|
| |
|
|
A348970
|
|
a(n) = A003959(n) - A129283(n), where A003959 is multiplicative with a(p^e) = (p+1)^e and A129283(n) is sum of n and its arithmetic derivative.
|
|
9
|
|
|
|
0, 0, 0, 1, 0, 1, 0, 7, 1, 1, 0, 8, 0, 1, 1, 33, 0, 9, 0, 10, 1, 1, 0, 40, 1, 1, 10, 12, 0, 11, 0, 131, 1, 1, 1, 48, 0, 1, 1, 54, 0, 13, 0, 16, 12, 1, 0, 164, 1, 13, 1, 18, 0, 57, 1, 68, 1, 1, 0, 64, 0, 1, 14, 473, 1, 17, 0, 22, 1, 15, 0, 204, 0, 1, 14, 24, 1, 19, 0, 230, 67, 1, 0, 80, 1, 1, 1, 96, 0, 75, 1, 28
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
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
|
|
|
|
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
|
(PARI)
A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
A003959(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1]++); factorback(f); };
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
