A072594 In prime factorization of n replace multiplication with bitwise logical 'xor'.

1, 2, 3, 0, 5, 1, 7, 2, 0, 7, 11, 3, 13, 5, 6, 0, 17, 2, 19, 5, 4, 9, 23, 1, 0, 15, 3, 7, 29, 4, 31, 2, 8, 19, 2, 0, 37, 17, 14, 7, 41, 6, 43, 11, 5, 21, 47, 3, 0, 2, 18, 13, 53, 1, 14, 5, 16, 31, 59, 6, 61, 29, 7, 0, 8, 10, 67, 17, 20, 0, 71, 2, 73, 39, 3, 19, 12, 12, 79, 5, 0, 43, 83, 4

Offset: 1

Author

Reinhard Zumkeller, Jun 23 2002

Keywords

• nonn
• nice

Comments

n is prime iff a(n)=n;

for primes p, k>0: a(p^k)=p*(k mod 2);

for m>1: a(m^2)=0, see A072595.

a(A127812(n)) = n and a(m) <> n for m < A127812(n).

Examples

			a(35) = a(5*7) = a(5) 'xor' a(7) = '101' xor '111' = '010' = 2.
		

Links

• R. Zumkeller, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A072591, A072593, A027746, A178910.

Programs

Haskell

import Data.Bits (xor)
a072594 = foldl1 xor . a027746_row :: Integer -> Integer
-- Reinhard Zumkeller, Nov 17 2012

Mathematica

a[n_] := BitXor @@ Flatten[ Table[ First[#], {Last[#]} ]& /@ FactorInteger[n] ]; Table[a[n], {n, 1, 84}] (* Jean-François Alcover, Mar 11 2013 *)

PARI

a(n)=if(n==1, return(1)); my(f=factor(n),t); for(i=1,#f~, if(f[i,2]%2, t=bitxor(t,f[i,1]))); t \\ Charles R Greathouse IV, Aug 28 2016

Python

from sympy import factorint
from operator import _xor_
from functools import reduce
def a(n): return reduce(_xor_, (f for f in factorint(n, multiple=True))) if n > 1 else 1
print([a(n) for n in range(1, 85)]) # Michael S. Branicky, May 31 2025