Re: Minima of Phi(m) over m

Below is a copy of an answer to my posting
about phi(m)/m from Gerd Verbouwe (Belgium).
For technical reasons, he could not get it
posted here, and he e-mailed it to me the day
of my posting, on June 30, 2005. After a long
delay due to emergencies, I checked his answer,
and with his agreement, I submit it for posting
here.

Jean-Claude Evard
Department of Mathematics
Western Kentucky University
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First, I recall the notation that I used:
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Phi(m) = [The Euler`s function]
= [The number of integers k
such that 0<k<m and gcd(k,m)=1].

r(m) = [The ratio phi(m) over m].

primorial = product of the smallest primes:
2# = 2 primorial = 2,
3# = 3 primorial = 2*3 = 6,
5# = 5 primorial = 2*3*5 = 30,
7# = 7 primorial = 2*3*5*7 = 210,
........
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Here is a copy of the answer from
Gerd Verbouwe of June 30, 2005:
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Sorry for replying personally, but my post
to the sci.math.research newsgroup didn't
come through. About

>Conjecture 1. The relative minimum values of r(m) occur at the points
>where m is a primorial, and the sequence of values of r(m) at these
>points is strictly decreasing.

Isn't observing that
r(m) = Phi(m) / m = \prod_{primes p_i | m} (1 - 1/p_i) (*)
enough?

Say p# < m < q# (p prime, q the next prime larger than p).

[a] The number of distinct prime factors of m is less than
or equal to \pi(p) = the number of prime factors of p#.
(because p# is the smallest number with \pi(p) prime factors)
(or "the number of distinct prime factors of m is strictly smaller
than \pi(q)".)
[b] Since all factors in (*) are less than one; if a prime P|m,
P > p occurs, it can be replaced by a smaller one, hence
no minimum in this case.
[c] Hence (*) is a relative minimum, when as much factors
(1-1/prime) as possible occur, and since 1-1/P > 1-1/p for
P>p, one needs the first N primes… . Of course these are
"local" minima, since, e.g. r(2* 17#) = r(17#) and (*)
shows immediately that it is strictly decreasing.

(Also,
>Conjecture 2. The relative maximum values of r(m) occur at the points
>where m is a prime, the sequence of values of r(m) at these points is
>strictly increasing, and its limit when m goes to infinity is 1.
becomes clear: it's maximal when only one factor (1-1/p) appears,
this happens first for the prime p itself; the limit of these is of course 1.)

Am I missing something here? gv
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I think that this proof is 100% OK. I have not checked
my third conjecture yet, and I intend to come back to
all this during next summer. Jean-Claude Evard

.