I am trying to figure out an efficient way to produce all the numbers in a given range for which, their bitwise AND with a positive integer (say x) gives the same number; that means k & x = k.
Is there a better way to do it so that I do not need to go through all the numbers from 0 to n and check for it?
This answer shows how to find all values of $x$ satisfying the equation for fixed $k$. The OP asked for all values of $k$ satisfying the equation for fixed $x$, but the same methods work for both problems.
First of all, we have $k \& x = k$ if $x_i = 1$ whenever $k_i = 1$. If we think of $k,x$ as sets whose elements are the 1 bits, then the equation becomes $k \cap x = k$, which is equivalent to $k \subseteq x$, and so if $i \in k$ (i.e., $k_i=1$) then $i \in x$ (i.e., $x_i=1$).
In order to generate all solutions $x$ in the range $0,\ldots,n$, it suffices to generate all solutions in increasing order; we can stop once reaching a solution which is larger than $n$. How do all solutions look like? Let the 0 bits of $k$ be $i_0 < i_1 < \cdots < i_\ell$. Then the solutions are $$ k + \sum_{j=0}^\ell b_j 2^{i_j}, $$ where $b_j \in \{0,1\}$; we can replace addition equivalently with bitwise OR. Moreover, if we think of $b_\ell\ldots b_0$ as a binary number, then the mapping from $b_\ell \ldots b_0$ to $k + \sum_j b_j 2^{i_j}$ is monotone: as we go over all values of $b_\ell \ldots b_0$ in increasing order, the values of $k + \sum_j b_j 2^{i_j}$ appear in increasing order. This leads to the following algorithm:
