A248763 Greatest k such that k^3 divides n!

1, 1, 1, 2, 2, 2, 2, 4, 12, 12, 12, 24, 24, 24, 360, 1440, 1440, 1440, 1440, 2880, 60480, 60480, 60480, 120960, 604800, 604800, 1814400, 3628800, 3628800, 3628800, 3628800, 14515200, 479001600, 479001600, 479001600, 958003200, 958003200, 958003200

Offset: 1

Author

Clark Kimberling, Oct 14 2014

Keywords

• nonn
• easy

Comments

Every term divides all its successors.

Examples

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

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1000
• Rafael Jakimczuk, On the h-th free part of the factorial, International Mathematical Forum, Vol. 12, No. 13 (2017), pp. 629-634.

Crossrefs

Cf. A000142, A053150, A145642, A248762.

Programs

Mathematica

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 *)
f[p_, e_] := p^Floor[e/3]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n!]; Array[a, 40] (* Amiram Eldar, Sep 01 2024 *)

PARI

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

Formula

From Amiram Eldar, Sep 01 2024: (Start)

a(n) = A053150(n!).

a(n) = (n! / A145642(n))^(1/3) = A248762(n)^(1/3).

log(a(n)) = (1/3)*n*log(n) - (log(3)+1)*n/3 + o(n) (Jakimczuk, 2017). (End)