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A180499 n^3 + n-th cubefree number.

%I A180499 #26 Aug 13 2024 02:24:46
%S A180499 2,10,30,68,130,222,350,521,739,1011,1343,1741,2211,2759,3392,4114,
%T A180499 4932,5852,6880,8022,9284,10673,12193,13852,15654,17606,19714,21985,
%U A180499 24423,27035,29827,32805,35975,39343
%N A180499 n^3 + n-th cubefree number.
%C A180499 First differs from n^3 + n (A034262) at n=8 because 8 is the first positive integer which is not cubefree.
%C A180499 What cubes appear in this sequence?
%C A180499 No cubes appear in this sequence: the n-th cubefree number is asymptotically zeta(3)*n, putting members of this sequence strictly between n^3 and (n+1)^3. (Lacking effective error bounds this actually only shows that there are finitely many.) - _Charles R Greathouse IV_, Jan 21 2011
%H A180499 Charles R Greathouse IV, <a href="/A180499/b180499.txt">Table of n, a(n) for n = 1..10000</a>
%F A180499 a(n) = n^3 + A004709(n) = A000578(n) + A004709(n).
%e A180499 a(8) = 8^3 + 8th number that is not divisible by any cube > 1 = 8^3 + 9 = 521.
%t A180499 #[[1]]+#[[2]]^3&/@Module[{cf=Select[Range[50],Max[FactorInteger[#][[All,2]]] < 3&]},Thread[{cf,Range[Length[cf]]}]] (* _Harvey P. Dale_, Jun 28 2020 *)
%o A180499 (Python)
%o A180499 from sympy import mobius, integer_nthroot
%o A180499 def A180499(n):
%o A180499     def f(x): return n+x-sum(mobius(k)*(x//k**3) for k in range(1, integer_nthroot(x,3)[0]+1))
%o A180499     m, k = n, f(n)
%o A180499     while m != k:
%o A180499         m, k = k, f(k)
%o A180499     return n**3+m # _Chai Wah Wu_, Aug 12 2024
%Y A180499 Cf. A000578, A004709, A034262, A161203.
%K A180499 nonn,easy
%O A180499 1,1
%A A180499 _Jonathan Vos Post_, Jan 20 2011