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Old 2015-08-30, 10:41   #23
ATH
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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.
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Old 2015-08-30, 13:15   #24
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Quote:
Originally Posted by ATH View Post
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.
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Old 2015-08-30, 13:45   #25
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Quote:
Originally Posted by R.D. Silverman View Post
Another idiot who does not listen.
only people who don't get called names have a reason to listen to you at this point.
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Old 2015-08-30, 18:08   #26
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so T.Rex you're basically checking when values in http://oeis.org/search?q=1%2C5%2C21%...lish&go=Search +2 are prime.
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Old 2015-08-30, 18:52   #27
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Quote:
Originally Posted by science_man_88 View Post
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.
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Old 2015-09-02, 09:26   #28
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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
}
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Old 2015-09-02, 18:13   #29
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Quote:
Originally Posted by primus View Post
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.
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Old 2015-09-02, 18:23   #30
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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.
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Old 2015-09-02, 18:43   #31
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Quote:
Originally Posted by Batalov View Post
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.
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Old 2015-09-02, 18:53   #32
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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.
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Old 2015-09-02, 19:01   #33
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Quote:
Originally Posted by T.Rex View Post
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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