[seqfan] Re: 5, 20, 120, 540, 6480, ...
Robert Gerbicz
robert.gerbicz at gmail.com
Wed Mar 30 23:22:35 CEST 2022
"I'm not so sure that it couldn't happen that a prime gap of the form
2p, with some medium-sized prime, could occur before, or more often, than
the factor p in p(n)+p(n+1), at a certain point."
That is pretty unlikely, even using weak conjectures. But use sharper
conjecture to simplify a little:
p(n+1)-p(n)<log(p(n)^2)
if q is in the denominator appears for the first time and:
p(n+1)-p(n)=2*q=log(p(n))^2 large prime gap occurred [this is the "worst"
case], then
p(n)=exp(sqrt(2*p(n)), the problem is that up to N=p(n) we would see that
the exponent of q in the numerator is likely in the range
exp(sqrt(2*p(n)))/p(n)^1.5 and this goes to infinity.
A numerical example: 2*191=10726905041-10726904659 [this is a maximal prime
gap], and of course up to this ~1e^10 limit you'd see many times that
p(n+1)+p(n) is divisible by 191.
M. F. Hasler <oeis at hasler.fr> ezt írta (időpont: 2022. márc. 30., Sze,
22:47):
> I just noticed that :
> X = { n such that p(n+1)+p(n) is not divisible by p(n+1) - p(n) }
> = { 4, 6, 8, 9, 11, 12, 14, 15, 16, 18, 19, 21, 22, 23, 24, 25, 27, 29,
> 30, 31, 32, ... }
> =?= A049579 : Prime subscripts for which residue of (prime(n)-1)!+1
> modulo prime(n)+2 equals 1.
> (equality is only apparent, but not obviously / proven to be true:
> I don't see immediately why this is the same, because the prime(n+1) does
> not at all enter the definition of a(n) in the second case...)
>
> Also notice that according to the first terms of X, it is rather rare that
> ( p(n+1)+p(n) ) / ( p(n+1) - p(n) )
> is non-integer, but later this is almost always the case! For that reason,
> I'm not so sure that it couldn't happen that a prime gap of the form
> 2p, with some medium-sized prime, could occur before, or more often, than
> the factor p in p(n)+p(n+1), at a certain point.
>
> -Maximilian
>
> On Wed, Mar 30, 2022 at 1:29 PM Robert Gerbicz <robert.gerbicz at gmail.com>
> wrote:
>
> > Hi !
> >
> > See:
> >
> >
> https://terrytao.wordpress.com/2016/03/14/biases-between-consecutive-primes/
> > and from that page: https://arxiv.org/pdf/1603.03720.pdf conjecture 1.1
> > what they have in conjecture 1.1 is that for consecutive p1,p2 primes you
> > will see p2+p1 is divisible by q more often than p2-q1.
> > [because in the latter case p2==p1==a mod q, while in the other case p2
> > and p1 are in different residue classes if q>2].
> > You'd still need effective constants on that conjecture's bounds, but at
> > least we see why this should be true,
> > notice also that p(n+1)-p(n) is "small", so you could prove that the
> > product is an integer up to a pretty large bound, just factorize the
> terms
> > using prime up to L, if p(n+1)-p(n)<=L is true for all n<=N.
> >
> >
> > Tomasz Ordowski <tomaszordowski at gmail.com> ezt írta (időpont: 2022.
> márc.
> > 30., Sze, 17:07):
> >
> > > Dear readers!
> > >
> > > Let a(n) = Product_{k=1..n}
> (prime(k+1)+prime(k))/(prime(k+1)-prime(k)).
> > > Conjecture: a(n) is an integer for every natural n.
> > > Is it known or provable?
> > >
> > > Best regards,
> > >
> > > Thomas Ordowski
> > >
> > >
> > > <#m_-7681193058414376069_DAB4FAD8-2DD7-40BB-A1B8-4E2AA1F9FDF2>
> > >
> > > --
> > > Seqfan Mailing list - http://list.seqfan.eu/
> > >
> >
> > --
> > Seqfan Mailing list - http://list.seqfan.eu/
> >
>
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