Sum over (divisors+-constant)

[Leroy Quet]

I just submitted these 2 sequences to the EIS:

%S A000001 0,1,-1,0,0,-1,1,-1,-1,1,1,-3
%N A000001 sum{k|n,k>=2} mu(k-1),
mu() is Moebius function.
%e A000001 12's divisors >=2 are 2, 3, 4, 6, and 12.
So a(12) = mu(1)+mu(2)+mu(3)+mu(5)+mu(11) =
1-1-1-1-1 = -3.
%Y A000001 A008683
%O A000001 1
%K A000001 ,more,sign,

%S A000001 -1,-2,-1,-3,0,-3,-1,-3,0,-2,-1,-5
%N A000001 sum{k|n} mu(k+1),
mu() is Moebius function.
%e A000001 12's divisors are 1, 2, 3, 4, 6, and 12.
So a(12) = mu(2)+mu(3)+mu(4)+mu(5)+mu(7)+mu(13) =
-1-1+0-1-1-1 = -5.
%Y A000001 A008683
%O A000001 1
%K A000001 ,more,sign,

I email seqfan because this idea can be used to generate a new class of 
sequences.

a(n) =

sum{k|n} b(k+c),

where c is a constant integer,

and the sum is over those k > |c| if c is negative,

and {b(k)} is an integer sequence, such as (say) the Euler phi function.


Heck, we can generalize:

a(n) = sum{k|n} b(d(k)),

where {d(k)} is a sequence of positive integers.

Hmmm...not too spectacular.
:)
 
thanks,
Leroy Quet