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**[SOLVED] infinite set of coprimes**

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

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- Thread starter kureta
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- #1

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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

- #2

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The primes, maybe? :P

- #3

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

- #4

CRGreathouse

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

- #5

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

- #6

CRGreathouse

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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?

- #7

matt grime

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

- #8

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

- #9

matt grime

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- #10

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a(n) = 5*2^(2n) +5*2^n + 1 is my guess. Prove or disprovedoes any one know such a set AND its defining formula which gives us the nth term.

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

- #11

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Forget this too, If n = 8 2^n equals 9 and 2^2n = 4 mod 11.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

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

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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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