Here is a partial answer. For those who want more information on this subject, Will Jagy has been kind enough to forward appropriate emails to me, to which I respond. I ask that you send requests about this to him to forward to me, until such time as I get a public email address set up.
For me, the key bit in the argument above is $E(t)$. I put it in because I intended to follow up with a post that ran through the argument again, except using only odd numbers (as per Westzynthius's footnote), and then certain thin sets ($S(k)$, the integers relatively prime to $P_k$: I have not seen this idea of applying thin sets to this sum elsewhere) . For example, using not the integers but instead the set $S(2)$ of numbers which are of the form $6m + 1$ or $6m - 1$, we can run through the argument again, except we replace $x/t$ by $\rho*x/t$ and $E(t)$ by $E_2(x/t)$. $rho$ \rho$ is the density (= 1/3) of $S(2)$ in the integers, and $E_2(x/t)$ is a more complicated error function which has a maximum value of 4/3. When the dust settles, we get for large $n$ a value for $x$ close to $Q * 2^{n-2}$.
Of course the idea of using even thinner sets now occurs. To set that up, suppose that $f(k)$ is an increasing function of $k$ to be constrained soon. For notational convenience, set $M(k)$ to be the maximum of the function $E_k(y)$ as $y$ ranges over all real numbers. Let us assume:
(subexponential growth in $k$ of $E_k(y)$ ) $M(k)*f(k) < 2^k$ .
Now with this assumption on $E_k$, the argument runs as follows: set up $I_0$ again with an inclusion exclusion argument, but count subsets $J_t$ of $S(k)$, the integers relatively prime to the $k$th primorial which are multiples of $t$. $t$ now ranges over the divisors of $P_n/P_k$. We get a sum of $2^{n-k}$ terms card($J_t$), each of which is replaced by a linear approximation as before, but now I change notation slightly and use the error function $E_k(y)$, and replace $Q$ by $Q_n$. I collect terms as before to get
$I_0 >= x/Q_n - \sum_{t \mid (P_n/P_k)} E_k(x/t)$ .
The smaller I can make this last sum, the better an upper bound I can get on $x$.
Using the growth assumption, I can bound the sum by $2^{n-k}*M(k)$, which is $2^n/f(k)$. I can then get an upper bound of $2^n/f(n-1) * Q_n$ on $x$ just by assuming the worst case values as well as the rather mild growth assumption.
However, it gets better than that. One thing I do know is that $E_k(y)$ is bounded by 1 when $y$ is less than 2. So for many $k$ and many large $t$, I can safely choose $x$ so that $E_k(x/t)$ is bounded by 1 for many $t$, so the sum looks like $2^{n-k} + D*(M(k) - 1)$, where $D$ can be much smaller than $2^{n-k}$.
The problem is that I do not know $M(k)$ or $E_k(y)$ that well. It is likely that $E_k(y)$ not only satisfies the subexponential growth assumption but also that $M(k)$ is bounded by a low degree polynomial in $k$. If this stronger assumption is true, then $x$ will also be bounded by a low degree polynomial in $n$. However, I want something like the growth assumption to hold so that I can comfortably choose $x$.
Now that I have committed myself, I will grind through the calculations to come
up with an explicit bound. I predict that $x <= n^2$, that is,
that the maximum gap in the sequence $S(n)$ is no bigger than $n^2$.
Now to try proving it.
Gerhard "Ask Me About System Design" Paseman, 2010.12.14
