A260728 Bitwise-OR of the exponents of all 4k+3 primes in the prime factorization of n.

0, 0, 0, 1, 0, 0, 1, 1, 0, 2, 0, 1, 1, 0, 1, 1, 0, 0, 2, 1, 0, 1, 1, 1, 1, 0, 0, 3, 1, 0, 1, 1, 0, 1, 0, 1, 2, 0, 1, 1, 0, 0, 1, 1, 1, 2, 1, 1, 1, 2, 0, 1, 0, 0, 3, 1, 1, 1, 0, 1, 1, 0, 1, 3, 0, 0, 1, 1, 0, 1, 1, 1, 2, 0, 0, 1, 1, 1, 1, 1, 0, 4, 0, 1, 1, 0, 1, 1, 1, 0, 2, 1, 1, 1, 1, 1, 1, 0, 2, 3, 0, 0, 1, 1, 0, 1, 0, 1, 3, 0, 1, 1, 1, 0, 1, 1, 0, 2, 1, 1, 1

Offset: 0

Author

Antti Karttunen, Aug 12 2015

Keywords

• nonn

Comments

A001481 (numbers that are the sum of 2 squares) gives the positions of even terms in this sequence, while its complement A022544 (numbers that are not the sum of 2 squares) gives the positions of odd terms.

If instead of bitwise-oring (A003986) we added in ordinary way the exponents of 4k+3 primes together, we would get the sequence A065339. For the positions where these two sequences differ see A260730.

Examples

			For n = 21 = 3^1 * 7^1 we compute A003986(1,1) = 1, thus a(21) = 1.
For n = 63 = 3^2 * 7^1 we compute A003986(2,1) = A003986(1,2) = 3, thus a(63) = 3.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A000035, A001481, A022544, A003986, A020639, A028234, A032742, A067029, A097706, A229062, A260730.

Cf. also A267113, A267116, A267099.

Differs from A065339 for the first time at n=21, where a(21) = 1, while A065339(21)=2.

Programs

Mathematica

Table[BitOr @@ (Map[Last, FactorInteger@ n /. {p_, } /; MemberQ[{0, 1, 2}, Mod[p, 4]] -> Nothing]), {n, 0, 120}] (* _Michael De Vlieger, Feb 07 2016 *)

Python

from functools import reduce
from operator import or_
from sympy import factorint
def A260728(n): return reduce(or_,(e for p, e in factorint(n).items() if p & 3 == 3),0) # Chai Wah Wu, Jun 28 2022

Scheme

(define (A260728 n) (cond ((< n 3) 0) ((even? n) (A260728 (/ n 2))) ((= 1 (modulo (A020639 n) 4)) (A260728 (A032742 n))) (else (A003986bi (A067029 n) (A260728 (A028234 n)))))) ;; A003986bi implements bitwise-or (see A003986).

Formula

If n < 3, a(n) = 0; thereafter, for any even n: a(n) = a(n/2), for any n with its smallest prime factor (A020639) of the form 4k+1: a(n) = a(A032742(n)), otherwise [when A020639(n) is of the form 4k+3] a(n) = A003986(A067029(n),a(A028234(n))).

Other identities. For all n >= 0:

A229062(n) = 1 - A000035(a(n)). [Reduced modulo 2 and complemented, the sequence gives the characteristic function of A001481.]

a(n) = a(A097706(n)). [The result depends only on the prime factors of the form 4k+3.]

a(n) = A267116(A097706(n)).

a(n) = A267113(A267099(n)).