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A055772
Square root of largest square dividing n!.
19
1, 1, 1, 2, 2, 12, 12, 24, 72, 720, 720, 1440, 1440, 10080, 30240, 120960, 120960, 725760, 725760, 7257600, 7257600, 79833600, 79833600, 958003200, 4790016000, 62270208000, 186810624000, 2615348736000, 2615348736000, 15692092416000
OFFSET
1,4
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..500
FORMULA
a(n) = A000188(n!) = sqrt(A008833(n!)) = sqrt(A055071(n)).
n! = a(n)^2*A055204(n) = a(n)^2*A007913(n!).
n! = (A000188(n!)^2)*A055229(n!)*A055231(n!).
log(a(n)) ~ n*log(n)/2. - David Radcliffe, Oct 17 2014
EXAMPLE
For n=6, 6! = 720 = 144*5 so a(6) = sqrt(144) = 12.
MAPLE
a:= proc(n)
local r, F, t;
r:= n!;
F:= ifactors(r)[2];
mul(t[1]^floor(t[2]/2), t=F)
end proc:
seq(a(n), n= 1 .. 100); # Robert Israel, Oct 19 2014
MATHEMATICA
Table[Last[Select[Sqrt[#]&/@Divisors[n!], IntegerQ]], {n, 30}] (* Harvey P. Dale, Oct 08 2012 *)
(Sqrt@Factorial@Range@30)/.Sqrt[_]->1 (* Morgan L. Owens, May 04 2016 *)
f[p_, e_] := p^Floor[e/2]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n!]; Array[a, 40] (* Amiram Eldar, Jul 26 2024 *)
PROG
(PARI) a(n)=core(n!, 2)[2] \\ Charles R Greathouse IV, Apr 03 2012
(Python)
from collections import Counter
from math import prod
from sympy import factorint
def A055772(n): return prod(p**(e>>1) for p, e in sum((Counter(factorint(i)) for i in range(2, n+1)), start=Counter()).items()) # Chai Wah Wu, Sep 24 2025
KEYWORD
nonn
AUTHOR
Labos Elemer, Jul 12 2000
STATUS
approved