A248763 -id:A248763 - cp's OEIS frontend

A145642 Cubefree part of n!.

1, 2, 6, 3, 15, 90, 630, 630, 210, 2100, 23100, 34650, 450450, 6306300, 28028, 7007, 119119, 2144142, 40738698, 101846745, 230945, 5080790, 116858170, 350574510, 70114902, 1822987452, 1822987452, 6380456082, 185033226378, 5550996791340

Offset: 1

Artur Jasinski, Oct 15 2008

nonn

Chai Wah Wu, Table of n, a(n) for n = 1..500
Rafael Jakimczuk, On the h-th free part of the factorial, International Mathematical Forum, Vol. 12, No. 13 (2017), pp. 629-634.
Eric Weisstein's World of Mathematics, Cubefree Part.

Cf. A000142 (n!), A050985 (cubefree part), A248762, A248763.

CF. A055204 (squarefree part of n!).

CubefreePart[n_Integer?Positive] := Times @@ Power @@@ ({#[[1]], Mod[ #[[2]], 3]} & /@ FactorInteger[n]); Table[CubefreePart[n! ], {n, 1, 40}]

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
import math
def A145642(n):
    return 1 if n <=1 else reduce(mul,[p**(e % 3) for p,e in factorint(math.factorial(n)).items()])
# Chai Wah Wu, Feb 04 2015

a(n) = A050985(A000142(n)). - Michel Marcus, Nov 07 2013
From Amiram Eldar, Sep 01 2024: (Start)
a(n) = n! / A248762(n) = n! / A248763(n)^3.
log(a(n)) = log(3) * n + o(n) (Jakimczuk, 2017). (End)

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)

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A145642 Cubefree part of n!.

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A248762 Greatest cube that divides n!.

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