[seqfan] Re: (prime(n)*prime(n+1)) mod prime(n+2) is odd
Giovanni Resta
giovanni.resta at iit.cnr.it
Wed Jun 7 11:32:48 CEST 2017
On 06/07/2017 01:54 AM, zak seidov via SeqFan wrote:
> (prime(n)*prime(n+1)) mod prime(n+2) is odd.
> n = 1, 2, 5, 7, 10, 14, 15, 23, 29, 46, 61.
> Finite? Full?
Probably finite and full.
If you write
prime(n)=p, prime(n+1)= p + g and prime(n+2)=p+g+h,
(i.e., g and h are the gaps between prime(n), prime(n+1), and prime(n+2),
then
the quotient of p(p+g) divided by p+g+h is
p-h and the remainder is g*h+h^2.
In general the value of g*h+h^2 is even.
To be odd, we need that g*h+h^2 > p+g+h=prime(n+2), so that
the real remainder is not g*h+h^2 but
g*h+h^2 - prime(n+2), which is odd.
But this could reasonably happen only when the primes involved
are small, and thus the gaps between them are large in comparison.
When the primes are larger, we know that, very roughly speaking, we can
expect the gaps to be about O((log p)^2), so the expression g*h+h^2
can't be larger than prime(n+2).
I'm not up-to-date about results on prime gaps, so I don't know
if this can be proved. Probably it can be shown it is implied
by one of the many conjectures about prime gaps.
Giovanni
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