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Feb 16, 2017 at 22:10 comment added Gerry Myerson There is a discussion of "prime-producing cubics" in a paper by Mott and Rose, available at math.fsu.edu/~aluffi/archive/paper134.ps
Feb 16, 2017 at 21:43 answer added Cooper Gates timeline score: 1
Aug 14, 2015 at 12:24 vote accept joro
Aug 13, 2015 at 16:32 answer added Igor Rivin timeline score: 8
Aug 13, 2015 at 12:51 comment added joro @FedorPetrov Thank you, your expectation is Bateman–Horn conjecture and is answer to Q2. I don't see how to compute the infinite product, can you? Finite product to 10^4 gives $0.0060077084$ modulo errors.
Aug 13, 2015 at 11:37 comment added Fedor Petrov For $f(x)$ of degree $d$ I expect that $F(n)$ tends to $d^{-1}\prod_p (1-n_p/p)/(1-1/p)$, where $n_p$ denotes the number of roots of polynomial $f(x)$ modulo $p$.
Aug 13, 2015 at 11:08 comment added joro @FedorPetrov What ratio F(n) do you expect for $f(x)=6x^2+1$ and $f(x)=6x^3+1$? Do you expect $primorial(k)x^3+1$ to give increasing ratio as $k$ gets larger? Limited numerical evidence suggests for $k=7$ the ratio is smaller at 10^n.
Aug 13, 2015 at 10:56 comment added Fedor Petrov Degree 3 makes values of order $n^3$, that is, probability of being prime becomes 3 times less, where $3=\log(n^3)/\log(n)$. Small primes give another factor. Total factor may become about $1.5$, why not?
Aug 13, 2015 at 10:21 comment added joro @FedorPetrov You might be right, but shouldn't the degree decrease the ratio enough for large values?
Aug 13, 2015 at 10:19 comment added Fedor Petrov For polynomial, say, $f(x)=6x+1$, such ratio $F(n)$ tends to 3. I mean that the reason is likely in behaviour of $f$ modulo small primes.
Aug 13, 2015 at 10:19 history edited joro CC BY-SA 3.0
Added data for 10^8
Aug 13, 2015 at 9:56 history asked joro CC BY-SA 3.0