Moebius Mu function

[Pieter Moree]

Dear list,

Ed Pegg and myself are talking about different densities below.

Ed Pegg is consideirng the density of INTEGERS n for which say
\mu(n)=1.

I am considering the density of PRIMES p for which \mu(p-1)=1 say.

It's actually a statement
equivalent with the prime number theorem that
the density equals 3/Pi^2 in the integer case.

To be more
precise: we have \sum_{n\le x: \mu(n)=1}1=3x/Pi^2+o(x) as x tends
to infinity.

It's elementary to show that \sum_{n\le x}\mu(n)^2=6x/Pi^2+o(x)
as x tends to infinity. By an equivalent form with the PNT we
have \sum_{n\le x}\mu(n)=o(x). Combination of these two results
yields the claims in the INTEGER case.

(I am not concerned with getting sharp error terms here, which
is possible.)

Now the PRIME number case (once more, different formulation...)
p is used to indicate primes.

Theorem (Leon Mirsky): \sum_{p\le x}\mu(p-1)^2=Ax/\log x +o(x/\log x)
as x tends to infinity, with A Artin's constant.
Amer. Math. Monthly 56 (1949), 17-19.

Conjecture: \sum_{p\le x, \mu(p-1)=1}1=(A/2)x/\log x + o(x/\log x)
\sum_{p\le x, \mu(p-1)=-1}1=(A/2)x/\log x + o(x/\log x)

Note that the conjecture implies Mirsky's result.

The idea is that if p is a prime such that p-1 is squarefree there is
no preference of having an even number of factors over having an odd
number of factors, i.e. p-1 should in this sense behave as an
`random integer'.

Profs Elliott and Friedlander informed me that they regard this
conjecture as quite deep.

Bests,
Pieter Moree

>>An unsolved question regarding this assertion is: does the set
>>of primes for which \mu(p-1)=1 (say) has a density, d(1) ?
>>
>>
> Yes... 3/Pi^2
>
>>It is a result that goes back to Mirsky that the set of primes p
>>for which p-1 is squarefree has density A, where A denotes the
>>Artin constant (A=\prod_q (1-1/(q(q-1)), q running over all primes).
>> Numerically A=0.3739558136....
>>
>>
> The density is 6/Pi^2.  See my article.
>
>>Conjecture: d(-1)=d(1)=A/2.
>>
>>
> d(-1)=d(1)=3/Pi^2
>
> --Ed Pegg Jr.