# [seqfan] Re: prime conjecture

RGWv rgwv at rgwv.com
Tue Dec 28 14:54:02 CET 2010

```?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) there is at least one prime in the interval
> [k*n!+1, k*n!+n*n].
>
>
> I've checked that this conjecture works for all n<=18, but I cannot prove
> 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
>
>
>
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>

Yes, that is very likely to be true.
My tighter bound: for every n>1 and 1<=k<=n there is a prime in
[k*n!+1,k*n!+3*n*log(n)^2] interval. (this is true for n<=100, proved by
PARI-GP),

n!<=n^n so log(n!)<=n*log(n) (By Stirling formula we can't get much better.)
>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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