the Mobius function

[Max]

Let n=2^k*m, where m is odd. Then

SUM_{d|n}(-1)^(n/d)*mobius(d) = SUM_{d1|2^k} SUM_{d2|m} (-1)^(n/(d1*d2))*mobius(d1*d2) =
  (since mobius is multiplicative and d2 is odd)
SUM_{d1|2^k} (-1)^(2^k/d1) mobius(d1) * SUM_{d2|m} mobius(d2).

The latter factor is 0 unless m=1. Hence, the statement is true if m>1.

Now let m=1. If n>2, then k>1 and

SUM_{d1|2^k} (-1)^(2^k/d1) mobius(d1) = 
  (all signs are plus except for d1=2^k)
SUM_{d1|2^(k-1)} mobius(d1) - mobius(2^k) = 0 - 0 = 0.

Max

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 
>
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>  
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