A366524 -id:A366524 - cp's OEIS frontend

A366523 Largest square divisor of n which is <= sqrt(n).

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 9, 1, 1, 4, 1, 1, 1, 4, 1, 9, 1, 4, 1, 1, 1, 4, 1, 1, 9, 4

Offset: 1

Ilya Gutkovskiy, Oct 11 2023

nonn

Andrew Howroyd, Table of n, a(n) for n = 1..10000

Cf. A008833, A033676, A366524.

Table[Last[Select[Divisors[n], # <= Sqrt[n] && IntegerQ[Sqrt[#]] &]], {n, 100}]

a(n) = {my(m=1); fordiv(sqrtint(n/core(n)), d, if(d^4 <= n, m=max(m,d))); m^2} \\ Andrew Howroyd, Oct 11 2023

A365837 Largest proper square divisor of n, for n >= 2; a(1) = 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 9, 1, 4, 1, 1, 1, 4, 1, 1, 9, 4, 1, 1, 1, 16, 1, 1, 1, 9, 1, 1, 1, 4, 1, 1, 1, 4, 9, 1, 1, 16, 1, 25, 1, 4, 1, 9, 1, 4, 1, 1, 1, 4, 1, 1, 9, 16, 1, 1, 1, 4, 1, 1, 1, 36, 1, 1, 25, 4, 1, 1, 1, 16, 9, 1, 1, 4, 1, 1, 1, 4, 1, 9, 1, 4, 1, 1, 1, 16, 1, 49, 9, 25

Offset: 1

Ilya Gutkovskiy, Oct 17 2023

nonn

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A008833, A032742, A063928, A085392, A089384, A093803, A366524.

f:= proc(n) local F, t;
  if issqr(n) then
    n/min(numtheory:-factorset(n))^2
  else
    F:= ifactors(n)[2];
    mul(t[1]^(2*floor(t[2]/2)),t=F)
  fi
end proc:
f(1):= 1:
map(f, [$1..100]); # Robert Israel, Nov 20 2023

Join[{1}, Table[Last[Select[Divisors[n], # < n && IntegerQ[Sqrt[#]]  &]], {n, 2, 100}]]
f[p_, e_] := p^(2*Floor[e/2]); a[n_] := Module[{fct = FactorInteger[n]}, Times @@ f @@@ fct/If[AllTrue[fct[[;; , 2]], EvenQ], fct[[1, 1]]^2, 1]]; Array[a, 100] (* Amiram Eldar, Oct 19 2023 *)

a(n) = if (n==1, 1, my(d=divisors(n)); vecmax(select(issquare, Vec(d, #d-1)))); \\ Michel Marcus, Oct 17 2023

from math import prod
from sympy import factorint
def A365837(n):
    if n<=1: return 1
    f = factorint(n)
    return prod(p**(e&-2) for p, e in f.items())//(min(f)**2 if all(e&1^1 for e in f.values()) else 1) # Chai Wah Wu, Oct 20 2023

cp's OEIS Frontend

A366523 Largest square divisor of n which is <= sqrt(n).

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