## Re: [PrimeNumbers] Digest Number 2962

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• ... This is recursive function sequence used in the Pollard Rho factoring. ... With f(x) = x**2 - 2, p = 3, every term after the initial term will have
Message 1 of 1 , May 24, 2010
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> 5. a puzzle
> Posted by: "Andrey Kulsha" Andrey_601@... andrey_601
> Date: Mon May 24, 2010 6:00 am ((PDT))
>
> Let f(x) be a quadratic polynomial with integer coefficients:
>
> f(x) = ax^2+bx+c
>
>
> Let's define an integer sequence starting with prime p:
>
> p, f(p), f(f(p)), f(f(f(p))), ...
>

This is recursive function sequence used in the Pollard Rho factoring.

>
> Taking f(x) = x^2 - 2 and p = 3, we have 4 consecutive distinct prime terms:
>
> 3, 7, 47, 2207 (the next term is composite)
>

With f(x) = x**2 - 2, p = 3,

every term after the initial term will have remainder 1, when divided by 3.

Since initial term is odd, every successive term will be odd.

Which terms will be divisible by 5?

Initial term has remainder 3 when divided by 5.
Next term has remainder 2 when divided by 5.
Next term has remainder 2 when divided by 5.

All successive terms have remainder 2 when divided by 5.

Initial term has remainder 3 when divided by 7.
Next term is divisible by 7. But it is 7, so we permit it to stay.
Next term has remainder 5 when divided by 7.
Next term has remainder 2 when divided by 7.
Next term has remainder 2 when divided by 7.

All successive terms have remainder 2 when divided by 7.

>
> With f(x) = x^2 - 2x + 2 and the same p = 3 we have 5 consecutive distinct prime terms (which are Fermat primes btw):
>
> 3, 5, 17, 257, 65537 (and the next is composite again)
>
>
> Taking Euler's f(x) = x^2 + x + 41, p = 2 gives us another chain of length 5:
>
> 2, 47, 2297, 5278547, 27863063709797 (next is divisible by 173)
>
>
> Is there a sequence with 6 or more consecutive distinct prime terms?

Yes, of course there is.

For example, I predict that

for some integer b, 0 < b < 210,

f(x) = x**2 + b x - 2

will generate a sequence with 6 or more consecutive distinct prime terms.

>
> Best regards,
>
> Andrey
>
>

Kermit
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