cp's OEIS Frontend

A053150 Cube root of largest cube dividing n.

%I A053150 #52 Mar 29 2025 10:52:01
%S A053150 1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,1,3,1,1,1,1,2,1,1,
%T A053150 1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,1,1,1,1,3,1,2,1,1,1,1,1,1,1,4,1,1,1,1,
%U A053150 1,1,1,2,1,1,1,1,1,1,1,2,3,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1
%N A053150 Cube root of largest cube dividing n.
%C A053150 This can be thought as a "lower 3rd root" of a positive integer. Upper k-th roots were studied by Broughan (2002, 2003, 2006). The sequence of "upper 3rd root" of positive integers is given by A019555. - _Petros Hadjicostas_, Sep 15 2019
%H A053150 Antti Karttunen, <a href="/A053150/b053150.txt">Table of n, a(n) for n = 1..10000</a>
%H A053150 Henry Bottomley, <a href="http://fs.gallup.unm.edu/Bottomley-Sm-Mult-Functions.htm">Some Smarandache-type multiplicative sequences</a>.
%H A053150 Kevin A. Broughan, <a href="https://doi.org/10.4064/aa101-2-2">Restricted divisor sums</a>, Acta Arithmetica, 101(2) (2002), 105-114.
%H A053150 Kevin A. Broughan, <a href="http://ijpam.eu/contents/2003-5-3/2/2.pdf">Relationship between the integer conductor and k-th root functions</a>, Int. J. Pure Appl. Math. 5(3) (2003), 253-275.
%H A053150 Kevin A. Broughan, <a href="http://nzjm.math.auckland.ac.nz/images/d/d6/Relaxations_of_the_ABC_Conjecture_using_integer_k%27th_roots.pdf">Relaxations of the ABC Conjecture using integer k'th roots</a>, New Zealand J. Math. 35(2) (2006), 121-136.
%H A053150 Vaclav Kotesovec, <a href="/A053150/a053150.jpg">Graph - the asymptotic ratio</a>.
%F A053150 Multiplicative with a(p^e) = p^[e/3]. - _Mitch Harris_, Apr 19 2005
%F A053150 a(n) = A008834(n)^(1/3) = sqrt(A000189(n)/A000188(A050985(n))).
%F A053150 Dirichlet g.f.: zeta(3s-1)*zeta(s)/zeta(3s). - _R. J. Mathar_, Apr 09 2011
%F A053150 Sum_{k=1..n} a(k) ~ Pi^2 * n / (6*zeta(3)) + 3*zeta(2/3) * n^(2/3) / Pi^2. - _Vaclav Kotesovec_, Jan 31 2019
%F A053150 a(n) = Sum_{d^3|n} phi(d). - _Ridouane Oudra_, Dec 30 2020
%F A053150 G.f.: Sum_{k>=1} phi(k) * x^(k^3) / (1 - x^(k^3)). - _Ilya Gutkovskiy_, Aug 20 2021
%t A053150 f[list_] := list[[1]]^Quotient[list[[2]], 3]; Table[Apply[Times, Map[f,FactorInteger[n]]], {n, 1, 81}] (* _Geoffrey Critzer_, Jan 21 2015 *)
%t A053150 Table[SelectFirst[Reverse@ Divisors@ n, IntegerQ[#^(1/3)] &]^(1/3), {n, 105}] (* _Michael De Vlieger_, Jul 28 2017 *)
%t A053150 f[p_, e_] := p^Floor[e/3]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* _Amiram Eldar_, Sep 18 2020 *)
%o A053150 (PARI) A053150(n) = { my(f = factor(n), m = 1); for (k=1, #f~, m *= (f[k, 1]^(f[k, 2]\3)); ); m; } \\ _Antti Karttunen_, Jul 28 2017
%o A053150 (PARI) a(n) = my(f = factor(n)); for (k=1, #f~, f[k,2] = f[k,2]\3); factorback(f); \\ _Michel Marcus_, Jul 28 2017
%o A053150 (Python)
%o A053150 from math import prod
%o A053150 from sympy import factorint
%o A053150 def A053150(n): return prod(p**(q//3) for p, q in factorint(n).items()) # _Chai Wah Wu_, Aug 18 2021
%Y A053150 Cf. A000188 (inner square root), A019554 (outer square root), A019555 (outer third root), A053164 (inner 4th root), A053166 (outer 4th root), A015052 (outer 5th root), A015053 (outer 6th root).
%Y A053150 Cf. A000189, A000190, A008834, A008835, A015051, A061704.
%K A053150 easy,nonn,mult
%O A053150 1,8
%A A053150 _Henry Bottomley_, Feb 28 2000
%E A053150 More terms from _Antti Karttunen_, Jul 28 2017