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

[Richard Guy]

How does this connect (if at all) with the Mertens
conjecture ?     R.

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.
>>
>> 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/
>
>