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.

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

Offset: 1

Author

Antti Karttunen, Nov 05 2021

Keywords

• nonn

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

Crossrefs

Cf. A003415, A003959, A129283, A211991, A348029, A348507, A348971, A348972, A348973, A348974.

Cf. A008578 (positions of zeros), A001358 (positions of ones).

Programs

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 *)

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); };
A348970(n) = (A003959(n) - (n+A003415(n)));

Formula

a(n) = A003959(n) - A129283(n) = A003959(n) - (n+A003415(n)).

a(n) = A348029(n) - A211991(n).

a(n) = A348507(n) - A003415(n).

For all n >= 1, a(A001358(n)) = 1.