Discovering Primes: The Relationship Between p and 2p + 2

[Sariaht]

P_c2 + 4

I found that there is a prime in between p and 2p + 2.

I also found that if p is a prime and p + q is a prime, then there should be a prime between p + q and p + 2q + 2, if this number is included.

 

[Muzza]

You might want to replace "speach" (sic) with "number"... I'm guessing that the Anglos might get a bit confused by that Swedish-ism ;)

 

[matt grime]

does the proof run like this:

the case of p a prime itself is true and there is a famous proof by Erdos that there if p is prime then the next prime is less than 2p.

if r is not prime, let p be the largest prime in the set 0,1,..,r (assume r not 1 where the result is trivial), then there is a prime between p and 2p, call it q. this prime cannot be less than r and we see

r<q<2p<2r<2r+2

sorry, I seem to have improved your bound in that.


haven't thought abuot the second claim you make

 

[Sariaht]

------------------/\---------------------
-----------------/\/\--------------------
----------------/\/\/\-------------------
---------------/\/\/\/\------------------
--------------/\/\/\/\/\-----------------
-------------/\/\/\/\/\/\----------------

In this pattern, the slasch-counter includes a prime in every new line. my theory is that if you draw the lines in the right way, two lines in a row can form primes.

In other words, if p is a prime and p + q is a prime, then there is a prime between p+q and p+2q+2

Maybe i should skip 2 in this equation?

--------/\---------2
-------/-----------3
------/\-----------5
-------/\----------7
------/\/\---------11
---------/\--------13
--------/\/\-------17
-----------/\------19
----------/\/\-----23
---------/\/\/\----29
--------/\---------31
Not really the same pattern here. This one ends at thirtyone.

Nope, i don't think so, and that's why i thought i was the first guy to come up with the equation.

 

[matt grime]

I must apologize in that Erdos's proof mentioned above does not presume p is a prime, it shows that given any p there is a prime between p and 2p (bertrand's postulate)

erdos's proof examines binomial coeffs, if you care.

 

[Sariaht]

Originally posted by matt grime
I must apologize in that Erdos's proof mentioned above does not presume p is a prime, it shows that given any p there is a prime between p and 2p (bertrand's postulate)

erdos's proof examines binomial coeffs, if you care.

47 is a prime p.

59 (p + q) is another prime.

But is there a prime between p + q and p + 2q (59 and 81) the last number included?

 

[matt grime]

Well, yes, but I didn't say anything about that.

I doubt the second conjecture you made is true, not least because it fails at 29/31 both 33 and 35 being composite.

 

[Sariaht]

Originally posted by matt grime
Well, yes, but I didn't say anything about that.

I doubt the second conjecture you made is true, not least because it fails at 29/31 both 33 and 35 being composite.

Your right. What if q is not a prime?

29 and 37 works fine.

 

[matt grime]

Even more unlikely, I'd've said. Let me rewrite it:

let s and t be prime (s<t and not differing by2) then you conjecture that there is a prime between t and t +(t-s)+2 which is a very unlikely proposition, as it must be true for all s and t.

In particular since there is a prime between [t/2]-1 and t, get that for any prime t there is a prime between t and [t/2]+3, which is a considerable improvement on Russell's postulate. That of course isn't evidence it is wrong, try and find some literature about it. But you claimed to have a proof of it so post that. Small counter examples don't spring to mind, but it it does imply since there 19 and 23 are primes differing by 4, the next is no more than 6 away from 23, the one after that no more than 8 after that, then 10 and so on. This seems very unlikely, and that for of thinking occurs every time we get primes differing by 4, eg 97 and 103.

 

[Sariaht]

If there is a prime between a and 2a, then there is a prime between between diff(p,q) and 2diff(p,q). let's say the funktion we use to get primes has an inverted funktion which has a derivata, able to have many values for one special variable, only no primes. Then there could possibly be a prime between the prime p + q and p + 2q (included), p being a prime and q not. Ofcourse that's farfetched.

But since delta x is f(p-q), and x is p and q, well, you must admit I tried.

 

[matt grime]

Originally posted by Sariaht
If there is a prime between a and 2a, then there is a prime between between diff(p,q) and 2diff(p,q). let's say the funktion we use to get primes has an inverted funktion which has a derivata, able to have many values for one special variable, only no primes. Then there could possibly be a prime between the prime p + q and p + 2q (included), p being a prime and q not. Ofcourse that's farfetched.

But since delta x is f(p-q), and x is p and q, well, you must admit I tried.

1) what function we use to get primes.

2) it will be defined on the integers and therefore continuity and differentiation are only formal concepts at best.

 

[Sariaht]

Originally posted by matt grime
1) what function we use to get primes.

2) it will be defined on the integers and therefore continuity and differentiation are only formal concepts at best.

What about this then (instead):

If p + 2 is a prime, then p + 4 cannot be (except for 3,5,7).

If that’s true, If p + 4 is a prime, p + 8 cannot be (except for 3,7,11)

Etc.

If p + 2ⁿ is a prime, then p + 2^{n + 1} cannot be (except for one case).

 

[matt grime]

But that's trivial for p prime greater than 5, since p is 1 or two mod three, p+2 then is 0 or 1 mod three. to both be prime it must be that p is 2 mod 3 and p+2 is 1 mod 3, and then 3 divides p+4. the case for arbitrary n is safely left to the reader.

 

[Sariaht]

It works for every n, i checked in the primelist. Your proof is excellent. Is this ever used?
Perhaps not. by the way, can you inverse the primefunktion? It sounds like a hard task.

 

[matt grime]

My proof is just the observation that given three consecutive odd numbers one of the must be divisilbe by three. the second part can be extended - if p is prime (greater than 5) and n and m are an odd an even number resp then at most one of p+2^n and p+2^m are prime, or more accurately if p is any odd number then at least on of those mentionend is divisible by three, that's all. you could do similar things: for any 4 consecutive odd numbers at least one is divisble by five and so on.

what is the 'primefunktion' you want to invert?

 

[Sariaht]

The one that Hurkyl talked about for a while.

That one that gives you all the primes.

 

[matt grime]

Originally posted by Sariaht
The one that Hurkyl talked about for a while.

That one that gives you all the primes.

well that isn't in thsi thread and I've not seen it, i presume (from a quick search) you mean the function p(n) from N to N which sends n to the n'th prime. This cannot be invertible since it is not a bijection. A function is invertible iff it is a bijection - a formal inverse of any suirjection exists but it is not a FUNCTION necessarily in that it it sends elements to sets of elements not a unique element as a function would need to do.

 

[Sariaht]

I've read about a reversed primefunction though, giving the number of the prime. I bet the inverse did or will tie together some lose ends in the primedistribution. There is some small equation lying there. Definitely.

 

[matt grime]

Originally posted by Sariaht
I've read about a reversed primefunction though, giving the number of the prime. I bet the inverse did or will tie together some lose ends in the primedistribution. There is some small equation lying there. definitely.

I think I'd say there was a very big function there.