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Let's say I have a number line of integers from 1 to 5 (5 total numbers). There are 4 pairs of evenly-distanced numbers: $(1, 3), (2, 4), (3, 5), (1, 5)$. Furthermore, there are ${5 \choose 2} = 10$ possible pairs of numbers. So we observe the proportion of evenly-distanced numbers is $\frac{4}{10}$.

In general, given a number from 1 to $n$ (where $n > 1$), what is the proportion of evenly-distanced pairs? Other comments:

  • We can't pair a number with itself. So $(1, 1)$ doesn't count.
  • Order doesn't change a pair. So $(1, 5)$ is not a different pair from $(5, 1)$.

I know there are ${n \choose 2}$ possible pairs, but I am having a tough time figuring out how to count the number of evenly-distanced pairs for a general $n$.

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    $\begingroup$ Notice your example interval has $3$ odd numbers and $2$ even numbers, and the total number of evenly-distanced pairs is $4=3+1=\binom32+\binom22$ ... can you generalize? $\endgroup$ Mar 10, 2022 at 8:14

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Firstly I apologize for my inexperience because I'm new here, I hope I will provide an appropriate answer for you.

Let $n$ be a whole number. Then there are two cases.

Case 1 ($n$ is even). If $n$ is an even whole number, then $n=2k$ for some whole number $k$. So there are only these pairs of evenly-distanced numbers:

$(1,3), (2,4), ..., (2k-3,2k-1), (2k-2,2k)$ [$2k-2$ pairs with $2$-distanced] $(1,5), (2,6), ..., (2k-5,2k-1), (2k-4,2k)$ [$2k-4$ pairs with $4$-distanced] $(1,7), (2,8), ..., (2k-7,2k-1), (2k-6,2k)$ [$2k-6$ pairs with $6$-distanced]

$\vdots$

$(1,2k-3), (2,2k-2), (3,2k-1), (4,2k)$ [$4$ pairs with $(2k-4)$-distanced] $(1,2k-1), (2,2k)$ [$2$ pairs with $(2k-2)$-distanced]

Here it follows that the number we are looking for is the sum of even integers from $2$ to $2k-2$, i.e. $2 + 4 + ... + (2k-4) + (2k-2) = k^2-k$.

Now we come to the other case.

Case 2 ($n$ is odd). If $n$ is an odd whole number, then $n = 2k+1$ for some whole number $k$. So there are only these pairs of evenly-distanced numbers:

$(1,3), (2,4), ..., (2k-2,2k), (2k-1,2k+1)$ [$2k-1$ pairs with $2$-distanced] $(1,5), (2,6), ..., (2k-4,2k), (2k-3,2k+1)$ [$2k-3$ pairs with $4$-distanced] $(1,7), (2,8), ..., (2k-6,2k), (2k-5,2k+1)$ [$2k-5$ pairs with $6$-distanced]

$\vdots$

$(1,2k-3), (2,2k-2), (3,2k-1), (4,2k), (5,2k+1)$ [$5$ pairs with $(2k-4)$-distanced] $(1,2k-1), (2,2k), (3,2k+1)$ [$3$ pairs with $(2k-2)$-distanced] $(1,2k+1)$ [$1$ pair with $2k$-distanced]

Here it follows that the number we are looking for is the sum of odd integers from $1$ to $2k-1$, i.e. $1 + 3 + ... + (2k-3) + (2k-1) = k^2$.

Finally, the answer is $(n/2)^2-(n/2)$ when $n$ is even, and $((n-1)/2)^2$ when $n$ is odd.

I hope you will be pleased with the answer, good luck with the rest of your life!

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