prime producing polynomial
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See this polynomial
f(n) = n^2 + n + 41 assume n is a positive integer I once received a standing ovation for a presentation on this topic at a 3 day math conference. It was at Salishan Oregon USA, at a conference for community college math teachers. I have done some community college math teaching. I hope you find this interesting. Regards, Matt 
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Here is a list of many algebraic factorization s to find cases when
f(n) = n^2 + n + 41 is a composite number. I used a data table from a Maple calculation to list numbers when f(n) is a composite number. Then I used the method of 3 point quadratic curve fit to list parabolas. The parabolas are parametric and for all integers on these parabolic curves, f(n) is a composite number. (There are no graphs in this file.) look Matt 
project to date
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Hi again all,
Here is a 4 page writeup with all the important points to date. Regards, Matt C Anderson 
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Hi All,
Here is some numerical evidence that there are infinitely many x such that x^2+x+41 is a prime number. See the attached graph. Regards, Matt 
Hi All,
I asked about this polynomial x^2+x+41 on mathoverflow.net see [URL="https://mathoverflow.net/questions/395359/whatareprimenumbervaluesofthetrinomialqnn2n41assumingn"]https://mathoverflow.net/questions/395359/whatareprimenumbervaluesofthetrinomialqnn2n41assumingn[/URL] Regards, Matt 
Responses at MathOverflow cover most of the ground. In particuar, numerical evidence doesn't address questions of infinitude.
One point  related to one of the responses  is that p is a prime factor of f(x) = x[sup]2[/sup] + x + 41 for some positive integer x when f(x) (mod p) splits into linear factors. This means that the discriminant 163 is a quadratic residue (mod p), which means [thanks to quadratic reciprocity!] that p is a quadratic residue (mod 163). The smallest prime p which is a quadratic residue (mod 163) is p = 41. Thus, f(x), x positive integer, is never divisible by any prime less than 41. 
What is the [URL="https://en.wikipedia.org/wiki/Natural_density"]natural density[/URL] of [URL="https://oeis.org/A056561"]A056561[/URL]? Is it zero? Or positive?

Hi sweety439 and all,
I learned a new phrase, "natural density" from Wikipedia. It is an open problem whether or not f(n) = n^2+n+41 is a prime number for an infinite number of positive integers n. If f(n) is prime only a finite number of times, then the natural density of f(n) as n goes to infinity would be 0. Also, it is possible, that even if f(n) is prime an infinite number of times, the natural density could still be 0. I wrote this Maple Code, and found some data points. > # A056561 from OEIS.org Numbers n such that n^2+n+41 is prime. > # n^2+n+41 is a prime number for 0<=n<=39. > > count := 0; > for n to 1000 do if isprime(n^2+n+41) then count := count+1; print("n making n^2+n+41 prime", n, "natural density", evalf(count/n)) end if; end do; Let f(n) = n^2+n+41. What is the natural density of f as a gets large? Assume 'n' is a nonnegative integer. My data from Maple calculations  n Natural density 39 1 100 0.86 1,000 0.58 10,000 0.41 My guess is that the natural density is greater than zero. Regards, Matt 
[QUOTE=MattcAnderson;588475]My guess is that the natural density is greater than zero.[/QUOTE]according to the BatemanHorn conjecture, the number of n less than or equal to X for which n[sup]2[/sup] + n + 41 is prime is asymptotically c*Li(X) for a positive constant c; this is asymptotically c*X/ln(X). This indicates a natural density of 0, though the decrease in density up to X decreases very slowly with X.
If N > 163 the density of n for which n[sup]2[/sup] + n + 41 is not divisible by any prime < N is [tex]\frac{162}{163}\prod_{p<N,\(\frac{p}{163}\)=+1}\(1  \frac{2}{p}\)[/tex]. where [tex]\(\frac{p}{163}\)[/tex] is the quadratic character of p (mod 163). This accounts for half the primes. It can be shown that the product tends to 0 as N increases without bound. 
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