Can sieve operations be compared in this way? Suppose we consider the set of integers in the interval $I_n=[2^n+1,2^{n+1}]$, and we pare this set down in two similar ways with the sieves $ab+a+b=k$ and $2ab+a+b=k$ for $k\in I_n$.  These sieves work out as $(a+1)b+a=k\to k\equiv -1\pmod{a+1}$ and $(2a+1)b+a=k\to k\equiv a\pmod{2a+1}$.  Transforming into sums naively, we get
$$\sum_{a=1}^{2^{(n+1)/2}}\left[2^n\over a+1\right]+O(1)\tag 1$$
$$\sum_{a=1}^{2^{(n+1)/2}}\left[2^n\over 2a+1\right]+O(1)\tag 2$$
On the surface, it seems obvious that $(1)$ will always cover at least as many elements as $(2)$ of the interval $I_n$ for any given $n$.  In particular, it seems that for every $2a+1$ there is an $a+1$ to match, but for $a=1$ we get $2^{n-1}$ in $(1)$ that is not present in $(2)$.  If we then make all the assumptions in favor of $(2)$ for the $O(1)$ terms, we get a value of $(1)-(2)=2^{n-1}-2^{(n+1)/2}$.
Is it logical to use these results to say that for $n\geq 3$, our sieve $ab+a+b$ always cancels at least as many elements as the sieve $2ab+a+b$ of any given interval $I_n$?  Is there any way to improve this argument, other than using non-naive versions of these sums?
 A: It seems the answers are "yes" and "yes".
In particular, we can improve the naive sums by noting that they should be taken over the primes, and then accounting for the overlaps that naturally occur.  Interestingly, we still have to account for the $O(1)$ terms in essentially all of these same numbers because of these overlaps, so the "naive" sums are really a pretty good start.
But if we want $(1)$ to map onto $(2)$ within any given interval, we have to start with saying that for $a+1$ in $(1)$ equal to $2a+1$ in $(2)$, we have almost-equal counts, and the $O(1)$ terms can be assumed to fall in favor of $(2)$, and then we have $a=1$ in $(1)$ to capture the rest.  But this count of $2^{n-1}$ is naive, and must account for overlaps with the rest of the primes in $(1)$.  Fortunately this accounting works out so that we have $2^{n-1}\over (n-1)\log 2$ as the non-overlapping portion, and this quantity is still asymptotically greater than $2^{(n+1)/2}$, the count of $O(1)$ terms.
Here's a take on this that looks at the twin primes:
https://drive.google.com/file/d/0B5KQFxR7gj3JdXpWX2pBa2RHMDA/view?usp=sharing
