A072438 Remove prime factors of form 4*k+1.

1, 2, 3, 4, 1, 6, 7, 8, 9, 2, 11, 12, 1, 14, 3, 16, 1, 18, 19, 4, 21, 22, 23, 24, 1, 2, 27, 28, 1, 6, 31, 32, 33, 2, 7, 36, 1, 38, 3, 8, 1, 42, 43, 44, 9, 46, 47, 48, 49, 2, 3, 4, 1, 54, 11, 56, 57, 2, 59, 12, 1, 62, 63, 64, 1, 66, 67, 4, 69, 14, 71, 72, 1, 2, 3, 76, 77, 6, 79, 16, 81, 2

Offset: 1

Author

Reinhard Zumkeller, Jun 17 2002

Keywords

• nonn
• mult
• easy

Comments

a(n) <= n; a(a(n)) = a(n).

All factors p^m of a(n) are of the form p=2 or p=4*k+3.

Examples

			a(90) = a(2*3*3*5) = a(2*(4*0+3)^2*(4*1+1)^1) = 2*3^2*1 = 18.
		

Links

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

Crossrefs

Cf. A072436, A004144, A002144, A002145.

Programs

Maple

a:= n-> mul(`if`(irem(i[1], 4)=1, 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] == 1, p^e, 1], {pe, FactorInteger[n]}];
Array[a, 100] (* Jean-François Alcover, May 29 2019 *)

PARI

a(n) = my(f=factor(n)); for (i=1, #f~, if ((f[i,1] % 4) == 1, f[i,1] = 1)); factorback(f); \\ Michel Marcus, Jun 09 2014

Python

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

Formula

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