I can do
DivisorSum[12, DivisorSum[#, # &] &]
without any problem--the answer is 55. But how do I use the divisors of the outer sum in the function for the inner sum? I have no problem with
Qij[p2_] := DivisorSum[p2, MoebiusMu[p2/#] Fij[#] &]/p2;
DivisorSum[per, Qij[#] &]
but if I try to eliminate Qij with
DivisorSum[per,DivisorSum[#, MoebiusMu[#/#1] Fij[#1] &]/#&]
where # is the outer divisor and #1 is the inner divisor, it does not work. Is there a way to distinguish the variables?
