What is the probability for $p\equiv m \pmod n $ with $p\in\mathbb{P}\ \wedge \ (m At least i have a conjecture, that this question is a reasonable one. What do you think?
The probability for $p\equiv m \pmod n $ with $p\in\mathbb{P}\ \wedge \ (m
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number-theory
prime-numbers
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0I would guess $1/n$. Primes numbers in a global sense seem to have a perfect statistical nature, but at a finite/local level "they are like weeds in the garden path ... you never know when the next one will spring up." – 2017-02-10
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0What if the probabiliy doesn't rely on $p$ ? – 2017-02-10
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0Sorry to take 30 mins to respond ... I was enjoying a logic question about people liking people who had heard about them. ... I would like to revise my answer slightly to $1/(n-1)$ ... I had not thought about $m=0$ ... I suppose we could write a computer experiment & see statistically if this is close. ... What is your conjecture ? – 2017-02-10
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0I have changed my mind again $1/ \phi(n)$ if m & n are coprime & $0$ otherwise... go on what do you think ? – 2017-02-10
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0I agree. The probability for m,n coprime is $\frac{1}{\zeta(2)}$. So the probability for $p\equiv m \pmod n$ should be $\large \frac{6}{\pi ^{2}\ \phi(n)}$. – 2017-02-11