## Re: [PrimeNumbers] Digest Number 3737

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• On 7/27/2013 4:54 AM, Werner ... For any finite collection of primes, p1,p2,p3,...p_m, a polynomial,p(x), with rational coefficients, can be generated such
Message 1 of 1 , Jul 27, 2013
On 7/27/2013 4:54 AM,

Werner

wrote:
> 1a. Re: Polynomials
> Posted by: "Alexander"werner.sand@... werner.sand
> Date: Fri Jul 26, 2013 5:45 am ((PDT))
>
>
>
> I sent my reply wrong, I'm afraid, so once more:
>
> I collect (each case n=1,2,3..100, I cannot say per cent):
>
> n²-n+41 (Euler) generates 86 primes
> n²-n+27941 generates 77 primes
> ¼* (n^5-133n^4+6729n^3 -158379n^2+1720294n-6823316) generates only 64 primes
>
> I checked n²-n+a up to a=10^6, no better result. Can anybody compute further? Can a better result actually be expected?
>
> Werner

For any finite collection of primes, p1,p2,p3,...p_m, a polynomial,p(x),
with

rational coefficients,
can be generated such p(x) = p_x.

a0 + a1 x + a2 x^2 + ... a_m x^m = p_x

a0 + a1 (x+1) + a2 (x+1)^2 + ... + a_m (x+1)^m = p_(x+1)

By taking first differences,

a1 + a2 (2x + 1) + a3 (3x^2 + 3x + 1) + ... + a_m( (x+1)^m - x^m)
= p_(x+1) - p_x

a1 + a2 (2(x+1) + 1) + a3 (3(x+1)^2 + 3(x+1) + 1) + ... + a_m( (x+2)^m -
(x+1)^m)
= p_(x+2) - p_(x+1)

Then take second differences, third differences, etc

until you reach a linear equation to be solved for a_m.

Then solve previous equations for a_(m-1), a(m-2), etc until you solve

a0 + a1 x + a2 x^2 + ... a_m x^m = p_x

for a0.

Kermit

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