## 24344Re: [PrimeNumbers] polynomial complexity conjecture

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• Aug 1, 2012
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> 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

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

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