Generate first all integers from $1$ to $2^{11}-1$, it takes $2^{11}-2=2046$ sums, with $x_k=x_{k-1}+x_0$ for $k=1,2,\dots,2^{11}-2$, where by trivial induction $x_k=k+1$. Then, write any number $n$ in base $2^{11}$ as $n=a_0+a_12^{11}+a_22^{22}+\dots$, where since $1000000=11\cdot90909+1$, there are $90910$ such $a_i$ (already generated since they are all less than $2^{11}$. Construct $n$ as follows: sum $a_{90909}$ with itself, then sum the result with itself, and so on, do so $11$ times, then add $a_{90908}$, thus obtaining $a_{90909}2^{11}+a_{90908}$. Proceed likewise to obtain $a_{90909}2^{22}+a_{90908}2^{11}+a_{90907}$, and continue doing so until obtaining $n$. Each time that we add a new $a_i$ costs us $12$ sums ($11$ to multiply by $2^{11}$, one more to add the $a_i$), since there are $90909$ $a_i$ appart from $a_{90909}$, it takes us $90909\cdot12=1090908<1100000-2048$ sums, and we are done.