MathGroup Archive 2010

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Re: Could you prove this proposition:the

  • To: mathgroup at smc.vnet.net
  • Subject: [mg107281] Re: [mg107269] Re: [mg107156] Could you prove this proposition:the
  • From: danl at wolfram.com
  • Date: Mon, 8 Feb 2010 03:33:19 -0500 (EST)
  • References: <c724ed861002030412k2f8008a1x8ce30b426991a812@mail.gmail.com>

>
> On 7 Feb 2010, at 05:11, a boy wrote:
>
>> It has been proved that there exists at least a prime in the interval
>> (n,2n).
>
> There are of course much better results than this one (Bertrand's
> postulate), although they are either asymptotic or true for some n
> larger than some fixed positive integer. But they all of this type, in
> other words they do not  Prime[k] in their statements. If you want to
> see what results with Prime[k] look like you can see here:
>
> http://en.wikipedia.org/wiki/Prime_gap
>
> Look for Rankin's result on the lower bound for the prime gap. It's a
> lower bound and is vastly more complicated than what you are proposing
> (although your conjecture is almost certainly weaker than Andrica's
> conjecture on the same page).
>
> Anyway: good luck.
>
> Andrzej Kozlowski

Well, it's not entirely hopeless, or at least not obviously so. The "Upper
bounds" section of that Wikipedia page suggests that the conjecture under
consideration (to wit, that prime(k+1)-prime(k)<k for all k) should hold
at least for all sufficiently large k.

Specifically, I think this follows from the work of Hoheisel and later
refinements such as Ingham's. In particular, the gap is shown to b
eeventually less than prime(k)^(4/5) (I use 4/5 as a value strictly
greater than 3/4). I believe the Prime Number Theorem guarantees that this
in turn is less than k for sufficiently large k; see

http://en.wikipedia.org/wiki/Prime_number_theorem

In particular see the error term bounds in comparing primepi(k) to
logintegral(k). I could be mistaken but I think they will imply the needed
inequality relating less-than-one powers of prime(k) to k.

Daniel Lichtblau
Wolfram Research


>> p[i+1]-p[i]<=i  iff there exists at least a prime in the interval
> (n,n+Pi(n)]
>> This is an improvement for the upper bound of prime gap, so I think it
> is not very difficult.
>> For the simpleness and elegance of the form p[i+1]-p[i]<=i, I think
> someone can prove this. We should be more optimistic!
>>
>> On Sat, Feb 6, 2010 at 10:11 PM, Andrzej Kozlowski <akoz at mimuw.edu.pl>
> wrote:
>> > I think it is not difficult to prove the proposition,but I can't do
> this still.
>>
>> You think or you hope? I think it is going to be extremely difficult
> to prove it and the reason is that nothing of this kind has been proved
> even though other people also have computers and eyes. There are some
> very weak asymptotic results and there are conjectures, for which the
> only evidence comes from numerical searches. The best known is Andrica's
> conjecture which states that  Sqrt[Prime[i+1]]-Sqrt[Prime[i]]<1 and
> appears to be stronger than yours, but nobody has any idea how to prove
> that. In fact, nobody can prove that
> Limit[Sqrt[Prime[n+1]]-Sqrt[Prime[n]],n->Infinity]=0 (this has been
> open since 1976), and in fact there is hardly any proved statement of
> this kind. So what is the reason for your optimism?
>>
>> Andrzej Kozlowski
>>
>>
>> On 6 Feb 2010, at 12:10, a boy wrote:
>>
>> > Yes,I want the proof of the fact that p[i+1]-p[i]<=i.
>> > I think it is not difficult to prove the proposition,but I can't do
> this still.
>> > If he or she give me a proof , I will be very happy and appreciate
> him or her!
>> >
>> > On Sat, Feb 6, 2010 at 6:50 PM, Andrzej Kozlowski
> <akoz at mimuw.edu.pl> wrote:
>> > Oh, I see. You meant you want the proof of the fact that
> p[i+1]-p[i]<=i? I misunderstood your question I thought you wanted to
> see the trivial deduction of the statement you had below that.
>> >
>> > But, considering that practically nothing is known about upper
> bounds on prime number gaps p[i+1]-p[i] in terms of i (all known results
> involve bounds in terms of p[i] and these are only asymptotic), this
> kind of proof would be a pretty big result so, in the unlikely event any
> of us could prove it, would you except him or her just to casually post
> it here?  ;-)
>> >
>> > Andrzej Kozlowski
>> >
>> >
>> >
>> > On 6 Feb 2010, at 08:47, a boy wrote:
>> >
>> > > When I was observing the prime gaps, I conjectured
>> > > p[i+1]-p[i]<=i
>> > >
>> > > This means there is at least a prime between the interval
> (n,n+Pi(n)].  I verified this by Mathematica and searched in web, but I
> can't prove this yet.
>> > >
>> > > On Sat, Feb 6, 2010 at 4:17 AM, Andrzej Kozlowski
> <akoz at mimuw.edu.pl> wrote:
>> > > Hmm... this is a little weird - how come you know this if you
> can't prove it? This is one of those cases where knowing something is
> essentially the same as proving it... but anyway:
>> > >
>> > > p[n]-p[1] = (p[n]-p[n-1]) + (p[n-1]-p[n-2]) + ... + (p[2]-p[1])
> <= (n-1)+ (n-2) + ... + 1 == (n-1) n/2
>> > >
>> > > hence
>> > >
>> > > p[n]<= p[1]+ (n-1)n/2 = 2 + (n-1)n/2
>> > >
>> > > Andrzej Kozlowski
>> > >
>> > >
>> > > On 4 Feb 2010, at 12:27, a boy wrote:
>> > >
>> > > > Hello!
>> > > > By my observation, I draw a conclusion: the i-th prime gap
>> > > > p[i+1]-p[i]<=i
>> > > > Could you give me a simple proof for the proposition?
>> > > >
>> > > > p[i+1]-p[i]<=i  ==>  p[n]<p[1]+1+2+..+ n-1=2+n(n-1)/2
>> > > >
>> > > > Mathematica code:
>> > > > n = 1;
>> > > > While[Prime[n + 1] - Prime[n] <= n, n++]
>> > > > n
>> > > >
>> > > > Clear[i];
>> > > > FindInstance[Prime[i + 1] - Prime[i] > i && 0 < i, {i}, =
> Integers]




  • Prev by Date: Re: Re: DeleteDuplicates is too slow?
  • Next by Date: Re: Re: Could you prove this proposition:the i-th prime gap p[i+1]-p <=i
  • Previous by thread: Re: Derivative of Interpolation function is highly-oscillatory
  • Next by thread: Test function argument with VectorQ and NumericQ