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a(n) is the Euler totient function phi applied to the n-th cubefree number.
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%I #26 Aug 06 2024 02:11:46

%S 1,1,2,2,4,2,6,6,4,10,4,12,6,8,16,6,18,8,12,10,22,20,12,12,28,8,30,20,

%T 16,24,12,36,18,24,40,12,42,20,24,22,46,42,20,32,24,52,40,36,28,58,16,

%U 60,30,36,48,20,66,32,44,24,70,72,36,40,36,60,24,78,40

%N a(n) is the Euler totient function phi applied to the n-th cubefree number.

%C The analogous sequence with squarefree numbers is A049200.

%H Amiram Eldar, <a href="/A358039/b358039.txt">Table of n, a(n) for n = 1..10000</a>

%H Zhu Weiyi, <a href="https://fs.unm.edu/SNJ/OnTheCubeFreeNumberSequence.pdf">On the cube free number sequences</a>, Smarandache Notions J., Vol. 14 (2004), pp. 199-202.

%F a(n) = A000010(A004709(n)).

%F Sum_{k=1..n} a(k) = (c/(2*zeta(3)))*n^2 + O(n^(3/2+eps)), where c = Product_{p prime} (1 - (p+1)/(p^3+p^2+1)) = 0.62583324412633345811... (Weiyi, 2004).

%F From _Amiram Eldar_, Oct 09 2023: (Start)

%F Sum_{n>=1} 1/(A004709(n)*a(n)) = Product_{p prime} (1 + (p^2+1)/((p-1)*p^3)) = 2.14437852780769816048... .

%F Sum_{n>=1} 1/a(n)^2 = Product_{p prime} (1 + (p^2+1)/((p-1)^2*p^2)) = 3.26032746607943673536... . (End)

%t EulerPhi[Select[Range[100], Max[FactorInteger[#][[;; , 2]]] < 3 &]]

%o (Python)

%o from sympy import mobius, integer_nthroot, totient

%o def A358039(n):

%o def f(x): return n+x-sum(mobius(k)*(x//k**3) for k in range(1, integer_nthroot(x,3)[0]+1))

%o m, k = n, f(n)

%o while m != k:

%o m, k = k, f(k)

%o return totient(m) # _Chai Wah Wu_, Aug 06 2024

%Y Cf. A000010, A002117, A004709, A049200, A358040.

%K nonn

%O 1,3

%A _Amiram Eldar_, Oct 29 2022