[SOLVED] infinite set of coprimes

kureta

does any one know an infinite set of coprimes except for the elements of sylvester's sequence. S(n)=S(n-1)*(S(n-1)-1)+1, with s(0)=2

Moo Of Doom

The primes, maybe? :P

kureta

yes. thanks but... well... nevermind

CRGreathouse

The primes = 1 mod 4? The noncomposites? {2*3, 5*7, 11*13, 17*19, ...}?

kureta

does any one know such a set AND its defining formula which gives us the nth term.

CRGreathouse

The primes have many defining formulas, a fair number of which use only basic operations (say, addition, multiplication, factorials, and sine). To what end do you want this?

matt grime

I don't have a reference to hand, but you can consider things like 2^n - 1. Indeed that is one proof that there are infintitely many primes.

Note 2^n - 1 = 2*(2^{n-1} -1) + 1, proving that they are coprime.

kureta

2^2-1=3 and 2^4-1=15 and 15/3=5 so they are not all coprimes am i wrong?

matt grime

Sorry, my mistake - consecutive terms are coprime, not every term. Duh. But there is something to do with things like this that demonstrates infintely many primes by producing coprimes. I don't have the reference to hand (I read it in 'Proofs from the Book' by Aigner and Ziegler).

ramsey2879

a(n) = 5*2^(2n) +5*2^n + 1 is my guess. Prove or disprove
One thing that is certain, they can only be divisible by a prime ending in 1 or 9

ramsey2879

Forget this too, If n = 8 2^n equals 9 and 2^2n = 4 mod 11.
5*(4+9) + 1 = 66. also there are other powers of 2 with the same residue mod 11 so those values for n give a(n) which are not coprime.
Maybe take n to be prime for the n in a(n) or something similar.

kureta

ramsey2879 your reply was the kind that i was looking for. thanks. but taking n to be prime makes this formula useless for me. because my purpose is to find a set of coprimes generated by a simple function . and you should see "a different approach to primes" thread for the reason of my asking for such a set.