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A248821 Number of cubes that divide 1!*2!*3!*...*n!.

%I A248821 #11 Oct 27 2023 20:45:41
%S A248821 1,1,1,2,6,10,36,64,220,468,1024,2052,7590,16224,50400,142800,246240,
%T A248821 510300,2261952,3545856,14152320,40986000,68428800,178293960,
%U A248821 784274400,1526805504,2782080000,9307872000,15858633600,28225260000,143730892800,225167040000
%N A248821 Number of cubes that divide 1!*2!*3!*...*n!.
%H A248821 Alois P. Heinz, <a href="/A248821/b248821.txt">Table of n, a(n) for n = 1..1000</a> (first 400 terms from Clark Kimberling)
%e A248821 a(5) counts these cubes that divide 34560:  1^3, 2^3, 3^3, 4^3, 6^3, 12^3.
%p A248821 b:= proc(n) option remember; add(i[2]*x^numtheory[pi](i[1]),
%p A248821       i=ifactors(n)[2])+`if`(n=1, 0, b(n-1))
%p A248821     end:
%p A248821 c:= proc(n) option remember; b(n)+`if`(n=1, 0, c(n-1)) end:
%p A248821 a:= n->(p->mul(iquo(coeff(p, x, i), 3)+1, i=1..degree(p)))(c(n)):
%p A248821 seq(a(n), n=1..30);  # _Alois P. Heinz_, Oct 16 2014
%t A248821 z = 40; p[n_] := Product[k!, {k, 1, n}];
%t A248821 f[n_] := f[n] = FactorInteger[p[n]];
%t A248821 r[m_, x_] := r[m, x] = m*Floor[x/m]
%t A248821 u[n_] := Table[f[n][[i, 1]], {i, 1, Length[f[n]]}];
%t A248821 v[n_] := Table[f[n][[i, 2]], {i, 1, Length[f[n]]}];
%t A248821 t[m_, n_] := Apply[Times, 1 + r[m, v[n]]/m]
%t A248821 m = 3; Table[t[m, n], {n, 1, z}] (* A248821 *)
%Y A248821 Cf. A000178, A000578, A248784, A248822, A248823.
%K A248821 nonn,easy
%O A248821 1,4
%A A248821 _Clark Kimberling_, Oct 15 2014