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In Bateman and Horn's paper* the authors say that "in some sense" the probability of a large number n being prime is about $1/\log n.$

For a set of polynomials $f_1(x),f_2(x),...,f_k(x),$ the the probability all are prime for random n would be about $1/(h_1\log n)\cdot 1/(h_2\log n)\cdot...\cdot 1/(h_k\log n,)$ in which $h_k $ is the degree of $f_k(x),$ and so the number of times that a set of such polynomials for $2\leq n < N $ are prime will be prime is about

$$\left((1/h_1)\cdot(1/h_2)\cdot...\cdot(1/h_k)\right)\cdot C\cdot\sum_2^N (\log n)^{-k} .$$

The authors explain $C(f_1,f_2,...,f_k)$ as a correction to account for the problem that numbers $f_k(n)$ are not really random. Fine.

Now C consists of ratios $r_p/s_p$ in which $r_p$ is "the chance that for a random n none of the integers $f_1(n),f_2(n),...,f_k(n)$ is divisible by p," and $s_p$ is "the chance that none of the integers in a random k-tuple is divisible by [a prime] p."

Despite the somewhat special setting I think this is an elementary argument and I do not see it. If we are trying to estimate the odds that a set of polynomials is prime for random n, why is

(chance that for random n none of f is divisible by p)/(chance that no integers in random k-tuple is divisible by p)

the right correction?

Thank you for any insight. The paragraph at top of page 364 in the cited article makes the heuristic argument but I hope the above is enough to answer the question.


Edit. Maybe it's just conditional probability? If B is P(n coprime to p) and A is P(f(x) coprime to p) then P(all f(x) coprime) = P(B)P(A|B)? Something along these lines?

*Bateman, Horn, 'A Heuristic Asymptotic Formula Concerning the Distribution of Prime Numbers,' 1962, Math. of Computation 16, 363-367.

share|improve this question
    
I do not understand the argument, perhaps because of problems with parentheses. The third line for example doesn't look right, indeed doesn't look like a probability. –  André Nicolas Jul 6 '13 at 0:36
    
@AndréNicolas: Thank you I corrected line 3 and added subscripts. If this question is still not clear please let me know. I may not have summarized well enough. –  daniel Jul 6 '13 at 4:37
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