cp's OEIS Frontend

A178645 a(n) = sum of divisors d of n such that d^k is not equal to n for any k >= 1.

%I A178645 #12 Jun 12 2018 21:14:58
%S A178645 0,1,1,1,1,6,1,5,1,8,1,16,1,10,9,9,1,21,1,22,11,14,1,36,1,16,10,28,1,
%T A178645 42,1,29,15,20,13,49,1,22,17,50,1,54,1,40,33,26,1,76,1,43,21,46,1,66,
%U A178645 17,64,23,32,1,108,1,34,41,49,19,78,1,58,27,74,1,123,1,40,49,64,19,90,1,106,28,44,1,140,23,46,33,92,1,144,21,76,35,50,25,156,1,73,57,107
%N A178645 a(n) = sum of divisors d of n such that d^k is not equal to n for any k >= 1.
%H A178645 Antti Karttunen, <a href="/A178645/b178645.txt">Table of n, a(n) for n = 1..65537</a>
%H A178645 <a href="/index/Su#sums_of_divisors">Index entries for sequences related to sums of divisors</a>
%F A178645 a(n) = A000203(n) - A175067(n).
%F A178645 a(1) = 0, a(p) = 1, a(pq) = p+q+1, a(pq...z) = [(p+1)*(q+1)*…*(z+1)] - (pq…z), for p, q = primes, k = natural numbers, pq...z = product of k (k > 2) distinct primes p, q, ..., z.
%e A178645 For n = 16, set of such divisors is {1, 8}; a(16) = 1+8=9.
%e A178645 For n = 90, which is not a perfect power (A001597), the only divisor d for which d^k = 90 is 90 itself, with k=1, thus a(90) = A001065(90) = A000203(90) - 90 = 144. - _Antti Karttunen_, Jun 12 2018
%o A178645 (PARI)
%o A178645 A175070(n) = if(!ispower(n),0,sumdiv(n,d,if((d>1)&&(d<n)&&((d^valuation(n,d))==n),d,0)));
%o A178645 A178645(n) = (sigma(n) - (A175070(n) + n)); \\ _Antti Karttunen_, Jun 12 2018
%Y A178645 Cf. A000203, A001065, A175067, A175070.
%K A178645 nonn
%O A178645 1,6
%A A178645 _Jaroslav Krizek_, Dec 25 2010
%E A178645 Term a(90) corrected from 204 to 144 by _Antti Karttunen_, Jun 12 2018