OFFSET
1,4
COMMENTS
From Bernard Schott, May 25 2021: (Start)
Answer: Yes, can be proved when n = Product_{i=1..k} p_i^e_i with n' = n * Sum_{i=1..k} (e_i/p_i) and cototient(n) = n * (1 - Product_{i=1..k} (1 - 1/p_i).
a(n) = 0 iff n is in A008578 (1 together with the primes).
a(n) = 1 iff n is in A006881 (squarefree semiprimes) (End).
LINKS
Antti Karttunen, Table of n, a(n) for n = 1..10000
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537
MATHEMATICA
Array[If[# < 2, 0, # Total[#2/#1 & @@@ FactorInteger[#]]] - # + EulerPhi[#] &, 96] (* Michael De Vlieger, May 24 2021 *)
PROG
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, May 23 2021
STATUS
approved