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 2015-08-30, 10:41 #23 ATH Einyen     Dec 2003 Denmark C7C16 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.
2015-08-30, 13:15   #24
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by ATH 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.
Another idiot who does not listen.

2015-08-30, 13:45   #25
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by R.D. Silverman Another idiot who does not listen.
only people who don't get called names have a reason to listen to you at this point.

 2015-08-30, 18:08 #26 science_man_88     "Forget I exist" Jul 2009 Dumbassville 26·131 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.
2015-08-30, 18:52   #27
T.Rex

Feb 2004
France

92210 Posts

Quote:
 Originally Posted by science_man_88 so T.Rex you're basically checking when values in http://oeis.org/search?q=1%2C5%2C21%...lish&go=Search +2 are prime.
Oh ?! Yes.

 2015-09-02, 09:26 #28 primus   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,n-1,s=s^2-2); s==6 }
2015-09-02, 18:13   #29
T.Rex

Feb 2004
France

2×461 Posts

Quote:
 Originally Posted by primus My unproven solution : For n>10 and n is even . Code: PPT(n)= { my(s=Mod(6,(2^n+5)/3)); for(i=1,n-1,s=s^2-2); s==6 }
So, seed 6 seems to work too. Another Universal initial value.

 2015-09-02, 18:23 #30 T.Rex     Feb 2004 France 16328 Posts Someone has posted a set of 4 much higher-level conjectures k2^n ± c , which make use of Chebitchev functions that are connected to x^2-2 . 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^n-c; my(s=Mod(2*polchebyshev(k,1,a/2),N)); for(i=1,n-1, s=s^2-2); if(!(s==2*polchebyshev(ceil(c/2),1,a/2)) && isprime(N),print(n))) } I've done a quick check of second conjecture with: Code: for(k=1,100,for(c=1,100, c2=Mod(c,8);if(c2==3||c2==5,print1(k);CEk2c(k,c,500)))) No counter-example found ! 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.
2015-09-02, 18:43   #31
T.Rex

Feb 2004
France

2×461 Posts

Quote:
 Originally Posted by Batalov There is no point in checking 'positives'! The proof of necessity was never a problem. Check all negatives.
Yes. Probably. However, there is no proof yet that, if N is prime, it always verifies the property. It's good to know.
Moreover, he's close to the latest element of the sequence: 786441 . Greatest values would be new findings.

 2015-09-02, 18:53 #32 T.Rex     Feb 2004 France 39A16 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 p-1 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_{p-1} = -S_1 (mod W_p). I think he can prove it one way: If W_p is prime, then S_{p-1} = -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.
2015-09-02, 19:01   #33
paulunderwood

Sep 2002
Database er0rr

2×7×281 Posts

Quote:
 Originally Posted by T.Rex 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_{p-1} = -S_1 (mod W_p). I think he can prove it one way: If W_p is prime, then S_{p-1} = -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."
I really hope "W_p is prime " does imply "S_p = S_2 (mod W_p)", otherwise several years of calculation on a dozen or so computers would seem to be in vain.

Last fiddled with by paulunderwood on 2015-09-02 at 19:09

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