Minimal Rotate-Right-Double numbers in base n
Python 2, 77 76 70 bytes



f=lambda n,m=1,i=1:i&lt;m and f(n,m,i*n)or(m+m%n*i)/n-m*2and f(n,m+1)or m


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Input: base n
Output: The smallest integer m satisfies the requirement.

Saved 6 bytes thanks to @Bubbler!

This is a brute-force search that starts at m = 1 and works its way up. Will run out of recursion limit if the actual solution is too large.

For each m, i keeps track of the current power of n, which is increased until i&gt;m.