[seqfan] Re: Inequalities for Euler's phi(n) and the Riemann hypothesis

jean-paul allouche jean-paul.allouche at imj-prg.fr
Tue Dec 20 16:55:43 CET 2016


Dear Vladimir, dear all

May be there is something to do in the same spirit
with Robin's criterion for the Riemann hypothesis
(for this criterion see, e.g.,
http://www.numdam.org/item?id=JTNB_2007__19_2_357_0
)

best
jean-paul

Le 20/12/16 à 12:58, Vladimir Shevelev a écrit :
> Dear SeqFans,
>
> Define P=e^gamma*loglog(n), where gamma is
> Euler's constant.
> In 1909, Landau proved that for each eps>0,
> there exist infinitely many n for which phi(n)< n/P',
> where in P' e^gamma is replaced by e^(gamma-eps).
> In 1983  J.-L. Nicolas beautifully strengthened Landau's
> result showing that there exist infinitely many n for
> which phi(n)<n/P.  In connection with this I submitted
> A279229  which lists numbers n for which phi(n) < n/P.
> It begins 3, 4, 5, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, ...
> Denote, further, c(n) = (n/phi(n) - P)*sqrt(log(n)).
> In 2012 Nicolas proved that the Riemann
> hypothesis is equivalent to the inequality: for n>=2,
> c(n)<=c(N), where N is the product of the first 66
> primes, such that c(N)=4.0628356921...
> Let n be in A279229, such that c(n)>0. I submitted
> sequence A279291: {floor(c(n))} beginning with
> 1, 1, 0, 2, 1, 0, 1, 2, 0, 0, 0, 1, 0, 0, 1, 0, 0, 2,...
> By Nicolas' result, assuming the R. H., we see that
>   a(n)<=4.
> On the other hand, I conjecture that a(n)<=4 is an
> absolute result which is independent of the validity
> of the R. H. If this conjecture is true, then the
> statement that the R. H. is false is equivalent to the
> existence of n for which c(n) is in interval (c(N),5).
>
> I was surprised that neither A279229 nor A279291
> were not in OEIS.
>
> All remarks are welcome.
>
> Best regards,
> Vladimir
>
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