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A293439
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Number of odious exponents in the prime factorization of n.
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11
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0, 1, 1, 1, 1, 2, 1, 0, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 1, 1, 1, 2, 0, 2, 1, 3, 1, 0, 2, 2, 2, 2, 1, 2, 2, 1, 1, 3, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 1, 2, 1, 2, 2, 1, 3, 1, 2, 2, 0, 2, 3, 1, 2, 2, 3, 1, 1, 1, 2, 2, 2, 2, 3, 1, 2, 1, 2, 1, 3, 2, 2, 2, 1, 1, 3, 2, 2, 2, 2, 2, 1, 1, 2, 2, 2, 1, 3, 1, 1, 3
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OFFSET
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1,6
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LINKS
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FORMULA
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a(n) >= 0, with equality if and only if n is an exponentially evil number (A262675).
a(n) <= A001221(n), with equality if and only if n is an exponentially odious number (A270428).
Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B + C), where B is Mertens's constant (A077761) and C = Sum_{p prime} f(1/p) = -0.12689613844142998028..., where f(x) = 1/2 - x - ((1-x)/2) * Product_{k>=0} (1-x^(2^k)). (End)
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EXAMPLE
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For n = 2 = 2^1, the only exponent 1 is odious (that is, has an odd Hamming weight and thus is included in A000069), so a(2) = 1.
For n = 24 = 2^3 * 3^1, the exponent 3 (with binary representation "11") is evil (has an even Hamming weight and thus is included in A001969), while the other exponent 1 is odious, so a(24) = 1.
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MATHEMATICA
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a[n_] := Total@ ThueMorse[FactorInteger[n][[;; , 2]]]; a[1] = 0; Array[a, 100] (* Amiram Eldar, May 18 2023 *)
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PROG
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(PARI) A293439(n) = vecsum(apply(e -> (hammingweight(e)%2), factorint(n)[, 2]));
(Python)
from sympy import factorint
def A293439(n): return sum(1 for e in factorint(n).values() if e.bit_count()&1) # Chai Wah Wu, Nov 23 2023
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CROSSREFS
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Differs from A144095 for the first time at n=24.
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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