OFFSET
1,8
COMMENTS
There are no negative terms. We prove this by induction over the prime factorization of n, showing that A348507(n) >= A003415(n) for all values of n >= 1. At n=1, both sequences have value 0, and at the primes both sequences obtain the value 1, so the base cases hold. We know that A348507(n)-(n/p) = (p+1)*A348507(n/p) for all prime factors p of n (see comment in A348507). With the arithmetic derivative we obtain respectively that A003415(n) = A003415(p*(n/p)) = A003415(p)*(n/p) + p*A003415(n/p) = (n/p) + p*A003415(n/p), for any prime factor p of n. Now A348507(p*(n/p)) >= A003415(p*(n/p)) iff A348507(p*(n/p)) - (n/p) >= A003415(p*(n/p)) - (n/p), that is, iff (p+1)*A348507(n/p) >= p*A003415(n/p), which indeed follows by the induction hypothesis, which assumes that A348507(x) >= A003415(x) for all proper divisors x of n.
LINKS
Antti Karttunen, Table of n, a(n) for n = 1..10000
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537
FORMULA
MATHEMATICA
d[0] = d[1] = 0; d[n_] := n*Plus @@ ((Last[#]/First[#]) & /@ FactorInteger[n]); f[p_, e_] := (p + 1)^e; a[1] = 0; a[n_] := Times @@ f @@@ FactorInteger[n] - n - d[n]; Array[a, 100] (* Amiram Eldar, Nov 05 2021 *)
PROG
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Nov 05 2021
STATUS
approved