[seqfan] Re: prime conjecture

[Robert Gerbicz]

2010/12/29 Dmitry Kamenetsky <dmitry.kamenetsky at rsise.anu.edu.au>

> Hi Robert,
>
> Huge thanks for your input! Your bound is much tighter and I really like
> how
> you've shown that there is only a finite number of failures. Since
> 3*n*log(n)^2 is quite small compared to k*n! I wonder if we can use it to
> find large primes? Either way we should consider how to turn this into a
> sequence.
>
> Cheers,
> Dmitry
>
> ----------------original message-----------------
> From: "Robert Gerbicz" robert.gerbicz at gmail.com
> To: "Sequence Fanatics Discussion list" seqfan at list.seqfan.eu
> Date: Tue, 28 Dec 2010 12:26:38 +0100
> -------------------------------------------------
>
>
> > 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
> >>
> >>
> >>
> >>
> >> _______________________________________________
> >>
> >> 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!+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.
> >
> > _______________________________________________
> >
> > Seqfan Mailing list - http://list.seqfan.eu/
> >
>
>
> One possible sequence could be in a triangle way:
a(n,k)=nextprime(k*n!+1)-k*n! (for n>=1 and 1<=k<=n).

"how you've shown that there is only a finite number of failures"
As I said this is only a heuristic, not a proof. Let p(n,k)=the probability
that [k*n!+1,k*n!+3*n*log(n)^2] contains no prime, then p(n,k)=O(1/n^3), so
sum(n=2,infinity,k=1,n,p(n,k))=sum(n=2,infinity,O(1/n^2))<infinity, then use
http://en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemma from it the
probability that infinitely many of them occur is 0, in other words: by 1
probability we get finite number of failures.