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A365172 The sum of divisors d of n such that gcd(d, n/d) is a square.

%I A365172 #7 Aug 25 2023 08:40:01
%S A365172 1,3,4,5,6,12,8,9,10,18,12,20,14,24,24,21,18,30,20,30,32,36,24,36,26,
%T A365172 42,28,40,30,72,32,45,48,54,48,50,38,60,56,54,42,96,44,60,60,72,48,84,
%U A365172 50,78,72,70,54,84,72,72,80,90,60,120,62,96,80,85,84,144,68
%N A365172 The sum of divisors d of n such that gcd(d, n/d) is a square.
%C A365172 The number of these divisors is A365171(n).
%H A365172 Amiram Eldar, <a href="/A365172/b365172.txt">Table of n, a(n) for n = 1..10000</a>
%F A365172 Multiplicative with a(p^e) = (p^(e+2) - 1)/(p^2 - 1) if e is even, and (1 + p^(2*floor((e+1)/4) + 1))*(p^(2*floor(e/4)+2) - 1)/(p^2 - 1) if e is odd.
%F A365172 a(n) <= A000203(n), with equality if and only if n is squarefree (A005117).
%F A365172 a(n) >= A034448(n), with equality if and only if n is not a biquadrateful number (A046101).
%F A365172 Sum_{k=1..n} a(k) ~ c * n^2, where c = 1/(2 * Product_{p prime} (1 - 1/p^2 + 1/p^3 - 1/p^5)) = 0.696082796052... .
%t A365172 f[p_, e_] := If[EvenQ[e], (p^(e + 2) - 1)/(p^2 - 1), (1 + p^(2*Floor[(e + 1)/4] + 1))*(p^(2*Floor[e/4] + 2) - 1)/(p^2 - 1)]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
%o A365172 (PARI) a(n) = {my(f = factor(n), p, e); prod(i = 1, #f~, p = f[i,1]; e = f[i,2]; if(e%2, (1 + p^(2*((e+1)\4)+1))*(p^(2*(e\4)+2) - 1)/(p^2 - 1), (p^(e+2) - 1)/(p^2 - 1)));}
%Y A365172 Cf. A000203, A005117, A034448, A035316, A046101, A365171, A365174.
%K A365172 nonn,easy,mult
%O A365172 1,2
%A A365172 _Amiram Eldar_, Aug 25 2023