Register to reply

Infinite set of coprimes

by kureta
Tags: coprimes, infinite, solved
Share this thread:
kureta
#1
Feb2-07, 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(n-1)*(S(n-1)-1)+1, with s(0)=2
Phys.Org News Partner Science news on Phys.org
'Office life' of bacteria may be their weak spot
Lunar explorers will walk at higher speeds than thought
Philips introduces BlueTouch, PulseRelief control for pain relief
Moo Of Doom
#2
Feb2-07, 11:03 AM
P: 367
The primes, maybe? :P
kureta
#3
Feb2-07, 11:06 AM
P: 12
yes. thanks but... well... nevermind

CRGreathouse
#4
Feb2-07, 07:16 PM
Sci Advisor
HW Helper
P: 3,682
Talking Infinite set of coprimes

The primes = 1 mod 4? The noncomposites? {2*3, 5*7, 11*13, 17*19, ...}?
kureta
#5
Feb2-07, 11:15 PM
P: 12
does any one know such a set AND its defining formula which gives us the nth term.
CRGreathouse
#6
Feb3-07, 04:22 PM
Sci Advisor
HW Helper
P: 3,682
Quote Quote by kureta View Post
does any one know such a set AND its defining formula which gives us the nth term.
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
#7
Feb5-07, 04:46 AM
Sci Advisor
HW Helper
P: 9,396
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
#8
Feb5-07, 05:18 AM
P: 12
2^2-1=3 and 2^4-1=15 and 15/3=5 so they are not all coprimes am i wrong?
matt grime
#9
Feb5-07, 12:42 PM
Sci Advisor
HW Helper
P: 9,396
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
#10
Feb5-07, 09:20 PM
P: 894
Quote Quote by kureta View Post
does any one know such a set AND its defining formula which gives us the nth term.
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
#11
Feb6-07, 08:43 AM
P: 894
Quote Quote by ramsey2879 View Post
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
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
#12
Feb6-07, 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.


Register to reply

Related Discussions
Infinite Group Has Infinite Subgroups Calculus & Beyond Homework 7
Infinite Series Calculus & Beyond Homework 2
Infinite capacitors Introductory Physics Homework 2
Infinite series help Calculus & Beyond Homework 2
Are prime numbers infinite? General Math 7