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Imagine that you have a fraction, where both the numerator and denominator take values in the set $\{1,2,\dots, n\}$. Let us assume that the fraction is smaller then and or eqal to 1. The question is how many unique values can you get with every single combination in relation to $n$ and is there a function to describe it?

So far, I think that there is a relation to prime numbers but I do not know what it is. Here are the values for $n = 1,2,\dots,20$: \begin{align*} & 1=1, 2=\mathbf{2}, 3=\mathbf{5}, 4=\mathbf{7}, 5=\mathbf{11}, 6=\mathbf{13}, 7=\mathbf{19}, \\ & 8=\mathbf{23}, 9=\mathbf{29}, 10=33, 11=\mathbf{43}, 12=\mathbf{47}, 13=\mathbf{59}, 14=65, \\ & 15=\mathbf{73}, 16=81, 17=\mathbf{97}, 18=\mathbf{103}, 19=121, 20=129, \end{align*} where the emphasized numbers are primes.

jurcistan
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    look up Farey sequences https://en.wikipedia.org/wiki/Farey_sequence – Will Jagy Feb 26 '24 at 19:38
  • I think your values are off by two - for example for $n = 3$ you should have $1/2, 1/3, 2/3$ (so $f(n) = 3$) and for $n = 4$ you should have in addition to these $1/6, 5/6$, so $f(n) = 5$. That said, take a look at https://oeis.org/A002088 . – Michael Lugo Feb 26 '24 at 20:08
  • @MichaelLugo For $n =3$, you are missing $1/1$. So $f(3) = 4, f(4) = 6, f(5) = 10$ like in your OEIS link. – Calvin Lin Feb 26 '24 at 22:23
  • @CalvinLin I thought about that but the OP said "let us assume that the fraction is smaller than 1". – Michael Lugo Feb 27 '24 at 14:30
  • @MichaelLugo True, the phrase "smaller than" is often ambiguous. Though then $f(1) = 0 , f(2) = 1$, so OP can decide how they want to fix the sequence. – Calvin Lin Feb 27 '24 at 19:07

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