a(m+1)=sum{k|m}mu(k)a(k), & Hidden Negatives

[Edwin Clark]

On Fri, 20 Aug 2004, Alec Mihailovs wrote:

> > From: Leroy Quet [mailto:qq-quet at mindspring.com]
> > Sent: Tuesday, August 17, 2004 3:04 PM
>  
> > Let a(1) = 1;
> > 
> > Let a(m+1) = sum{k|m} mu(k) a(k),
> > 
> > where mu is the Mobius (Moebius) function.
> > 
> 
> a:=proc(n) option remember;
> add(numtheory[mobius](i)*a(i),i in numtheory[divisors](n-1)) end:
> a(1):=1:
> seq(a(n),n=1..100);
> 
> 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, -1, 0, 0, 1, 1, 0, 1, 1, 0, 1,
> 1, 1, 0, 1, -1, 2, -1, 2, 0, 0, 0, 0, 1, 0, -1, -1, 1, 0, 1, 0, 0, 0, 0,
> 1, 1, 0, 1, 2, 0, 1, 1, 1, -1, -1, 0, 1, -1, 2, 3, 1, 0, 0, 1, 0, 0, 1,
> -1, 2, 1, 0, -1, 0, -1, -2, 2, -1, 1, 1, 0, 1, 1, 2, 1, 3, 0, 1, -1, -2,
> 0, 2, 2, 2, 1, 0, -1, 0
>  


The set {a(n) | n=1...10^5} contains all integers from -111 to 104 
including -111 and 104. So it is likely that all integers appear somewhere 
in the sequence? 

--Edwin