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MattcAnderson 2021-06-15 09:08

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.


MattcAnderson 2021-06-16 07:42

1 Attachment(s)
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.)



MattcAnderson 2021-08-15 21:38

project to date
1 Attachment(s)
Hi again all,

Here is a 4 page write-up with all the important points to date.


Matt C Anderson

MattcAnderson 2021-09-03 13:47

1 Attachment(s)
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.


MattcAnderson 2021-09-03 13:51

Hi All,

I asked about this polynomial x^2+x+41 on




Dr Sardonicus 2021-09-03 16:14

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.

sweety439 2021-09-04 14:22

What is the [URL=""]natural density[/URL] of [URL=""]A056561[/URL]? Is it zero? Or positive?

MattcAnderson 2021-09-23 14:05

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 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 non-negative 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.


Dr Sardonicus 2021-09-23 15:46

[QUOTE=MattcAnderson;588475]My guess is that the natural density is greater than zero.[/QUOTE]according to the Bateman-Horn 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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