# [seqfan] Re: prime conjecture

Dmitry Kamenetsky dmitry.kamenetsky at rsise.anu.edu.au
Tue Dec 28 15:17:59 CET 2010

Hi Bob,

It's funny that you mentioned that. I always felt that the bounds in A060715
are rather loose. So I tried to tighten them and hence came up with this
conjecture.

Dmitry

----------------original message-----------------
From: "RGWv" rgwv at rgwv.com
To: "Sequence Fanatics Discussion list" seqfan at list.seqfan.eu
Date: Tue, 28 Dec 2010 08:54:02 -0500
-------------------------------------------------

> ?You should probably look at A060715. Bob.
>
> -----Original Message-----
> From: Robert Gerbicz
> Sent: Tuesday, December 28, 2010 6:26 AM
> To: Sequence Fanatics Discussion list
> Subject: [seqfan] Re: prime conjecture
>
> 2010/12/28 Dmitry Kamenetsky dmitry.kamenetsky at rsise.anu.edu.au
>
>> Hello all,
>>
>> First of all Happy New Year to everyone! Now for some serious stuff. I
>> have
>> the following conjecture:
>>
>> For every n>=1 and k (1> [k*n!+1, k*n!+n*n].
>>
>>
>> I've checked that this conjecture works for all n> it
>> for all n. Can anyone help me? Also it would be nice to get a tighter
>> bound
>> on the upper range.
>>
>>
>> Sincerely,
>> Dmitry Kamenetsky
>>
>>
>>
>>
>> _______________________________________________
>>
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>>
>
> Yes, that is very likely to be true.
> My tighter bound: for every n>1 and 1 [k*n!+1,k*n!+3*n*log(n)^2] interval.
(this is true for n PARI-GP),
>
> n!<=n^n so log(n!) From prime number theorem the probability that a random
n is prime is about
> 1/log(n). So the probability that the above interval contains no prime is
> about (assuming that these numbers are random):
> (1-1/(n*log(n)))^(3*n*log(n)^2)~exp(-3*log(n))=1/n^3, we have n choices
> for
> k value when n is fixed. That means about 1/n^2 probability that we fail
on
> n. But sum(n=2,infinity,1/n^2) converges, from this we can expect only
> finite number of n,k pairs that my conjecture is false. Obviously it is
only
> a heuristic.
>
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>
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>
>
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>

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