A248765 -id:A248765 - cp's OEIS frontend

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

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)

A248781 Number of integers k^4 that divide n!

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 2, 4, 6, 6, 6, 6, 6, 6, 8, 8, 15, 15, 30, 30, 30, 30, 36, 36, 36, 48, 112, 112, 112, 112, 128, 128, 144, 216, 270, 270, 270, 270, 300, 300, 300, 300, 660, 792, 792, 792, 864, 1296, 1728, 1728, 3744, 3744, 4368, 4368, 4704, 4704, 4704

Offset: 1

Clark Kimberling, Oct 15 2014

			a(10) counts these divisors of 10!:  1, 16, 81, 256, 1296, 20736.

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

Cf. A000142, A055993, A248780, A248765.

z = 130; m = 4;
f[n_] := f[n] = FactorInteger[n!]; r[x_] := r[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]]}];
a[n_] := Apply[Times, 1 + r[v[n]]/m]
t = Table[a[n], {n, 1, z}] (* A248781 *)

a(n)=c=0;d=divisors(n!);for(i=1,#d,if(ispower(d[i])&&ispower(d[i])%4==0,c++));c+1
n=1;while(n<50,print1(a(n),", ");n++) \\ Derek Orr, Oct 20 2014

A248764 Greatest 4th power integer that divides n!

Original entry on oeis.org

1, 1, 1, 1, 1, 16, 16, 16, 1296, 20736, 20736, 20736, 20736, 20736, 20736, 331776, 331776, 429981696, 429981696, 268738560000, 268738560000, 268738560000, 268738560000, 4299816960000, 4299816960000, 4299816960000, 348285173760000, 13379723235164160000

Offset: 1

Clark Kimberling, Oct 14 2014

Every term divides all its successors.

			a(6) = 16 because 16 divides 6! and if k > 2 then k^4 does not divide 6!.

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

Cf. A000142, A008835, A248762, A248765, 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^(4*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]^(4*(f[i, 2]\4)));} \\ Amiram Eldar, Sep 01 2024

a(n) = n!/A248766(n).
From Amiram Eldar, Sep 01 2024: (Start)
a(n) = A008835(n!).
a(n) = A248765(n)^4. (End)

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

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A248781 Number of integers k^4 that divide n!

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A248764 Greatest 4th power integer that divides n!

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