Thanks for the further explanation, Citrix!
So, you are splitting the sequence k*2^n1 into multiple subsequences: (k*2^i)*(2^5760)^m1, where n=5760*m+i. And some of those subsequences (or quite many  depending on k) are eliminated by the covering set. This is essentially the idea behind Geoffrey Reynolds' sr1sieve and sr2sieve.
In other words you are just picking some "slices" of n (mod 5760) from the whole sequence. Summing up the weights of all subsequences should yield the weight of the original sequence.
If you concentrate on just one (or a few) of those subsequences, then it's obviously faster than testing the whole sequence. But typically one has to pick quite a few low weight k's to find at least one prime. Thus, on average it shouldn't make a difference whether to pick a few of the "whole" sequences, or a few hundreds of the ultralowweight subsequences (out of the same set of k).
To give just a simpler example for illustration:
Let's take k=3, e.g. 3*2^n1. After some appropriate sieving, there will be even and odd exponents surviving (since k is divisible by 3). We could therefore split the original sequence into even and odd exponents: 3*2^(2m)1 and 3*2^(2m+1)1 (= 3*4^m1 and 6*4^m1).
Then k=3 in base=4 is just a subset of k=3 in base=2. But if the goal is to test the whole set of k=3 in base=2 up to a given n, then there is no benefit from testing both k=3 and k=6 in base=4 up to m=n/2.
Last fiddled with by Thomas11 on 20130815 at 07:53
