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A248765 Greatest k such that k^4 divides n!

%I A248765 #8 Sep 01 2024 09:37:39
%S A248765 1,1,1,1,1,2,2,2,6,12,12,12,12,12,12,24,24,144,144,720,720,720,720,
%T A248765 1440,1440,1440,4320,60480,60480,60480,60480,120960,120960,241920,
%U A248765 1209600,3628800,3628800,3628800,3628800,7257600
%N A248765 Greatest k such that k^4 divides n!
%C A248765 Every term divides all its successors.
%H A248765 Clark Kimberling, <a href="/A248765/b248765.txt">Table of n, a(n) for n = 1..1000</a>
%H A248765 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 A248765 From _Amiram Eldar_, Sep 01 2024: (Start)
%F A248765 a(n) = A053164(n!).
%F A248765 a(n) = (n! / A248766(n))^(1/4) = A248764(n)^(1/4).
%F A248765 log(a(n)) = (1/4)*n*log(n) - (2*log(2)+1)*n/4 + o(n) (Jakimczuk, 2017). (End)
%e A248765 a(6) = 2 because 2^4 divides 6! and if k > 2 then k^4 does not divide 6!.
%t A248765 z = 40; f[n_] := f[n] = FactorInteger[n!]; r[m_, x_] := r[m, x] = m*Floor[x/m];
%t A248765 u[n_] := Table[f[n][[i, 1]], {i, 1, Length[f[n]]}];
%t A248765 v[n_] := Table[f[n][[i, 2]], {i, 1, Length[f[n]]}];
%t A248765 p[m_, n_] := p[m, n] = Product[u[n][[i]]^r[m, v[n]][[i]], {i, 1, Length[f[n]]}];
%t A248765 m = 4; Table[p[m, n], {n, 1, z}]  (* A248764 *)
%t A248765 Table[p[m, n]^(1/m), {n, 1, z}]   (* A248765 *)
%t A248765 Table[n!/p[m, n], {n, 1, z}]      (* A248766 *)
%t A248765 f[p_, e_] := p^Floor[e/4]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n!]; Array[a, 30] (* _Amiram Eldar_, Sep 01 2024 *)
%o A248765 (PARI) a(n) = {my(f = factor(n!)); prod(i = 1, #f~, f[i, 1]^(f[i, 2]\4));} \\ _Amiram Eldar_, Sep 01 2024
%Y A248765 Cf. A000142, A053164, A248762, A248764, A248766.
%K A248765 nonn,easy
%O A248765 1,6
%A A248765 _Clark Kimberling_, Oct 14 2014