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A248763
Greatest k such that k^3 divides n!
3
1, 1, 1, 2, 2, 2, 2, 4, 12, 12, 12, 24, 24, 24, 360, 1440, 1440, 1440, 1440, 2880, 60480, 60480, 60480, 120960, 604800, 604800, 1814400, 3628800, 3628800, 3628800, 3628800, 14515200, 479001600, 479001600, 479001600, 958003200, 958003200, 958003200
OFFSET
1,4
COMMENTS
Every term divides all its successors.
LINKS
Rafael Jakimczuk, On the h-th free part of the factorial, International Mathematical Forum, Vol. 12, No. 13 (2017), pp. 629-634.
FORMULA
From Amiram Eldar, Sep 01 2024: (Start)
a(n) = A053150(n!).
a(n) = (n! / A145642(n))^(1/3) = A248762(n)^(1/3).
log(a(n)) = (1/3)*n*log(n) - (log(3)+1)*n/3 + o(n) (Jakimczuk, 2017). (End)
EXAMPLE
a(4) = 2 because 2^3 divides 24 and if k > 2, then k^3 > 8 does not divide 24.
MATHEMATICA
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 = 3; Table[p[m, n], {n, 1, z}] (* A248762 *)
Table[p[m, n]^(1/m), {n, 1, z}] (* A248763 *)
Table[n!/p[m, n], {n, 1, z}] (* A145642 *)
f[p_, e_] := p^Floor[e/3]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n!]; Array[a, 40] (* Amiram Eldar, Sep 01 2024 *)
PROG
(PARI) a(n) = {my(f = factor(n!)); prod(i = 1, #f~, f[i, 1]^(f[i, 2]\3)); } \\ Amiram Eldar, Sep 01 2024
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Oct 14 2014
STATUS
approved