the Mobius function

[Emeric Deutsch]

Here is my own solution to the problem given below.
Define f by f(n)=(-1)^n and g by g(1)=-1, g(2)=2 and g(n)=0 for n>=3.
Then sum_{d|n}(g(d))=(-1)^n=f(n).
Indeed, if n is odd, then 2 is not a divisor, so the only contribution 
to this sum is g(1)=-1. If n is even, then both 1 and 2 are divisors, 
so the only contribution to the sum is g(1)+g(2)=-1+2=1.
Now, by the Mobius inversion theorem, we obtain 
 sum_{d|n}(-1)^(n/d)*mobius(d) = g(n), i.e. 0 for n>=3. 
Emeric


On Fri, 28 Jan 2005, Emeric Deutsch wrote:

> Dear Seqfans,
> It is well-known that for n>1 we have  
> 
>    SUM_{d|n}mobius(d) = 0.
> 
> Is it known or easy to see that for n>2 we have 
> 
>   SUM_{d|n}(-1)^(n/d)*mobius(d) = 0 ?
> 
> Thanks.
> Emeric 
> 
>