Sequence of primes

OP

13

My only breakthrough has been in realizing that all the primes must be distinct.

Reply 2

bobx2001

10

Have you tried using de Moivre?

Reply 3

OP

13

Proof by contradiction: Assume

p_n=5

, then

p_1p_2..p_{n-1}+1=2^a3^b5^c

Since

p_1=2, p_2=3

it is obvious that

a=0,\ b=0

. So

p_1p_2..p_{n-1}+1=5^c

ie

p_1p_2..p_{n-1}=5^c-1

.
So

p_1p_2..p_{n-1}

is a multiple of 4.

This is obviously a contradiction as the LHS contains only one multiple of 2.

QED

Reply 4

OP

13

Is my proof right, guys?

Reply 5

Chaoslord

16

let me get this straight, so P2 = 2, P3 = 2x3 and p4 = 2x3x5?

Reply 6

OP

13

Chaoslord

let me get this straight, so P2 = 2, P3 = 2x3 and p4 = 2x3x5?

p_2

is the greatest prime factor of 2+1 which is 3.

p_3

is the greatest prime factor of 2*3+1 which is 7.

p_4

is the greatest prime factor of 2*3*7+1 which is 43 and so on.

Reply 7

Chaoslord

16

thomaskurian89

p_2

is the greatest prime factor of 2+1 which is 3.

p_3

is the greatest prime factor of 2*3+1 which is 7.

p_4

is the greatest prime factor of 2*3*7+1 which is 43 and so on.

so P5 will be the multiplication of the first 5 primes +1?

mate this is mental xD

Reply 8

OP

13

Chaoslord

so P5 will be the multiplication of the first 5 primes +1?

p_5

is the greatest prime factor of the first 4 primes (of this sequence) + 1.

Reply 9

14

It seems pretty simple to me. Assume we have the first k primes in the sequence and so the product p(1).p(2)...p(k).

We must look for prime factors of p(1).p(2)...p(k)+1.

None of the primes in the list already fit the bill seeing as they don't divide 1.

It must be a prime not on the list.

By nature of the fact that the product on the left is made up of primes, no prime not on the list can divide the product. Also, no prime can divide 1.

Hence we are left with one option; the new prime must be equal to p(1)p(2)...p(k)+1.
The numbers generated are therefore an increasing sequence, hence once there exists a prime p(i)>=r, no j>i exists such that p(j)<r. With r=5, and i=3 say, it is shown.

Reply 10

18

thomaskurian89

QED

Looks good assuming you've done your each prime only occurs once (which is trivial)

Edenr - I don't think you get the question

Reply 11

14

SimonM

Looks good assuming you've done your each prime only occurs once (which is trivial)

Edenr - I don't think you get the question

Where am I going wrong?

EDIT: Ah, I see a flaw in my argument. Please feel free to disregard it :biggrin:

But I do get the question.

Reply 12

rnd

6

Edenr

Where am I going wrong?

For one thing p1p2p3p4...pn+1 is not always prime.

Reply 13

18

Edenr

Where am I going wrong?

p_1 p_2 \ldots p_k+1

is not necessarily prime for a start

Your assumption is also wrong

It really looks like a poor proof of the infinitude of primes

Reply 14

14

SimonM

p_1 p_2 \ldots p_k+1

is not necessarily prime for a start

Your assumption is also wrong

It really looks like a poor proof of the infinitude of primes

Yes, I noticed that flaw. Though my assumption (I assume you mean the one on the first line?) isn't wrong, it's merely the outlining of where I'm starting. Maybe it would be have been clearer if I'd said 'consider'.

Reply 15

18

Edenr

Yes, I noticed that flaw. Though my assumption (I assume you mean the one on the first line?) isn't wrong, it's merely the outlining of where I'm starting. Maybe it would be have been clearer if I'd said 'consider'.

Yes it is; at least it doesn't related to the problem in question

Reply 16

OP

13

SimonM

Looks good assuming you've done your each prime only occurs once (which is trivial)

Thanks.

Reply 17

14

SimonM

Yes it is; at least it doesn't related to the problem in question

Well it does, but it's really a moot point now anyway.

Reply 18

18

Edenr

Well it does, but it's really a moot point now anyway.

On closer inspection, what you've said is just ambiguous. I originally understood it to mean.

Assume the first k prime numbers are in the sequence
but you meant
Assume the first k numbers in the sequence

Reply 19