mersenneforum.org 14° Primality test and factorization of Lepore ( conjecture )
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2017-12-28, 07:35   #45
Alberico Lepore

May 2017
ITALY

2·32·29 Posts

Quote:
 Originally Posted by CRGreathouse If it depends on the number, it's not a constant. What kind of dependency does it have?
if I'm not mistaken if N = a ^ 2 + n * a without repetitions goes from 1 to a / (2 * n)

2017-12-28, 07:40   #46
LaurV
Romulan Interpreter

"name field"
Jun 2011
Thailand

233518 Posts

Quote:
 Originally Posted by Alberico Lepore if I'm not mistaken if N = a ^ 2 + n * a without repetitions goes from 1 to a / (2 * n)
Do you mean you have an algorithm which checks if N=a*(a+n) is prime?

2017-12-28, 08:34   #47
Alberico Lepore

May 2017
ITALY

2·32·29 Posts

Quote:
 Originally Posted by LaurV Do you mean you have an algorithm which checks if N=a*(a+n) is prime?
Give me a few minutes to think about it

2017-12-28, 11:48   #48
Alberico Lepore

May 2017
ITALY

2·32·29 Posts

Quote:
 Originally Posted by LaurV Do you mean you have an algorithm which checks if N=a*(a+n) is prime?
thank you,
you pointed out to me that I was wrong
it is 2 ^ X, X with repetitions
but I do not give up

 2017-12-30, 09:43 #49 Alberico Lepore     May 2017 ITALY 10128 Posts The algorithm that I show you is in X with repetitions, if you want to apply the variant of log_2, you will have X * log_2 () with X without repetitions. For Z we are interested if you admit solutions or not For others we must test whether the parent's product difference is the same but opposite sign from that of the child Example CHECK Z [1135793-2*a^2]=-Z , Z=-a*(a-n) , a^2+n*a=1135793 YES solution bad way , Z is positive [1135793-2*a^2]=Z , Z=-a*(a-n) , a^2+n*a=1135793 NO solution good way CHECK Q [1135793-2*a^2]=Z , Z=-a*(a-n), Z-2*(a)^2=-[a*[(a)-(-(a-n)-(a))]]=A , a^2+n*a=1135793 [1135793-2*a^2]=Z , Z=-a*(a-n), Z-2*(a-n)^2=-[(-(a-n))*[-(a-n)-(a-(-(a-n)))]]=B , a^2+n*a=1135793 [-[(a)-(-(a-n)-(a))]-a]=-((a-n)-(a)) , a^2+n*a=1135793 There are no solutions are different then Q = A -[a*[(a)-(-(a-n)-(a))]]=a*(-(3*a-n))=Q CHECK R Q-2*a^2=[a*(-(3*a-n))]-2*a^2=-[a*[a-(-(3*a-n)-a)]]=A , a^2+n*a=1135793 Q-2*(-(3*a-n))^2=-[(-(3*a-n))*[(-(3*a-n))-(a-(-(3*a-n)))]]=B , a^2+n*a=1135793 [-[a-(-(3*a-n)-a)]-a]=-[[(a)-(-(a-n)-(a))]-a] , a^2+n*a=1135793 There are no solutions are different then R = A -[a*[a-(-(3*a-n)-a)]]=a*(-5*a+n)=R CHECK T R-2*a^2=-[a*[a-((-5*a+n)-a)]]=A , a^2+n*a=1135793 R-2*(-5*a+n)^2=-[(-5*a+n)*[(-5*a+n)-(a-(-5*a+n))]]=B , a^2+n*a=1135793 [(-5*a+n)-a]=-[[a-((-5*a+n)-a)]-a] , a^2+n*a=1135793 TRUE they are the same, better to say they are opposite then T = B -[(-5*a+n)*[(-5*a+n)-(a-(-5*a+n))]]=(-5*a+n)*(-11*a+2*n)=T etc.etc. Does anyone confirm me the accuracy of the algorithm?

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