20090519, 12:25  #1 
Jun 2003
Suva, Fiji
2^{3}·3·5·17 Posts 
Sieve needed for a^(2^b)+(a+1)^(2^b)
Integers of the form i=[a^(2^b)]+[(a+1)^(2^b)], with a,b integers; provide very prime (or prp3) series with increasing a, as the possible prime factors of composite i are highly restricted in this series.
As far as I can make out, this series has not been explored before. I would think that an assault on the Lifchitz prp list could be made by a team from this site, but to do so would require a windows executable that quickly eliminates a's that give composite i at a given b. This should be relatively easy given the quality of sieve programmers here. What is needed is an understanding of the primes that appear as factors. I lack this understanding at present. Regards Robert Smith 
20090519, 13:18  #2 
Jun 2003
3^{4}×67 Posts 
So, you're proposing a Generalized Fermat Number with fixed b? AFAIK, Geoff had written a sieve for this, but (I think) it is limited to a^(2^b)+1 form. But pretty sure, he can hack it to fit your need.

20090519, 13:27  #3 
"Robert Gerbicz"
Oct 2005
Hungary
7×229 Posts 
It's easy to prove that if p=(prime)divisor of a^(2^b)+(a+1)^(2^b) then p1 is divisible by 2^(b+1).

20090519, 13:44  #4  
"Bob Silverman"
Nov 2003
North of Boston
7496_{10} Posts 
Quote:
fixed a, such primes will be quite rare. 

20090519, 14:46  #5 
(loop (#_fork))
Feb 2006
Cambridge, England
6454_{10} Posts 
The primes are rare for fixed A for the same reason that Fermat primes are rare, but the hope is that they will be relatively more common for fixed B than among a set of random numbers of the same size; for instance, n^128 + (n+1)^128 is prime for the twenty values <3000
31 37 65 191 255 287 359 786 836 1178 1229 1503 1601 1609 2093 2103 2254 2307 2471 2934 whilst Last fiddled with by fivemack on 20090519 at 15:46 
20090519, 15:10  #6  
"Bob Silverman"
Nov 2003
North of Boston
2^{3}×937 Posts 
Quote:
An interesting question that I am going to look into is: Does the sum (over all a,b) of the reciprocals of these primes converge? 

20090520, 09:58  #7 
"Bob Silverman"
Nov 2003
North of Boston
2^{3}×937 Posts 

20090520, 17:23  #8  
Jun 2003
Suva, Fiji
2^{3}·3·5·17 Posts 
Quote:
If a sieve is constructed using a formula to generate the possible factors of i, this would allow some very deep factoring indeed, lessening the need for prp3 tests, which is why I think the series interesting. The prp3 top 10000 allows us to look first at b=12 and produce prp3's quickly that qualify for the lower part of the 10000. But higher b should make the top of the list quake in their boots. 

20090520, 17:28  #9 
Jun 2003
Suva, Fiji
2^{3}·3·5·17 Posts 

20090520, 17:32  #10  
Jun 2003
Suva, Fiji
2^{3}×3×5×17 Posts 
Quote:


20090520, 18:54  #11 
Just call me Henry
"David"
Sep 2007
Liverpool (GMT/BST)
13577_{8} Posts 
this is a link to a list of the possible factors for b=12
it should be easily adjustable to other values of b if i have got it right then there is an amazingly small number of possible factors Last fiddled with by henryzz on 20090520 at 18:56 
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