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A248763 Greatest k such that k^3 divides n!

%I A248763 #9 Sep 01 2024 09:37:31
%S A248763 1,1,1,2,2,2,2,4,12,12,12,24,24,24,360,1440,1440,1440,1440,2880,60480,
%T A248763 60480,60480,120960,604800,604800,1814400,3628800,3628800,3628800,
%U A248763 3628800,14515200,479001600,479001600,479001600,958003200,958003200,958003200
%N A248763 Greatest k such that k^3 divides n!
%C A248763 Every term divides all its successors.
%H A248763 Clark Kimberling, <a href="/A248763/b248763.txt">Table of n, a(n) for n = 1..1000</a>
%H A248763 Rafael Jakimczuk, <a href="https://doi.org/10.12988/imf.2017.7542">On the h-th free part of the factorial</a>, International Mathematical Forum, Vol. 12, No. 13 (2017), pp. 629-634.
%F A248763 From _Amiram Eldar_, Sep 01 2024: (Start)
%F A248763 a(n) = A053150(n!).
%F A248763 a(n) = (n! / A145642(n))^(1/3) = A248762(n)^(1/3).
%F A248763 log(a(n)) = (1/3)*n*log(n) - (log(3)+1)*n/3 + o(n) (Jakimczuk, 2017). (End)
%e A248763 a(4) = 2 because 2^3 divides 24 and if k > 2, then k^3 > 8 does not divide 24.
%t A248763 z = 40; f[n_] := f[n] = FactorInteger[n!]; r[m_, x_] := r[m, x] = m*Floor[x/m];
%t A248763 u[n_] := Table[f[n][[i, 1]], {i, 1, Length[f[n]]}];
%t A248763 v[n_] := Table[f[n][[i, 2]], {i, 1, Length[f[n]]}];
%t A248763 p[m_, n_] := p[m, n] = Product[u[n][[i]]^r[m, v[n]][[i]], {i, 1, Length[f[n]]}];
%t A248763 m = 3; Table[p[m, n], {n, 1, z}]  (* A248762 *)
%t A248763 Table[p[m, n]^(1/m), {n, 1, z}]   (* A248763 *)
%t A248763 Table[n!/p[m, n], {n, 1, z}]      (* A145642 *)
%t A248763 f[p_, e_] := p^Floor[e/3]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n!]; Array[a, 40] (* _Amiram Eldar_, Sep 01 2024 *)
%o A248763 (PARI) a(n) = {my(f = factor(n!)); prod(i = 1, #f~, f[i, 1]^(f[i, 2]\3));} \\ _Amiram Eldar_, Sep 01 2024
%Y A248763 Cf. A000142, A053150, A145642, A248762.
%K A248763 nonn,easy
%O A248763 1,4
%A A248763 _Clark Kimberling_, Oct 14 2014