A008834 -id:A008834 - cp's OEIS frontend

A336590 Numbers k such that k/A008834(k) is squarefree, where A008834(k) is the largest cube dividing k.

1, 2, 3, 5, 6, 7, 8, 10, 11, 13, 14, 15, 16, 17, 19, 21, 22, 23, 24, 26, 27, 29, 30, 31, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 46, 47, 48, 51, 53, 54, 55, 56, 57, 58, 59, 61, 62, 64, 65, 66, 67, 69, 70, 71, 73, 74, 77, 78, 79, 80, 81, 82, 83, 85, 86, 87, 88

Offset: 1

Amiram Eldar, Jul 26 2020

nonn

Numbers such that none of the exponents in their prime factorization is of the form 3*m + 2.
Cohen (1962) proved that for a given number k >= 2 the asymptotic density of numbers whose exponents in their prime factorization are of the forms k*m or k*m + 1 only is zeta(k)/zeta(2). In this sequence k = 3, and therefore its asymptotic density is zeta(3)/zeta(2) = 6*zeta(3)/Pi^2 = 0.7307629694... (A253905).
Also, numbers k whose number of divisors, A000005(k), is not divisible by 3, i.e., complement of A059269.

			6 is a term since 6 = 2^1 * 3^1 and 1 is not of the form 3*m + 2.
9 is not a term since 9 = 3^2 and 2 is of the form 3*m + 2.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Eckford Cohen, Arithmetical notes. III. Certain equally distributed sets of integers, Pacific Journal of Mathematics, No. 12, Vol. 1 (1962), pp. 77-84.
Eckford Cohen, Arithmetical Notes, XIII. A Sequal to Note IV, Elemente der Mathematik, Vol. 18 (1963), pp. 8-11.
L. G. Sathe, On a congruence property of the divisor function, American Journal of Mathematics, Vol. 67, No. 3 (1945), pp. 397-406.

Cf. A000005, A002117, A005117, A008834, A013661, A059269, A253905.

Union of A211337 and A211338.

Select[Range[100], Max[Mod[FactorInteger[#][[;;,2]], 3]] < 2 &]

A008833 Largest square dividing n.

Original entry on oeis.org

1, 1, 1, 4, 1, 1, 1, 4, 9, 1, 1, 4, 1, 1, 1, 16, 1, 9, 1, 4, 1, 1, 1, 4, 25, 1, 9, 4, 1, 1, 1, 16, 1, 1, 1, 36, 1, 1, 1, 4, 1, 1, 1, 4, 9, 1, 1, 16, 49, 25, 1, 4, 1, 9, 1, 4, 1, 1, 1, 4, 1, 1, 9, 64, 1, 1, 1, 4, 1, 1, 1, 36, 1, 1, 25, 4, 1, 1, 1, 16, 81, 1, 1, 4, 1, 1, 1, 4, 1, 9, 1, 4, 1, 1, 1, 16, 1

Offset: 1

N. J. A. Sloane

The Dirichlet generating function of the arithmetic function of the largest t-th power dividing n is zeta(s)*zeta(t*s-t)/zeta(s*t), here with t=2 and in A008834 and A008835 with t=3 and t=4, respectively. - R. J. Mathar, Feb 19 2011

Daniel Forgues, Table of n, a(n) for n = 1..100000
Henry Bottomley, Some Smarandache-type multiplicative sequences
R. J. Mathar, Survey of Dirichlet series of multiplicative arithmetic functions arXiv:1106.4038 [math.NT], 2011-2012, Remark 16.
Andrew Reiter, On (mod n) spirals, 2014, see also posting to Number Theory Mailing List, Mar 23 2014.
Eric Weisstein's World of Mathematics, Square part

Cf. A000188, A005563, A007913, A019554, A059897, A068310.

Cf. A008836, A358272, A001615.

a008833 n = head $ filter ((== 0) . (mod n)) $
   reverse $ takeWhile (<= n) $ tail a000290_list
-- Reinhard Zumkeller, Nov 13 2011

A008833 := proc(n)
    expand(numtheory:-nthpow(n,2)) ;
end proc:
seq(A008833(n), n=1..100) ;

a[n_] := First[ Select[ Reverse[ Divisors[n]], IntegerQ[Sqrt[#]]&, 1]]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Dec 12 2011 *)
f[p_, e_] := p^(2*Floor[e/2]); a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Jul 07 2020 *)

A008833(n)=n/core(n) \\ Michael B. Porter, Oct 17 2009

from sympy.ntheory.factor_ import core
def A008833(n): return n//core(n) # Chai Wah Wu, Dec 30 2021

a(n) = A000188(n)^2 = n/A007913(n). Cf. A019554.
Multiplicative with a(p^e) = p^(2[e/2]). - David W. Wilson, Aug 01 2001
Dirichlet g.f.: zeta(s)*zeta(2s-2)/zeta(2s). - R. J. Mathar, Oct 31 2011
a(n) = A005563(n-1) / A068310(n) for n > 1. - Reinhard Zumkeller, Nov 26 2011
Sum_{k=1..n} a(k) ~ Zeta(3/2) * n^(3/2) / (3*Zeta(3)). - Vaclav Kotesovec, Feb 01 2019
a(A059897(n,k)) = A059897(a(n), a(k)). - Peter Munn, Nov 30 2019
From Ridouane Oudra, May 11 2025: (Start)
a(n) = Sum_{d|n} lambda(d)*d*psi(n/d), where lambda = A008836 and psi = A001615.
a(n) = lambda(n) * Sum_{d|n} lambda(d)*d*phi(n/d).
a(n) = A008836(n) * A358272(n). (End)

A050985 Cubefree part of n.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 1, 9, 10, 11, 12, 13, 14, 15, 2, 17, 18, 19, 20, 21, 22, 23, 3, 25, 26, 1, 28, 29, 30, 31, 4, 33, 34, 35, 36, 37, 38, 39, 5, 41, 42, 43, 44, 45, 46, 47, 6, 49, 50, 51, 52, 53, 2, 55, 7, 57, 58, 59, 60, 61, 62, 63, 1, 65, 66, 67, 68, 69, 70, 71, 9, 73, 74, 75

Offset: 1

Eric W. Weisstein, Dec 11 1999

This is an unusual sequence in the sense that the 83.2% of the integers that belong to A004709 occur infinitely many times, whereas the remaining 16.8% of the integers that belong to A046099 never occur at all. - Ant King, Sep 22 2013

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harvey P. Dale)
Henry Bottomley, Some Smarandache-type multiplicative sequences.
Eric Weisstein's World of Mathematics, Cubefree Part.
Eric Weisstein's World of Mathematics, Dirichlet Generating Function.

Cf. A007913, A008834, A053165, A004709, A046099, A301596, A301597.

A050985 := proc(n)
    n/A008834(n) ;
end proc:
seq(A050985(n),n=1..40) ; # R. J. Mathar, Dec 08 2015

cf[n_]:=Module[{tr=Transpose[FactorInteger[n]],ex,cb},ex= tr[[2]]- Mod[ tr[[2]],3];cb=Times@@(First[#]^Last[#]&/@Transpose[{tr[[1]], ex}]);n/cb]; Array[cf,75] (* Harvey P. Dale, Jun 03 2012 *)
f[p_, e_] := p^Mod[e, 3]; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 07 2020 *)

a(n) = my(f=factor(n)); f[,2] = apply(x->(x % 3), f[,2]); factorback(f); \\ Michel Marcus, Jan 06 2019

from operator import mul
from functools import reduce
from sympy import factorint
def A050985(n):
    return 1 if n <=1 else reduce(mul,[p**(e % 3) for p,e in factorint(n).items()])
# Chai Wah Wu, Feb 04 2015

Multiplicative with p^e -> p^(e mod 3), p prime. - Reinhard Zumkeller, Nov 22 2009
Dirichlet g.f.: zeta(3s)*zeta(s-1)/zeta(3s-3). - R. J. Mathar, Feb 11 2011
a(n) = n/A008834(n). - R. J. Mathar, Dec 08 2015
Sum_{k=1..n} a(k) ~ Pi^6 * n^2 / (1890*Zeta(3)). - Vaclav Kotesovec, Feb 08 2019

A053150 Cube root of largest cube dividing n.

Original entry on oeis.org

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, 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, 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

Offset: 1

Henry Bottomley, Feb 28 2000

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

Antti Karttunen, Table of n, a(n) for n = 1..10000
Henry Bottomley, Some Smarandache-type multiplicative sequences.
Kevin A. Broughan, Restricted divisor sums, Acta Arithmetica, 101(2) (2002), 105-114.
Kevin A. Broughan, Relationship between the integer conductor and k-th root functions, Int. J. Pure Appl. Math. 5(3) (2003), 253-275.
Kevin A. Broughan, Relaxations of the ABC Conjecture using integer k'th roots, New Zealand J. Math. 35(2) (2006), 121-136.
Vaclav Kotesovec, Graph - the asymptotic ratio.

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).

Cf. A000189, A000190, A008834, A008835, A015051, A061704.

f[list_] := list[[1]]^Quotient[list[[2]], 3]; Table[Apply[Times, Map[f,FactorInteger[n]]], {n, 1, 81}] (* Geoffrey Critzer, Jan 21 2015 *)
Table[SelectFirst[Reverse@ Divisors@ n, IntegerQ[#^(1/3)] &]^(1/3), {n, 105}] (* Michael De Vlieger, Jul 28 2017 *)
f[p_, e_] := p^Floor[e/3]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 18 2020 *)

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

a(n) = my(f = factor(n)); for (k=1, #f~, f[k,2] = f[k,2]\3); factorback(f); \\ Michel Marcus, Jul 28 2017

from math import prod
from sympy import factorint
def A053150(n): return prod(p**(q//3) for p, q in factorint(n).items()) # Chai Wah Wu, Aug 18 2021

Multiplicative with a(p^e) = p^[e/3]. - Mitch Harris, Apr 19 2005
a(n) = A008834(n)^(1/3) = sqrt(A000189(n)/A000188(A050985(n))).
Dirichlet g.f.: zeta(3s-1)*zeta(s)/zeta(3s). - R. J. Mathar, Apr 09 2011
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
a(n) = Sum_{d^3|n} phi(d). - Ridouane Oudra, Dec 30 2020
G.f.: Sum_{k>=1} phi(k) * x^(k^3) / (1 - x^(k^3)). - Ilya Gutkovskiy, Aug 20 2021

More terms from Antti Karttunen, Jul 28 2017

A008835 Largest 4th power dividing n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 16, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 16, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 16, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 16, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 16, 81

Offset: 1

N. J. A. Sloane

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Henry Bottomley, Some Smarandache-type multiplicative sequences.

Cf. A000188, A000190, A008833, A008834, A053164, A053165.

with(numtheory): [ seq( expand(nthpow(i,4)),i=1..200) ];

Max@ Select[Divisors@ #, IntegerQ@ Power[#, 1/4] &] & /@ Range@ 81 (* Michael De Vlieger, Mar 18 2015 *)
f[p_, e_] := p^(e - Mod[e, 4]); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Aug 15 2023 *)

a(n) = {f = factor(n); for (i=1, #f~, f[i,2] = 4*(f[i,2]\4);); factorback(f);} \\ Michel Marcus, Mar 16 2015

from math import prod
from sympy import factorint
def A008835(n): return prod(p**(e&-4) for p, e in factorint(n).items()) # Chai Wah Wu, Aug 08 2024

a(n) = A000188(A000188(n))^4.
Multiplicative with a(p^e) = p^(4[e/4]). - Mitch Harris, Apr 19 2005
Dirichlet g.f.: zeta(s) * zeta(4s-4) / zeta(4s). - Álvar Ibeas, Feb 12 2015
Sum_{k=1..n} a(k) ~ zeta(5/4) * n^(5/4) / (5*zeta(5)) - 45*n/Pi^4. - Vaclav Kotesovec, Feb 03 2019
a(n) = n/A053165(n). - Amiram Eldar, Aug 15 2023
a(n) = A053164(n)^4. - Amiram Eldar, Sep 01 2024

Entry improved by comments from Henry Bottomley, Feb 29 2000

A053149 Smallest cube divisible by n.

Original entry on oeis.org

1, 8, 27, 8, 125, 216, 343, 8, 27, 1000, 1331, 216, 2197, 2744, 3375, 64, 4913, 216, 6859, 1000, 9261, 10648, 12167, 216, 125, 17576, 27, 2744, 24389, 27000, 29791, 64, 35937, 39304, 42875, 216, 50653, 54872, 59319, 1000, 68921, 74088, 79507, 10648

Offset: 1

Henry Bottomley, Feb 28 2000

Amiram Eldar, Table of n, a(n) for n = 1..10000
Henry Bottomley, Some Smarandache-type multiplicative sequences.

Cf. A000188, A000189, A007913, A008834, A019554, A048798, A050985.

Cf. A002117, A013667, A330596.

a[n_] := For[k = 1, True, k++, If[ Divisible[c = k^3, n], Return[c]]]; Table[a[n], {n, 1, 44}] (* Jean-François Alcover, Sep 03 2012 *)
f[p_, e_] := p^(e + Mod[3 - e, 3]); a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Aug 29 2019 *)
scdn[n_]:=Module[{c=Ceiling[Surd[n,3]]},While[!Divisible[c^3,n],c++];c^3]; Array[scdn,50] (* Harvey P. Dale, Jun 13 2020 *)

a(n) = {my(f = factor(n)); prod(i = 1, #f~, f[i,1]^(f[i,2] + (3-f[i,2])%3));} \\ Amiram Eldar, Oct 27 2022

a(n) = (n/A000189(n))^3 = A008834(n)*A019554(A050985(n))^3 = n*A050985(n)^2/A000188(A050985(n))^3.
a(n) = n * A048798(n). - Franklin T. Adams-Watters, Apr 08 2009
From Amiram Eldar, Jul 29 2022: (Start)
Multiplicative with a(p^e) = p^(e + ((3-e) mod 3)).
Sum_{n>=1} 1/a(n) = Product_{p prime} ((p^3+2)/(p^3-1)) = 1.655234386560802506... . (End)
Sum_{k=1..n} a(k) ~ c * n^4, where c = (zeta(9)/(4*zeta(3))) * Product_{p prime} (1 - 1/p^2 + 1/p^3) = A013667*A330596/(4*A002117) = 0.1559906... . - Amiram Eldar, Oct 27 2022

A353897 a(n) is the largest divisor of n whose exponents in its prime factorization are all powers of 2 (A138302).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 4, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 12, 25, 26, 9, 28, 29, 30, 31, 16, 33, 34, 35, 36, 37, 38, 39, 20, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 18, 55, 28, 57, 58, 59, 60, 61, 62, 63, 16, 65, 66, 67, 68

Offset: 1

Amiram Eldar, May 10 2022

			a(27) = 9 since 9 = 3^2 is the largest divisor of 27 with an exponent in its prime factorization, 2, that is a power of 2.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A138302, A353898.

Similar sequences: A000265, A007947, A008834, A055071, A350390.

f[p_, e_] := p^(2^Floor[Log2[e]]); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

Multiplicative with a(p^e) = p^(2^floor(log_2(e))).
a(n) = n if and only if n is in A138302.
Sum_{k=1..n} a(k) ~ c*n^2, where c = 0.4616988732... = (1/2) * Product_{p prime} (1 + Sum_{k>=1} (p^f(k) - p^(f(k-1)+1))/p^(2*k)), f(k) = 2^floor(log_2(k)) and f(0) = 0.

A062378 n divided by largest cubefree factor of n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 1, 1, 1, 1, 1, 16, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4, 9, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 2

Offset: 1

Henry Bottomley, Jun 18 2001

Numerator of n/rad(n)^2, where rad is the squarefree kernel of n (A007947), denominator: A055231. - Reinhard Zumkeller, Dec 10 2002

Antti Karttunen, Table of n, a(n) for n = 1..10000
Henry Bottomley, Some Smarandache-type multiplicative sequences.

Cf. A000189, A000578, A007948, A008834, A019555, A048798, A050985, A053149, A053150, A056551, A056552. See A003557 for squares and A062379 for 4th powers.
Differs from A073753 for the first time at n=90, where a(90) = 1, while A073753(90) = 3.
Cf. A007947, A055231.

f[p_, e_] := p^Max[e-2, 0]; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 07 2020 *)

a(n)=my(f=factor(n));prod(i=1,#f~,f[i,1]^max(f[i,2]-2,0)) \\ Charles R Greathouse IV, Aug 08 2013

(define (A062378 n) (/ n (A007948 n)))
(definec (A007948 n) (if (= 1 n) n (* (expt (A020639 n) (min 2 (A067029 n))) (A007948 (A028234 n)))))
;; Antti Karttunen, Nov 28 2017

a(n) = n / A007948(n).
a(n) = A003557(A003557(n)). - Antti Karttunen, Nov 28 2017
Multiplicative with a(p^e) = p^max(e-2, 0). - Amiram Eldar, Sep 07 2020
Dirichlet g.f.: zeta(s-1) * Product_{p prime} (1 - 1/p^(s-1) + 1/p^s - 1/p^(2*s-1) + 1/p^(2*s)). - Amiram Eldar, Dec 07 2023

A056552 Powerfree kernel of cubefree part of n.

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 1, 3, 10, 11, 6, 13, 14, 15, 2, 17, 6, 19, 10, 21, 22, 23, 3, 5, 26, 1, 14, 29, 30, 31, 2, 33, 34, 35, 6, 37, 38, 39, 5, 41, 42, 43, 22, 15, 46, 47, 6, 7, 10, 51, 26, 53, 2, 55, 7, 57, 58, 59, 30, 61, 62, 21, 1, 65, 66, 67, 34, 69, 70, 71, 3, 73, 74, 15, 38, 77

Offset: 1

Henry Bottomley, Jun 25 2000

			a(32) = 2 because cubefree part of 32 is 4 and powerfree kernel of 4 is 2.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Henry Bottomley, Some Smarandache-type multiplicative sequences.

Cf. A000189, A000578, A007947, A008834, A013664, A019555, A048798, A050985, A053149, A053150, A056551.

f[p_, e_] :=  p^If[Divisible[e, 3], 0, 1]; a[n_] := Times @@ (f @@@ FactorInteger[ n]); Array[a, 100] (* Amiram Eldar, Aug 29 2019 *)

a(n) = my(f=factor(n)); for (k=1, #f~, if (frac(f[k,2]/3), f[k,2] = 1, f[k,2] = 0)); factorback(f); \\ Michel Marcus, Feb 28 2019

a(n) = A007947(A050985(n)) = A019555(A050985(n)) = n/(A053150(n)*A000189(n)) = A019555(n)/A053150(n) = A056551(n)^(1/3).
If n = Product_{j} Pj^Ej then a(n) = Product_{j} Pj^Fj, where Fj = 0 if Ej is 0 or a multiple of 3 and Fj = 1 otherwise.
Multiplicative with a(p^e) = p^(if 3|e, then 0, else 1). - Mitch Harris, Apr 19 2005
Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(6)/2) * Product_{p prime} (1 - 1/p^2 + 1/p^3 - 1/p^4) = 0.3480772773... . - Amiram Eldar, Oct 28 2022
Dirichlet g.f.: zeta(3*s) * Product_{p prime} (1 + 1/p^(s-1) + 1/p^(2*s-1)). - Amiram Eldar, Sep 16 2023

A248762 Greatest cube that divides n!.

Original entry on oeis.org

1, 1, 1, 8, 8, 8, 8, 64, 1728, 1728, 1728, 13824, 13824, 13824, 46656000, 2985984000, 2985984000, 2985984000, 2985984000, 23887872000, 221225582592000, 221225582592000, 221225582592000, 1769804660736000, 221225582592000000, 221225582592000000

Offset: 1

Clark Kimberling, Oct 14 2014

Every term divides all its successors.

			a(4) = 8 because 8 divides 24 and if k > 2 then k^3 does not divide 24.

Clark Kimberling, Table of n, a(n) for n = 1..1000

Cf. A000142, A008834, A145642, A248763.

z = 40; f[n_] := f[n] = FactorInteger[n!]; r[m_, x_] := r[m, x] = m*Floor[x/m];
u[n_] := Table[f[n][[i, 1]], {i, 1, Length[f[n]]}];
v[n_] := Table[f[n][[i, 2]], {i, 1, Length[f[n]]}];
p[m_, n_] := p[m, n] = Product[u[n][[i]]^r[m, v[n]][[i]], {i, 1, Length[f[n]]}];
m = 3; Table[p[m, n], {n, 1, z}]  (* A248762 *)
Table[p[m, n]^(1/m), {n, 1, z}]   (* A248763 *)
Table[n!/p[m, n], {n, 1, z}]      (* A145642 *)
gk[n_]:=Select[Divisors[n!],IntegerQ[Surd[#,3]]&]; Max[#]&/@Array[gk,30] (* Harvey P. Dale, Sep 16 2021 *)
f[p_, e_] := p^(3*Floor[e/3]); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n!]; Array[a, 30] (* Amiram Eldar, Sep 01 2024 *)

a(n)=k=ceil((n!/2)^(1/3));while(n!%k^3,k--);k^3
vector(20,n,a(n)) \\ Derek Orr, Oct 19 2014

a(n) = {my(f = factor(n!)); prod(i = 1, #f~, f[i, 1]^(3*(f[i, 2]\3)));} \\ Amiram Eldar, Sep 01 2024

a(n) = n!/A145642(n).
From Amiram Eldar, Sep 01 2024: (Start)
a(n) = A008834(n!).
a(n) = A248763(n)^3. (End)

cp's OEIS Frontend

A336590 Numbers k such that k/A008834(k) is squarefree, where A008834(k) is the largest cube dividing k.

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A008833 Largest square dividing n.

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A050985 Cubefree part of n.

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A053150 Cube root of largest cube dividing n.

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A008835 Largest 4th power dividing n.

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Maple

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A053149 Smallest cube divisible by n.

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A353897 a(n) is the largest divisor of n whose exponents in its prime factorization are all powers of 2 (A138302).

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