A sequence with primes

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Rhombic
Posts: 58
Joined: Wed Jan 01, 2014 11:42 pm UTC

A sequence with primes

Many mad men have argued for the past 200 years that they have found a relation for the prime numbers, or a way to generate every single one. Of course, most of them were quite ridiculous and other gave some, but not all, of the primes. I have found one of these myself. In fact, I have checked for them until the prime number 47:
f(n)=pn2 - 2
BEING pn THE nTH PRIME NUMBER
That function gives a prime number up to pn = 47; I have not checked any more. Can anyone check it or something? Up to now every f(n) was prime.

PsiSquared
Posts: 126
Joined: Wed May 09, 2012 6:02 pm UTC

Re: A sequence with primes

112 – 2 = 119 is not prime (119 = 7x23)

Niether are f(17), f(23), f(31) or f(41).

WibblyWobbly
Can't Get No
Posts: 506
Joined: Fri Apr 05, 2013 1:03 pm UTC

Re: A sequence with primes

Sorry; last one had a small error:

f(5) = p52 - 2 = 112 - 2 = 121 - 2 = 119 = 17 * 7?

Edit: PsiSquared has it better; 17, 23, 31, 41, 53, 59, ... In fact, of the first 17 prime numbers, this formula fails to generate primes for 7 of them. Plus, there's quite a few it misses, even when it does work.

PsiSquared
Posts: 126
Joined: Wed May 09, 2012 6:02 pm UTC

Re: A sequence with primes

WibblyWobbly wrote:Sorry; last one had a small error:

f(5) = p52 - 2 = 112 - 2 = 121 - 2 = 119 = 17 * 7?

Oh, dear... you're right of-course. How on earth did I get 23?!

alessandro95
Posts: 109
Joined: Wed Apr 24, 2013 1:33 am UTC

Re: A sequence with primes

there are some prime generating polynomials you may find interesting if you didn't already know about them, of course they don't generate all the primes, and they don't even give ordered primes.
The primary reason Bourbaki stopped writing books was the realization that Lang was one single person.

Forest Goose
Posts: 377
Joined: Sat May 18, 2013 9:27 am UTC

Re: A sequence with primes

It's not overly hard to show that for a polynomial P with integer coefficients that it will be nonprime for infinitely many prime arguments:

For an integer x if q | P(x), then x = y mod q => q | P(y).

Using Dirichlet's Theorem: if q is prime, then for any t not divisible by q, there are infinitely many primes p so p = t mod q. Hence, q | P(p) for infinitely many prime p if q divides P(t).

It is well known that for any integer polynomial that it is composite infinitely often at integer arguments, thus, unless q | P(t) implies q | t for all primes q, we are done.

Supposing that the implication holds: P(0) must be nonzero (from our initial hypothesis), so q | P(t) => t = 0 mod q => q | P(0). Hence, there can only be finitely many such primes q. But, this means that composite values in Ps image must grow at an exponential rate (since they are products over a finite set). For x not 1, P(qx) is composite, but this sequence does not grow exponentially, and, thus, must introduce additional prime factors.

*I put this together on my phone at a laundromat, please forgive errors or poor writing.
Forest Goose: A rare, but wily, form of goose; best known for dropping on unsuspecting hikers, from trees, to steal sweets.