A183099 a(n) = sum of powerful divisors d (excluding 1) of n.

0, 0, 0, 4, 0, 0, 0, 12, 9, 0, 0, 4, 0, 0, 0, 28, 0, 9, 0, 4, 0, 0, 0, 12, 25, 0, 36, 4, 0, 0, 0, 60, 0, 0, 0, 49, 0, 0, 0, 12, 0, 0, 0, 4, 9, 0, 0, 28, 49, 25, 0, 4, 0, 36, 0, 12, 0, 0, 0, 4, 0, 0, 9, 124, 0, 0, 0, 4, 0, 0, 0, 129, 0, 0, 25, 4, 0, 0, 0, 28, 117, 0, 0, 4, 0, 0, 0, 12, 0, 9, 0, 4, 0, 0, 0, 60, 0, 49, 9, 129

Offset: 1

Author

Jaroslav Krizek, Dec 25 2010

Keywords

• nonn
• easy

Comments

a(n) = sum of divisors d of n from set A001694(m) - powerful numbers for m >=2.

Examples

			For n = 12, set of such divisors is {4}; a(12) = 4.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..16385
• Index entries for sequences related to sums of divisors.

Crossrefs

Cf. A000203, A001694, A183097, A183100.

Programs

Mathematica

f[p_, e_] := (p^(e+1)-1)/(p-1) - p; a[1] = 0; a[n_] := Times @@ f @@@ FactorInteger[n] - 1; Array[a, 100] (* Amiram Eldar, Aug 29 2023 *)

PARI

A183099(n) = (sumdiv(n, d, ispowerful(d)*d) - 1); \\ Antti Karttunen, Oct 07 2017

Formula

a(n) = A000203(n) - A183100(n) = A183097(n) - 1.

a(1) = 0, a(p) = 0, a(pq) = 0, a(pq...z) = 0, a(p^k) = ((p^(k+1)-1) / (p-1))-p-1, for p, q = primes, k = natural numbers, pq...z = product of k (k > 2) distinct primes p, q, ..., z.