cp's OEIS Frontend

A178636 If n = Product (p_i^k_i) for i = 1, ..., j then a(n) is the sum of the divisors d that are not in the set {1, p_1^k_1, p_2^k_2, ..., p_j^k_j}.

%I A178636 #13 Dec 11 2022 04:24:04
%S A178636 0,0,0,2,0,6,0,6,3,10,0,20,0,14,15,14,0,27,0,32,21,22,0,48,5,26,12,44,
%T A178636 0,61,0,30,33,34,35,77,0,38,39,76,0,83,0,68,63,46,0,104,7,65,51,80,0,
%U A178636 90,55,104,57,58,0,155,0,62,87,62,65,127,0,104,69,129,0,177,0,74,95,116,77,149,0,164,39,82,0,209,85,86,87,160,0,217,91,140,93,94,95,216,0,119,135,187
%N A178636 If n = Product (p_i^k_i) for i = 1, ..., j then a(n) is the sum of the divisors d that are not in the set {1, p_1^k_1, p_2^k_2, ..., p_j^k_j}.
%F A178636 a(n) = A000203(n) - A159077(n) = A167515(n) - 1.
%F A178636 a(1) = 0, a(p) = 0, a(pq) = pq, a(pq...z) = [(p+1)* (q+1)* ... *(z+1)] - [p+q+ ...+z] - 1, a(p^k) = (p^k-p)/(p-1), for p, q = primes, k = natural numbers, pq...z = product of k (k > 2) distinct primes p, q, ..., z.
%e A178636 For n = 12, set of divisors {1, p_1^k_1, p_2^k_2, ..., p_j^k_j}: {1, 3, 4}. Complement of divisors: {2, 6, 12}. a(12) = 2+6+12 = 20.
%K A178636 nonn
%O A178636 1,4
%A A178636 _Jaroslav Krizek_, Dec 25 2010
%E A178636 I edited the definition to fix the grammar and make it understandable.
%E A178636 a(100) corrected by _Georg Fischer_, Dec 10 2022