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A380400 The sum of unitary divisors of n that are perfect powers (A001597).

%I A380400 #12 Jan 25 2025 07:49:32
%S A380400 1,1,1,5,1,1,1,9,10,1,1,5,1,1,1,17,1,10,1,5,1,1,1,9,26,1,28,5,1,1,1,
%T A380400 33,1,1,1,50,1,1,1,9,1,1,1,5,10,1,1,17,50,26,1,5,1,28,1,9,1,1,1,5,1,1,
%U A380400 10,65,1,1,1,5,1,1,1,18,1,1,26,5,1,1,1,17,82
%N A380400 The sum of unitary divisors of n that are perfect powers (A001597).
%C A380400 First differs from A360720 at n = 72.
%C A380400 The number of unitary divisors of n that are perfect powers is A380398(n).
%H A380400 Amiram Eldar, <a href="/A380400/b380400.txt">Table of n, a(n) for n = 1..10000</a>
%H A380400 Vaclav Kotesovec, <a href="/A380400/a380400.jpg">Plot of Sum_{k=1..n} a(k)/n^(3/2) and Sum_{k=1..n} A360720(k)/n^(3/2), for n = 1..1000000</a>
%F A380400 a(n) = Sum_{d|n, gcd(d, n/d) == 1} d * [d in A001597], where [] is the Iverson bracket.
%F A380400 a(n) <= A360720(n).
%F A380400 a(n) = 1 if and only if n is squarefree (A005117).
%e A380400 a(4) = 5 since 4 have 2 unitary divisors that are perfect powers, 1 and 4 = 2^2, and 1 + 4 = 5.
%e A380400 a(72) = 18 since 72 have 3 unitary divisors that are perfect powers, 1, 8 = 2^3, and 9 = 3^2, and 1 + 8 + 9 = 18.
%t A380400 ppQ[n_] := n == 1 || GCD @@ FactorInteger[n][[;; , 2]] > 1; a[n_] := DivisorSum[n, # &, CoprimeQ[#, n/#] && ppQ[#] &]; Array[a, 100]
%o A380400 (PARI) a(n) = sumdiv(n, d, d * (gcd(d, n/d) == 1 && (d == 1 || ispower(d))));
%Y A380400 Cf. A001597, A005117, A077610, A358347, A360720, A380398.
%K A380400 nonn,easy
%O A380400 1,4
%A A380400 _Amiram Eldar_, Jan 23 2025