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

0,10

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.

Links

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

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)).

Example

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.

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 *)

Prog

(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).
(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

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.

Sequence in context: A356241 A091430 A362221 * A065339 A122434 A141571

Adjacent sequences: A260725 A260726 A260727 * A260729 A260730 A260731

Keyword

nonn

Author

Antti Karttunen, Aug 12 2015

Status

approved