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

[Alec Mihailovs]

> 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.
> 
> I get, by hand, the sequence beginning:
> 
> 1,1,0,1,0,1,1,0,0,1,1,1,1,1,0,1,0,...

The 12th term is wrong. It should be a(12)=a(1)-a(11)=1-1=0. Further,
a(14)=a(1)-a(13)=1-1=0, and a(15)=a(1)-a(2)-a(7)+a(14)=1-1-1+0=-1, then
a(16)=a(1)-a(3)-a(5)+a(15)=1-0-0+(-1)=0. Here are the first 100 terms
calculated in Maple, 

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
 
> I get that a(52) = 2, so not every term is either 0 or 1.
> 
> (I am betting that not every term is necessarily nonnegative either.)
> 
> Could someone extend this sequence?
> When is the first negative term?

Alec Mihailovs 
http://math.tntech.edu/alec/