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A293439
Number of odious exponents in the prime factorization of n.
11
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
1,6
FORMULA
Additive with a(p^e) = A010060(e).
a(n) = A007814(A293443(n)).
From Amiram Eldar, Sep 28 2023: (Start)
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)
EXAMPLE
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.
MATHEMATICA
a[n_] := Total@ ThueMorse[FactorInteger[n][[;; , 2]]]; a[1] = 0; Array[a, 100] (* Amiram Eldar, May 18 2023 *)
PROG
(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
CROSSREFS
Cf. A270428 (numbers such that a(n) = A001221(n)).
Differs from A144095 for the first time at n=24.
Sequence in context: A366078 A344652 A366074 * A144095 A362719 A076092
KEYWORD
nonn,easy
AUTHOR
Antti Karttunen, Nov 01 2017
STATUS
approved