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How many subsets are there of set $[2n]$ (for $n > 0$) where no two chosen numbers differ by exactly $n$?

I know that for consecutive numbers the answer would obey Fibonacci-like formula (How many subsets does the set $\{1, 2, \dots , n\}$ have that contain no two consecutive integers if $1$ and $n$ also count as consecutive?).

But is there any way to translate that into the case with $2n$ numbers without elements differing by $n$?

thefool
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    So, from each pair ${a,a+n}$ you pick at most one element to be in your subset. It should fall out pretty quickly from there. – Gerry Myerson Aug 11 '23 at 00:29

1 Answers1

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You have the following pairs in your set: $$(1,n+1), (2,n+2),\ldots, (n, 2n).$$ From each pair, you can choose one element, the other element, or neither. So you have $3^n$ total subsets.

PTrivedi
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