20230129, 11:47  #56 
Aug 2002
Buenos Aires, Argentina
2·761 Posts 
Running PRP on the cofactor requires 71 days in my computer. I think that it is too much for me.

20230203, 06:37  #57 
Einyen
Dec 2003
Denmark
110101111010_{2} Posts 

20230203, 10:47  #59  
Einyen
Dec 2003
Denmark
2×3×5^{2}×23 Posts 
Quote:
2*k*p^{2} + 1 instead of 2*k*p + 1 There are only 1 known example where the factor is prime: 2*674487*93077^{2} + 1 and 20 others found in this thread that are composite and 1 unknown status: (2^{281903623}1)/4929034977952529199403122399722497889153231 (see post #36) Last fiddled with by ATH on 20230203 at 10:50 

20230203, 13:07  #60 
Dec 2022
2^{3}·5·11 Posts 
So, the reasonable question is: how did you use the mersenne.ca data if its format was such an obstacle as alpertron saw?
And, yes, this topic turns out not really to be related to Wieferich primes, except in the trivial way that the whole number (as a divisor of itself) is on the list if and only if the exponent is one, as 2^p  1 = 1 <> 2^(p1) = 1, mod anything. 
20230203, 13:59  #61  
Einyen
Dec 2003
Denmark
110101111010_{2} Posts 
Quote:
https://www.mersenne.ca/export/ 

20230203, 14:14  #62 
Aug 2002
Buenos Aires, Argentina
2·761 Posts 
Notice that you have to sort the numbers by exponent. After you get n prime factors of 2^{k}  1 from mersenne.ca, you have to test which of the 2^{n + 1}  1 factors (we have to exclude the number 1) of 2^{k}  1 are congruent to 1 mod k^{2}.
Last fiddled with by alpertron on 20230203 at 14:22 
20230203, 16:36  #63  
Einyen
Dec 2003
Denmark
2·3·5^{2}·23 Posts 
Quote:
Don't you mean if there are n factors there are 2^{n}  1 options to check? I my program I treat it as a binary number if each factor is "on" or "off" in this product: f1*f2*f3...*fn Skip 0 because that would give the product 1, then make a loop from 1 to 2^{n}1 and multiply those factors that are "on" in the binary representation of that number. For example n=5: There are 5 ways to choose 1 factor, (5 choose 2) = 10 ways to choose 2 of 5 factors, (5 choose 3) = 10 ways to choose 3 of 5, (5 choose 4) = 5 and 1 way to choose all: 5+10+10+5+1 = 2^{5}1 Edit: Ah right for each of those above I check the product as well as (2^{p}1)/product and finally the number (2^{p}1) itself so a total of 2*(2^{n}1) + 1 = 2^{n+1} 1, you are right. Last fiddled with by ATH on 20230203 at 16:40 

20230204, 13:05  #64 
Aug 2002
Buenos Aires, Argentina
1522_{10} Posts 
Yesterday I started running PRP for (2^{281903623}1)/4929034977952529199403122399722497889153231
It will require about 70 days. 
20230204, 15:43  #65  
Einyen
Dec 2003
Denmark
2×3×5^{2}×23 Posts 
Quote:
If N=2^{p}1=a*b, and a is the "product" then b=N/a is the one we want to check mod p^{2}. If we calculate N (mod p^{2}) by modular exponentiation and the "product" a mod p^{2} then: N mod p^{2} = (a mod p^{2})*(b mod p^{2}) But we only want to know if (b mod p^{2}) is 1 or not, and the only way it can be 1 is if: N (mod p^{2}) = a (mod p^{2}) (where a is the current "product" in our loop checking all posibilities.) So we only need to calculate (2^{p}1) mod p^{2} once by modular exponentiation and then compare it to all the different products mod p^{2}. Last fiddled with by ATH on 20230204 at 15:45 

20230205, 13:44  #66 
Dec 2022
2^{3}×5×11 Posts 
I see (and alpertron has already gone through how that computation is best accomplished).
So is that "exponent,k" format newly added, after the search in which he said he couldn't use mersenne.ca's data? It is indeed not difficult with that. 
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