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A260737
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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.
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4
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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
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OFFSET
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1,10
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LINKS
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EXAMPLE
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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.
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PROG
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(Scheme, with Aubrey Jaffer's SLIB Scheme library)
(require 'factor)
(define (A260737 n) (let loop ((s 0) (pfs (factor n))) (cond ((or (null? pfs) (null? (cdr pfs))) s) (else (loop (fold-left (lambda (a p) (+ a (A101080bi (car pfs) p))) s (cdr pfs)) (cdr pfs))))))
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CROSSREFS
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Cf. A000961 (positions of the zeros), A261077 (positions of the ones).
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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