Re: Re: Could you prove this proposition:the i-th prime gap p[i+1]-p <=i
- To: mathgroup at smc.vnet.net
- Subject: [mg107282] Re: [mg107269] Re: [mg107156] Could you prove this proposition:the i-th prime gap p[i+1]-p <=i
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Mon, 8 Feb 2010 03:33:30 -0500 (EST)
- References: <c724ed861002030412k2f8008a1x8ce30b426991a812@mail.gmail.com> <201002041127.GAA29855@smc.vnet.net> <A725035C-2B94-425D-8644-FEE4081C4816@mimuw.edu.pl> <c724ed861002052347o335184celaa42b9629cddf85a@mail.gmail.com> <6B29BB02-5CB0-48CB-B4CE-98D8F6B18949@mimuw.edu.pl> <c724ed861002060310s6f102822k12d1037cff3ffd57@mail.gmail.com> <9A89E95D-25FF-41BB-A72F-9AAA9ED1B94F@mimuw.edu.pl> <c724ed861002062011i3f2cc58eud3d7c1614d6ffe68@mail.gmail.com> <201002071114.GAA25359@smc.vnet.net> <52082.98.212.144.5.1265569224.squirrel@webmail.wolfram.com>
Yes, you did not notice, I think, my last post in this thread (or maybe it has not appeared yet). Anyway, I wrote: Note also, that Selberg has shown that if the Riemann hypothesis is true then for almost all x the interval [x,x+ f(x) Log(x)^2] contains a prime, where f(x) is any function that tends to infinity with x. So in your case just take f(x)= PrimePi[x]*Log[x]^2. Since by the prime number theorem PrimePi[x] is asymptotically x/Log[x], f(x) is x*Log[x] which does tend to infinity with x. Andrzej On 7 Feb 2010, at 20:00, danl at wolfram.com wrote: >> >> 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] > >
- References:
- Could you prove this proposition:the i-th prime gap p[i+1]-p[i]<=i
- From: a boy <a.dozy.boy@gmail.com>
- Re: Could you prove this proposition:the i-th prime gap p[i+1]-p[i]<=i
- From: Andrzej Kozlowski <akoz@mimuw.edu.pl>
- Could you prove this proposition:the i-th prime gap p[i+1]-p[i]<=i