20180217, 17:58  #1 
"Sam"
Nov 2016
2^{3}·41 Posts 
Good sieve for Generalized Pierpoint primes
What is a good sieve program for Generalized Pierpoint primes which do not have the form k*b^n+1 where (k < 2^32) is small? That is, a sieve which works on primes of the form 2^a*p^b*q^c*r^d...(r_n)^(d_n) where p, q, r..., r_n are distinct odd primes and there are NO restrictions on the exponents a, b, c, d,... d_n, such as their size. Does the sieve work when trying to find primes of the form 2^a*p^b*q^c*r^d...(r_n)^(d_n) where all the primes (p, q, r,..., r_n) are fixed, and all the exponents for the primes are fixed (except only one, two, or even three primes).
For example, I found a prime of the form 2^a*3^b*5^c*7^d+1 with no definite ratio, pattern, or restrictions for the exponents a, b, c, d. Here is what my sieve file looked like: (I chose the fixed exponents for 2 and 3 randomly and the exponent ranges for 5 and 7 randomly) sieve.txt:  ABC2 2^1473*3^2731*5^$a*7^$b+1 a: from 1000 to 1000 b: from 400 to 500  Running the program up to b = 415, I only found one PRP (which was later proved prime): 2^1473*3^2731*5^1020*7^408+1  If I wanted to try higher fixed exponents and ranges, what would be a good sieve program to use so I know which numbers I should test? Thanks for help. 
20180217, 18:15  #2  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:


20180218, 13:26  #3 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
You can use quadratic reciprocity to help some all even powers clump to form a quadratc residue. All powers not in those, but divisble by 3 form cubic residues, 5 pentic residues ...

20180218, 17:52  #4 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Triple post
You can also show things like: 2^(8x+4)*3^(16y+8)*5^(16z+8)+1 are always divisible by 17. 
20180218, 19:06  #5 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2·2,969 Posts 
I am not sure that there is a particularly good example. polysieve from http://mersenneforum.org/showthread....lysieve&page=8 would probably sieve it although it is not what it was designed for.

20180218, 19:42  #6  
"Forget I exist"
Jul 2009
Dumbassville
20300_{8} Posts 
Quote:
Last fiddled with by science_man_88 on 20180218 at 19:51 

20180219, 01:30  #7  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
Last fiddled with by science_man_88 on 20180219 at 01:33 

20180219, 21:34  #8 
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
Code:
forprime(x=1,10,forprime(y=x+1,100, for(z=1,y1,if(lift(Mod(x,y)^z)==n1,print(x","y","z);next(2))))) 
20180224, 21:19  #9  
"Sam"
Nov 2016
101001000_{2} Posts 
Quote:
2^(8x+4) = 2^4 = 1 (mod 17) 3^(16y+8) = 3^8 = 1 (mod 17) 5^(16z+8) = 5^8 = 1 (mod 17) Add these up and you get (1)+(1)+1 = 1 (mod 17) Then adding 1, you get (1)+1 = 0 (mod 17) The congruence holds for any values of x, y, z. Here is another example: 2^(22x+11)*3^(31y)+1 cannot be prime for any integers x, y. Now as a quick exercise, show that this is true. 

20180224, 21:41  #10  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:


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