A248765 Greatest k such that k^4 divides n!

1, 1, 1, 1, 1, 2, 2, 2, 6, 12, 12, 12, 12, 12, 12, 24, 24, 144, 144, 720, 720, 720, 720, 1440, 1440, 1440, 4320, 60480, 60480, 60480, 60480, 120960, 120960, 241920, 1209600, 3628800, 3628800, 3628800, 3628800, 7257600

Offset

1,6

Comments

Every term divides all its successors.

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.

Formula

From Amiram Eldar, Sep 01 2024: (Start)

a(n) = A053164(n!).

a(n) = (n! / A248766(n))^(1/4) = A248764(n)^(1/4).

log(a(n)) = (1/4)*n*log(n) - (2*log(2)+1)*n/4 + o(n) (Jakimczuk, 2017). (End)

Example

a(6) = 2 because 2^4 divides 6! and if k > 2 then k^4 does not divide 6!.

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 = 4; Table[p[m, n], {n, 1, z}]  (* A248764 *)
Table[p[m, n]^(1/m), {n, 1, z}]   (* A248765 *)
Table[n!/p[m, n], {n, 1, z}]      (* A248766 *)
f[p_, e_] := p^Floor[e/4]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n!]; Array[a, 30] (* Amiram Eldar, Sep 01 2024 *)

Prog

(PARI) a(n) = {my(f = factor(n!)); prod(i = 1, #f~, f[i, 1]^(f[i, 2]\4)); } \\ Amiram Eldar, Sep 01 2024

Crossrefs

Cf. A000142, A053164, A248762, A248764, A248766.

Sequence in context: A058756 A237614 A257255 * A007039 A025248 A213170

Adjacent sequences: A248762 A248763 A248764 * A248766 A248767 A248768

Keyword

nonn,easy

Author

Clark Kimberling, Oct 14 2014

Status

approved