- Conjecture:

This polynomial expression is always zero for all triplets of consecutive prime numbers greater than 23.

Product[k Prime[n+1]+(m-k)*Prime[n-1]-m Prime[n],

{m,2,Log[n]^(5-Pi)},{k,1,m-1}]=0

It isinteresting to note that the order of complexity of this expression is less than

log[n] ^ 4 and hence is a expression with polynomial complexity.

Sincerely

Sebastian Martin Ruiz

[Non-text portions of this message have been removed] - I am pretty sure, but cannot be certain, that this was explained in Paul Underwood's recent reference to Richard's video, expounding and expanding Emil Artin's theories regarding the counting of congruencies, in relation to recipricosity.

If I am wrong, I apologise, but on the surface, it seems to be.

Sent from my iPad

On 1 Aug 2012, at 18:52, Sebastian Martin Ruiz <s_m_ruiz@...> wrote:

> Conjecture:

>

> This polynomial expression is always zero for all triplets of consecutive prime numbers greater than 23.

>

> Product[k Prime[n+1]+(m-k)*Prime[n-1]-m Prime[n],

> {m,2,Log[n]^(5-Pi)},{k,1,m-1}]=0

>

>

> It isinteresting to note that the order of complexity of this expression is less than

> log[n] ^ 4 and hence is a expression with polynomial complexity.

>

> Sincerely

>

> Sebastian Martin Ruiz

>

> [Non-text portions of this message have been removed]

>

>

[Non-text portions of this message have been removed] - I've been watching the video and are different things.

________________________________

De: "bobgillson@..." <bobgillson@...>

Para: Sebastian Martin Ruiz <s_m_ruiz@...>

CC: "primenumbers@yahoogroups.com" <primenumbers@yahoogroups.com>

Enviado: Miércoles 1 de agosto de 2012 20:41

Asunto: Re: [PrimeNumbers] polynomial complexity conjecture

I am pretty sure, but cannot be certain, that this was explained in Paul Underwood's recent reference to Richard's video, expounding and expanding Emil Artin's theories regarding the counting of congruencies, in relation to recipricosity.

If I am wrong, I apologise, but on the surface, it seems to be.

Sent from my iPad

On 1 Aug 2012, at 18:52, Sebastian Martin Ruiz <mailto:s_m_ruiz%40yahoo.es> wrote:

> Conjecture:

>

> This polynomial expression is always zero for all triplets of consecutive prime numbers greater than 23.

>

> Product[k Prime[n+1]+(m-k)*Prime[n-1]-m Prime[n],

> {m,2,Log[n]^(5-Pi)},{k,1,m-1}]=0

>

>

> It isinteresting to note that the order of complexity of this expression is less than

> log[n] ^ 4 and hence is a expression with polynomial complexity.

>

> Sincerely

>

> Sebastian Martin Ruiz

>

> [Non-text portions of this message have been removed]

>

>

[Non-text portions of this message have been removed]

[Non-text portions of this message have been removed] > Conjecture:

Rewriting the conjecture into a bit more readable form, we get the

> �

> This polynomial expression is always zero for all triplets of consecutive prime numbers greater than 23.

> �

> Product[k Prime[n+1]+(m-k)*Prime[n-1]-m Prime[n],

> {m,2,Log[n]^(5-Pi)},{k,1,m-1}]=0

following claim (where p[n] denotes n-th prime number):

For n >= 10, if we express (p[n+1] - p[n-1])/(p[n] - p[n-1]) as

reduced fraction, its numerator will not exceed Log(n)^(5-Pi).

In term of prime gaps (where g[n] := p[n+1] - p[n]), the claim is

equivalent to (g[n]+g[n-1])/gcd(g[n],g[n-1]) <= Log(n)^(5-Pi).

Finally, taking logarithms, it can be rewritten as

r(n) = Log[(g[n]+g[n-1])/gcd(g[n],g[n-1])] / Log[Log(n)]] <= 5-Pi.

The conjecture itself looks both strange and weak at the same time; it

asserts a quite strong property of two consecutive prime gaps, but since

it deals with two gaps rather than a single one, there is no obvious

relation to the other, well-known conjectures or theorems. For example,

on one hand, Cramer's conjecture on gaps runs a bit short of yours -- its

exponent is 2 and yours is smaller than that. On the other hand, it

applies to all gaps uniformly, while yours only talks about two, implying

the second one cannot be much bigger than the first.

Numerically, 5-Pi is a bit more than 1.8584 and the closest challengers

I've found so far are:

n = 5949:

p[n-1] = 58789

p[n] = 58831

p[n+1] = 58889

g[n-1] = 42

g[n] = 58

r(n) = Log(50)/Log(Log(5949)) =

1.8092074158688967510883160301868016764

n = 8040878

p[n-1] = 142414553

p[n] = 142414669

p[n+1] = 142414859

g[n-1] = 116

g[n] = 190

r(n) = Log(153)/Log(Log(8040878)) =

1.8184569954230012429937644772045398171

Still seems to be in the "safe" region and my gut feeling is that the

conjecture might be true. [yes, law of small numbers in practice ;-) ]

Peter

[Non-text portions of this message have been removed]