A260737 Sum of Hamming distances between binary representations of prime factors of n, summed over all nonordered pairs of primes present (with multiplicity) in the prime factorization of n.

0, 0, 0, 0, 0, 1, 0, 0, 0, 3, 0, 2, 0, 2, 2, 0, 0, 2, 0, 6, 1, 2, 0, 3, 0, 4, 0, 4, 0, 6, 0, 0, 1, 3, 1, 4, 0, 2, 3, 9, 0, 4, 0, 4, 4, 3, 0, 4, 0, 6, 2, 8, 0, 3, 3, 6, 1, 5, 0, 10, 0, 4, 2, 0, 1, 4, 0, 6, 2, 6, 0, 6, 0, 4, 4, 4, 2, 8, 0, 12, 0, 4, 0, 7, 2, 3, 4, 6, 0, 9, 2, 6, 3, 4, 3, 5, 0, 4, 2, 12, 0, 6, 0, 12, 4, 5, 0, 6, 0, 8, 3, 8, 0, 4, 2, 10, 6, 4, 3, 14

Offset: 1

Author

Antti Karttunen, Sep 22 2015

Keywords

• nonn
• base

Examples

			For n = 1 the prime factorization is empty, thus there is nothing to sum, so a(1) = 0.
For n = 6 = 2*3, a(6) = 1 because the Hamming distance between 2 and 3 is 1 as 2 = "10" in binary and 3 = "11" in binary.
For n = 10 = 2*5, a(10) = 3 because the Hamming distance between 2 and 5 is 3 as 2 = "10" in binary (extended with a leading zero to make it "010") and 5 = "101" in binary.
For n = 12 = 2*2*3, a(12) = 2 because the Hamming distance between 2 and 3 is 1, and the pair (2,3) occurs twice as one can pick either one of the two 2's present in the prime factorization to be a pair of a single 3. Note that the Hamming distance between 2 and 2 is 0, thus the pair (2,2) of prime divisors does not contribute to the sum.
For n = 36 = 2*2*3*3, a(36) = 4 because the Hamming distance between 2 and 3 is 1, and the prime factor pair (2,3) occurs four times in total. Note that the Hamming distance is zero between 2 and 2 as well as between 3 and 3, thus the pairs (2,2) and (3,3) do not contribute to the sum.
		

Links

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

Crossrefs

Cf. A101080.

Cf. A000961 (positions of the zeros), A261077 (positions of the ones).

Cf. A072594, A072595, A235488.

Cf. also A261079.