I’m looking for the minimum distance between any two members of the geometric progressions 2, 4, 8,… and 3,9,27,…
It seems like the pair of numbers which has the minimum distance is (2,3). Can you help me find a proof?
Also what if one of the sequences starts later? For example if the sequence starts at 8,16,32,… or higher.
The numbers in the series are all natural numbers, and that means that the distance between two must be an integer. The distance between $2$ and $3$ is $1$. So, for the answer to be anything else, we need the distance between them to be $0$, so:$$x^2=x^3$$This is only true when $x=0$ or $x=1$, neither of which are a term in either of the series. So, the minimum distance is $1$, which occurs at $2$ and $3$. Unless there is another pair with distance $1$, it is $2$ and $3$.