A072436 Remove prime factors of form 4*k+3.

1, 2, 1, 4, 5, 2, 1, 8, 1, 10, 1, 4, 13, 2, 5, 16, 17, 2, 1, 20, 1, 2, 1, 8, 25, 26, 1, 4, 29, 10, 1, 32, 1, 34, 5, 4, 37, 2, 13, 40, 41, 2, 1, 4, 5, 2, 1, 16, 1, 50, 17, 52, 53, 2, 5, 8, 1, 58, 1, 20, 61, 2, 1, 64, 65, 2, 1, 68, 1, 10, 1, 8, 73, 74, 25, 4, 1, 26, 1, 80, 1, 82, 1, 4, 85, 2, 29

Offset: 1

Author

Reinhard Zumkeller, Jun 17 2002

Keywords

• nonn
• mult
• easy

Comments

a(n) <= n; a(a(n)) = a(n); for all factors p^m of a(n): p=2 or p=4*k+1.

Examples

			a(90) = a(2*3*3*5) = a(2*(4*0+3)^2*(4*1+1)^1) = 2*1^2*5 = 10.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A072438, A072437, A002144, A002145, A065338.

Equals n / A097706(n).

Programs

Maple

a:= n-> mul(`if`(irem(i[1], 4)=3, 1, i[1]^i[2]), i=ifactors(n)[2]):
seq(a(n), n=1..100);  # Alois P. Heinz, Jun 09 2014

Mathematica

a[n_] := n/Product[{p, e} = pe; If[Mod[p, 4] == 3, p^e, 1], {pe, FactorInteger[n]}];
Array[a, 100] (* Jean-François Alcover, May 29 2019 *)

PARI

a(n) = my(f=factor(n)); for (k=1, #f~, if ((f[k,1] % 4) == 3, f[k,1]=1)); factorback(f); \\ Michel Marcus, May 08 2017

Python

from sympy import factorint
from operator import mul
def a(n):
    f = factorint(n)
    return 1 if n == 1 else reduce(mul, [1 if i%4==3 else i**f[i] for i in f])# Indranil Ghosh, May 08 2017

Formula

Multiplicative with a(p) = (if p==3 (mod 4) then 1 else p).