A248764 -id:A248764 - cp's OEIS frontend

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

Clark Kimberling, Oct 14 2014

Every term divides all its successors.

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

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.

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

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 *)

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

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)

A248766 Greatest 4th-power-free divisor of n!

Original entry on oeis.org

1, 2, 6, 24, 120, 45, 315, 2520, 280, 175, 1925, 23100, 300300, 4204200, 63063000, 63063000, 1072071000, 14889875, 282907625, 9053044, 190113924, 4182506328, 96197645544, 144296468316, 3607411707900, 93792704405400, 31264234801800, 22787343150, 660832951350

Offset: 1

Clark Kimberling, Oct 14 2014

			a(6) = 45 because 45 divides 6! and if k > 45 divides 6!, then h^4 divides 6!/k for some h > 1.

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.

Cf. A000142, A053165, A248764, A248765.

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^Mod[e, 4]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n!]; Array[a, 30] (* Amiram Eldar, Sep 01 2024 *)

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

a(n) = n!/A248764(n).
From Amiram Eldar, Sep 01 2024: (Start)
a(n) = A053165(n!).
log(a(n)) = 2*log(2) * n + o(n) (Jakimczuk, 2017). (End)

cp's OEIS Frontend

A248765 Greatest k such that k^4 divides n!

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