A293439 Number of odious exponents in the prime factorization of n.
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
Offset: 1
Examples
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.
Links
Crossrefs
Programs
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Mathematica
a[n_] := Total@ ThueMorse[FactorInteger[n][[;; , 2]]]; a[1] = 0; Array[a, 100] (* Amiram Eldar, May 18 2023 *)
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PARI
A293439(n) = vecsum(apply(e -> (hammingweight(e)%2), factorint(n)[, 2]));
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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
Formula
Additive with a(p^e) = A010060(e).
From Amiram Eldar, Sep 28 2023: (Start)
a(n) >= 0, with equality if and only if n is an exponentially evil number (A262675).
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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