A267113 Bitwise-OR of the exponents of all 4k+1 primes in the prime factorization of n.

0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 2, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 2, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 2, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 2, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1

Offset: 1

Author

Antti Karttunen, Feb 03 2016

Keywords

• nonn

Examples

			For n = 65 = 5 * 13 = (4+1)^1 * ((3*4)+1)^1, bitwise-or of 1 and 1 is 1, thus a(65) = 1.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A170818, A267116, A260728.

Cf. A004144 (indices of zeros), A009003 (of nonzeros).

Differs from both A046080 and A083025 for the first time at n=65, which here a(65) = 1.

Formula

a(n) = A267116(A170818(n)).

Other identities. For all n >= 0:

a(n) = a(A170818(n)). [The result depends only on the prime factors of the form 4k+1.]

a(n) <= A083025(n).