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Old 2023-01-26, 15:44   #12
Dr Sardonicus
 
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Feb 2017
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Quote:
Originally Posted by Andrew Usher View Post
<snip>
Note that I computed, as I discussed, the bisection V(14,1), not V(4,1) - it saves time and the odd terms are never used anyway.
<snip>
Speaking of "saving time", what about the time of people other than yourself? Your own time is yours to waste. In posting without doing your homework, however, you are wasting the time of others. In trying to defend your mistakes, you are also wasting the time of others.

As to "saving time" in running code: If I'm writing a Pari-GP script to do numerical verification, and the numbers are fairly small, I don't care if my script is total kludge. As long as it does what I intend, and runs in a reasonable length of time, I'm happy. If anything mathematically interesting turns up, then I'll either try to find a mathematical proof, write a faster routine, or turn the problem over to someone who's a better programmer than I am and/or has better software for the purpose (which is just about anybody else on this Forum).

As a case in point, it occurred to me to try a different Lucas test which also applies to p == 7 (mod 24). As an initial check, I ran a "verbose" version up to a very small limit, just to make sure it was doing what I wanted it to do:
Code:
? w=Mod(x+1,x^2-3);forstep(n=7,500,24,r=lift(lift(-(Mod(1,n)*w)^((n+1))));if(r==2,print(n" "lift(Mod(2,n)^((n-1)/2)))))
7 1
31 1
79 1
103 1
127 1
151 1
199 1
223 1
271 1
367 1
439 1
463 1
487 1
?
Reassured, I looked for pseudoprimes, using the limit 50000000 I used before:
Code:
? w=Mod(x+1,x^2-3);forstep(n=7,50000000,24,r=lift(lift(-(Mod(1,n)*w)^((n+1))));if(r==2&&!isprime(n),print(n" "factor(n)" "lift(Mod(2,n)^((n-1)/2)))))
?
Hmm. None to 50000000. I pushed on to 100000000:
Code:
? w=Mod(x+1,x^2-3);forstep(n=49999999,100000000,24,r=lift(lift(-(Mod(1,n)*w)^((n+1))));if(r==2&&!isprime(n),print(n" "factor(n)" "lift(Mod(2,n)^((n-1)/2)))))
?
OK, none up to 108. That's it for me looking for pseudoprimes WRT this test. I'm confident they exist. Some may already be known. But if someone wants to push a simple-minded search further, that someone will not be me.
Quote:
Originally Posted by Andrew Usher View Post
<snip>
The term 'Lucas pseudoprime' was avoided because of its vagueness
<snip>
"Was avoided" is passive voice.

So you instead adopted a term which already has a meaning different from your usage? Outstanding!

Quote:
Originally Posted by Andrew Usher View Post
<snip>
Even if that were a sincere question it would not be your business, and also out of place in a thread about actual mathematics
<snip>
Speaking of "actual mathematics," have you learned yet how to prove that for prime p == 7 (mod 24), L(p+1)/4 == 0 (mod p)?

Or how about this: Given that for prime p == 7 (mod 24) we have Lp+1 == 2 (mod p) (can you prove this?), it follows that L(p+1)/2 is either 2 or -2. (Do you know why?)

The result L(p+1)/4 == 0 (mod p) depends on the fact that L(p+1)/2 == -2 (mod p) always, rather than +2. (Can you prove this fact?)

Now, the argument used to prove L(p+1)/4 == 0 (mod p) for prime p == 7 (mod 24), simply does not work for composite numbers n == 7 (mod 24). Yet in some cases, as we have seen, the conclusion holds anyway. If it's not for the reason that insures its validity for primes, why?

Last fiddled with by Dr Sardonicus on 2023-01-26 at 15:49 Reason: Insert missing modifier "prime"
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Old 2023-01-27, 13:42   #13
Andrew Usher
 
Dec 2022

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The term 'Lucas-Lehmer pseudoprimes' seems not yet in use, and I gave an appropriate definition, so there should be no problem there. If OEIS were to add this list, they might well give it that, or some similar, title (they're not very consistent though).
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