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%I #8 Mar 16 2024 04:20:31
%S 1,3,4,5,6,12,8,1,10,18,12,20,14,24,24,1,18,30,20,30,32,36,24,4,26,42,
%T 1,40,30,72,32,1,48,54,48,50,38,60,56,6,42,96,44,60,60,72,48,4,50,78,
%U 72,70,54,3,72,8,80,90,60,120,62,96,80,1,84,144,68,90,96
%N The sum of the unitary divisors of n that are cubefree numbers (A004709).
%C The number of these divisors is A365498(n).
%D D. Suryanarayana, The number and sum of k-free intergers <= x which are prime to n, Indian J. Math., Vol. 11 (1969), pp. 131-139.
%H Amiram Eldar, <a href="/A371242/b371242.txt">Table of n, a(n) for n = 1..10000</a>
%H Francesco Pappalardi, <a href="http://www.mat.uniroma3.it/users/pappa/papers/allahabad2003.pdf">A survey on k-freeness</a>, Number Theory, Ramanujan Math. Soc. Lect. Notes Ser., Vol. 1 (2003), pp. 71-88.
%F Multiplicative with a(p^e) = p^e + 1 for e <= 2, and a(p^e) = 1 for e >= 3.
%F a(n) = 1 if and only if n is cubefull (A036966).
%F a(n) = A000005(n) if and only if n is squarefree (A005117).
%F a(n) <= A034448(n), with equality if and only if n is cubefree.
%F Dirichlet g.f.: zeta(s) * Product_{p prime} (1 + 1/p^(s-1) + 1/p^(2*s-2) - 1/p^(2*s-1) - 1/p^(3*s-2)).
%F Sum_{j=1..n} a(j) ~ c * n^2 / 2, where c = Product_{p prime} (1 - 1/p^3 + 1/(p^2 + p)) = 1.16545286600957717104.... .
%F In general, the formula above holds for the sum of unitary divisors of n that are k-free numbers (k >= 2) with c = Product_{p prime} (1 - 1/p^k + 1/(p^2 + p)) (Suryanarayana, 1969). If k = 2 then c = A065465. In the limit when k -> oo, c = A306633.
%t f[p_, e_] := If[e < 3, p^e + 1, 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
%o (PARI) a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i, 2] < 3, f[i, 1]^f[i, 2] + 1, 1));}
%Y Cf. A000005, A004709, A005117, A034448, A036966, A065465, A092261 (squarefree analog), A306633, A365498.
%K nonn,easy,mult
%O 1,2
%A _Amiram Eldar_, Mar 16 2024
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