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For $m,n \in \mathbb Z$, the probability that $m,n$ picked at random are coprime is $\approx \dfrac{6}{\pi^2}$

For $m,n \in 2\mathbb Z +1$, ran a C program over randomly picked $m,n$ and the coprime probability is $\approx \dfrac{8}{\pi^2}$

Question

Is there a way to derive $\dfrac{8}{\pi^2}$, similar to the Probability that two random numbers are coprime is $\frac{6}{\pi^2}$

Guessing a probability $\dfrac{3}{4}$ comes into play since $\dfrac{\dfrac{6}{\pi^2}}{\dfrac{3}{4}}$ is $\dfrac{8}{\pi^2}$

vengy
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    for $m, n\in\mathbb Z$, the probability that they don't share the factor $2$ is $\dfrac34$, whereas for $m,n\in2\mathbb Z+1$ it's $1$ – J. W. Tanner Feb 27 '23 at 01:18
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    As both numbers are odd, there is no need to consider whether they are both divisible by 2. So by looking at part 4 of the second answer in https://math.stackexchange.com/questions/64498/probability-that-two-random-numbers-are-coprime-is-frac6-pi2 we can remove the term in the product on the LHS corresponding to p=2. This term is $1-\frac{1}{2^2}=\frac{3}{4}$ and dividing through by this gives the answer. Notice that just as $\frac{6}{\pi^2}$ is the reciprocal of $\sum_{n=1}^{\infty} \frac{1}{n^2}$, $\frac{8}{\pi^2}$ is the reciprocal of $\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2}$. – SFA Feb 27 '23 at 01:27
  • See https://math.stackexchange.com/questions/1472592/probability-that-two-integers-chosen-at-random-from-an-arithmetic-progression-ar and https://math.stackexchange.com/questions/2098846/on-probability-of-coprime-integers – Gerry Myerson Feb 27 '23 at 02:26

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