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A382423 The number of exponents in the prime factorization of n-th biquadratefree number that are equal to 2.

%I A382423 #11 Mar 26 2025 11:38:24
%S A382423 0,0,0,1,0,0,0,0,1,0,0,1,0,0,0,0,1,0,1,0,0,0,0,1,0,0,1,0,0,0,0,0,0,2,
%T A382423 0,0,0,0,0,0,0,1,1,0,0,1,1,0,1,0,0,0,0,0,0,0,1,0,0,1,0,0,0,1,0,0,0,1,
%U A382423 0,0,1,1,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0
%N A382423 The number of exponents in the prime factorization of n-th biquadratefree number that are equal to 2.
%H A382423 Amiram Eldar, <a href="/A382423/b382423.txt">Table of n, a(n) for n = 1..10000</a>
%H A382423 Sourabhashis Das, Wentang Kuo, and Yu-Ru Liu, <a href="https://doi.org/10.1016/j.jnt.2024.08.007">On the number of prime factors with a given multiplicity over h-free and h-full numbers</a>, Journal of Number Theory, Vol. 267 (2025), pp. 176-201; <a href="https://arxiv.org/abs/2409.11275">arXiv preprint</a>, arXiv:2409.11275 [math.NT], 2024. See Theorem 1.2.
%F A382423 a(n) = A369427(A046100(n)).
%F A382423 a(n) = A382425(n) - A382424(n).
%F A382423 Sum_{A046100(k) <= x} a(k) = c * x + O(sqrt(x)/log(x)), where c = (1/zeta(4)) * Sum_{p prime} (p*(p-1)/(p^4-1)) = 0.26498866091940182979... (Das et al., 2025).
%F A382423 Sum_{k=1..n} a(k) ~ c * n, where c = Sum_{p prime} (p*(p-1)/(p^4-1)) = 0.28680338438307129... - _Vaclav Kotesovec_, Mar 25 2025 (according to the above formula)
%t A382423 f[k_] := Module[{e = If[k == 1, {}, FactorInteger[k][[;; , 2]]]}, If[AllTrue[e, # < 4 &], Count[e, 2], Nothing]]; Array[f, 150]
%o A382423 (PARI) list(kmax) = {my(e, is); for(k = 1, kmax, e = factor(k)[, 2]; is = 1; for(i = 1, #e, if(e[i] > 3, is = 0; break)); if(is, print1(#select(x -> x == 2, e), ", "))); }
%Y A382423 Cf. A013662, A046100, A369427, A376366, A382422, A382424, A382425.
%K A382423 nonn,easy
%O A382423 1,34
%A A382423 _Amiram Eldar_, Mar 25 2025