A382423 The number of exponents in the prime factorization of n-th biquadratefree number that are equal to 2.
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, 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, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0
Offset: 1
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
- Sourabhashis Das, Wentang Kuo, and Yu-Ru Liu, On the number of prime factors with a given multiplicity over h-free and h-full numbers, Journal of Number Theory, Vol. 267 (2025), pp. 176-201; arXiv preprint, arXiv:2409.11275 [math.NT], 2024. See Theorem 1.2.
Programs
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Mathematica
f[k_] := Module[{e = If[k == 1, {}, FactorInteger[k][[;; , 2]]]}, If[AllTrue[e, # < 4 &], Count[e, 2], Nothing]]; Array[f, 150] -
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), ", "))); }
Formula
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).
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)