if it was not positive for the known PRPs. 
20150830, 13:15  #24 
Nov 2003
2^{2}×5×373 Posts 

20150830, 13:45  #25 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 

20150830, 18:08  #26 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
so T.Rex you're basically checking when values in http://oeis.org/search?q=1%2C5%2C21%...lish&go=Search +2 are prime.

20150830, 18:52  #27  
Feb 2004
France
2·3^{3}·17 Posts 
Quote:


20150902, 09:26  #28 
Jul 2014
Montenegro
2·13 Posts 
My unproven solution :
For n>10 and n is even . Code:
PPT(n)= { my(s=Mod(6,(2^n+5)/3)); for(i=1,n1,s=s^22); s==6 } 
20150902, 18:13  #29 
Feb 2004
France
2×3^{3}×17 Posts 

20150902, 18:23  #30 
Feb 2004
France
396_{16} Posts 
Someone has posted a set of 4 much higherlevel conjectures k2^n ± c , which make use of Chebitchev functions that are connected to x^22 .
See: StackExchange : "Conjectured compositeness tests for N=k⋅2n±c" . Code:
CEk2c(k,c,g)= { for(n=2*c+1,g,a=6; N=k*2^nc; my(s=Mod(2*polchebyshev(k,1,a/2),N)); for(i=1,n1, s=s^22); if(!(s==2*polchebyshev(ceil(c/2),1,a/2)) && isprime(N),print(n))) } Code:
for(k=1,100,for(c=1,100, c2=Mod(c,8);if(c2==3c2==5,print1(k);CEk2c(k,c,500)))) At least for 2 cases, it is easy to show that conjecture 2) is equivalent to using cycles of the digraph. However, no proof at all. 
20150902, 18:43  #31  
Feb 2004
France
2·3^{3}·17 Posts 
Quote:
Moreover, he's close to the latest element of the sequence: 786441 . Greatest values would be new findings. 

20150902, 18:53  #32 
Feb 2004
France
2·3^{3}·17 Posts 
Here is what HC Williams and Mr Wagstaff had answered at the time I asked them some help:
Mr H.C. Williams : "Because we cannot easily factor p1 in your case, I am very doubtful that you will be able to prove your test. If you do, you will have done something very remarkable, indeed." Mr Wagstaff : "One of my PhD students is trying to prove Anton's conjecture. He has formulated a slightly simpler form: W_p = (2^p + 1)/3 is prime iff S_{p1} = S_1 (mod W_p). I think he can prove it one way: If W_p is prime, then S_{p1} = S_1 (mod W_p) (which implies Anton's S_p = S_2 (mod W_p)). Probably you can prove that, too. But he can't prove the converse." HC Williams was "very doubtful" and Mr Wagstaff had asked a student to search a complete proof. They never said that is is impossible. Williams said that, using usual technics, he sees no solution. Wagstaff spent some time of one his students. So, usual technics very probably cannot be used for building a proof. So we need someone who imagines another revolutionary technic, or someone who can prove that the conjecture is wrong. 
20150902, 19:01  #33  
Sep 2002
Database er0rr
3×1,291 Posts 
Quote:
Last fiddled with by paulunderwood on 20150902 at 19:09 

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