Moebius-Function-Based Sequences

[Leroy Quet]

"mu(k)" is the Moebius (Mobius) function,
defined by:

sum{k=1 to oo} mu(k)/k^r  = 1/zeta(r).

http://www.research.att.com/projects/OEIS?Anum=A008683


I do not believe this (interesting, in my opinion) sequence is in the EIS 
yet:

a(n) = smallest integer > n where
mu(n) = mu(a(n)).

6, 3, 5, 8, 7, 10, 11, 9, 12, 14, 13, 16, ...

-

Also, a related sequence:

b(n) = a(n) - n:

5, 1, 2, 4, 2, 4, 4, 1, 3, 4, 2, 4,...

In regards to {b(n)}, do the terms of this sequence approach a finite 
average?

ie. does x =
limit{m->oo} (1/m) sum{n=1 to m} b(n)

exist?

Does x have a closed form?

--

Now, the sequences of n's where
mu(n) = m(n+1) IS in the EIS.

http://www.research.att.com/projects/OEIS?Anum=A064148

But the actual values of mu(c(n)), where c(n) is the n_th term of 
A064148, is not in the EIS.

-1, 0, 1, 1, 0, 0, -1, -1 ,...

--

Finally, I do not believe the following is in the EIS either:

d(n) = Number of positive divisors k of n, where

mu(k) = 1 and  mu(n/k) = -1.

0, 1, 1, 0, 1, 0, 1, 0, 0, 0,...

I get the relation (hopefully correct):

4*d(n) + sum{k|n} mu(k)*mu(n/k) =

product{p|n} e(p,n),

where the product is over the distinct primes dividing n;
e(p,n) = 2 if p|n but p^2 does not divide n;
e(p,n) = 1 if p^2|n but p^3 does not divide n;
e(p,n) = 0 if p^3|n.

thanks,
Leroy Quet