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A385006 The sum of the biquadratefree divisors of n.

%I A385006 #9 Jun 15 2025 12:54:33
%S A385006 1,3,4,7,6,12,8,15,13,18,12,28,14,24,24,15,18,39,20,42,32,36,24,60,31,
%T A385006 42,40,56,30,72,32,15,48,54,48,91,38,60,56,90,42,96,44,84,78,72,48,60,
%U A385006 57,93,72,98,54,120,72,120,80,90,60,168,62,96,104,15,84,144
%N A385006 The sum of the biquadratefree divisors of n.
%C A385006 First differs from A365682 and A366992 at n = 32.
%C A385006 The number of these divisors is A252505(n), and the largest of them is A058035(n).
%H A385006 Amiram Eldar, <a href="/A385006/b385006.txt">Table of n, a(n) for n = 1..10000</a>
%F A385006 Multiplicative with a(p^e) = (p^min(e+1, 4) - 1)/(p - 1).
%F A385006 In general, the sum of the k-free (numbers that are not divisible by a k-th power larger than 1) divisors of n is multiplicative with a(p^e) = (p^min(e+1, k) - 1)/(p - 1).
%F A385006 Dirichlet g.f.: zeta(s) * zeta(s-1) /zeta(4*s-4).
%F A385006 In general, the sum of the k-free divisors of n has Dirichlet g.f.: zeta(s)*zeta(s-1)/zeta(k*s-k).
%F A385006 Sum_{k=1..n} a(k) ~ (15/(2*Pi^2)) * n^2.
%F A385006 In general, the sum of the k-free divisors of n has an average order (Pi^2/(12*zeta(k))) * n^2.
%t A385006 f[p_, e_] := (p^Min[e+1, 4] - 1)/(p - 1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
%o A385006 (PARI) a(n) = {my(f = factor(n), p, e); prod(i = 1, #f~, p = f[i,1]; e = f[i,2]; (p^min(e+1, 4) - 1)/(p - 1));}
%Y A385006 Cf. A046100, A048250, A058035, A073185, A252505, A365682, A366992.
%Y A385006 The sum of divisors d of n such that d is: A000593 (odd), A033634 (exponentially odd), A035316 (square), A038712 (power of 2), A048250 (squarefree), A072079 (3-smooth), A073185 (cubefree), A113061 (cube), A162296 (nonsquarefree), A183097 (powerful), A186099 (5-rough), A353900 (exponentially 2^n), A385005 (cubefull), this sequence (biquadratefree).
%K A385006 nonn,easy,mult
%O A385006 1,2
%A A385006 _Amiram Eldar_, Jun 15 2025