
#1
Feb207, 09:43 AM

P: 12

does any one know an infinite set of coprimes except for the elements of sylvester's sequence. S(n)=S(n1)*(S(n1)1)+1, with s(0)=2




#2
Feb207, 11:03 AM

P: 367

The primes, maybe? :P




#3
Feb207, 11:06 AM

P: 12

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




#4
Feb207, 07:16 PM

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P: 3,680

[SOLVED] infinite set of coprimes
The primes = 1 mod 4? The noncomposites? {2*3, 5*7, 11*13, 17*19, ...}?




#5
Feb207, 11:15 PM

P: 12

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




#6
Feb307, 04:22 PM

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#7
Feb507, 04:46 AM

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P: 9,398

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^{n1} 1) + 1, proving that they are coprime. 



#8
Feb507, 05:18 AM

P: 12

2^21=3 and 2^41=15 and 15/3=5 so they are not all coprimes am i wrong?




#9
Feb507, 12:42 PM

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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).




#10
Feb507, 09:20 PM

P: 891

One thing that is certain, they can only be divisible by a prime ending in 1 or 9 



#11
Feb607, 08:43 AM

P: 891

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. 



#12
Feb607, 09:50 AM

P: 12

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.



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