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]
>
>

```

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