cp's OEIS Frontend

A382422 The product of exponents in the prime factorization of the biquadratefree numbers.

%I A382422 #8 Mar 25 2025 10:11:51
%S A382422 1,1,1,2,1,1,1,3,2,1,1,2,1,1,1,1,2,1,2,1,1,1,3,2,1,3,2,1,1,1,1,1,1,4,
%T A382422 1,1,1,3,1,1,1,2,2,1,1,2,2,1,2,1,3,1,3,1,1,1,2,1,1,2,1,1,1,2,1,1,1,6,
%U A382422 1,1,2,2,1,1,1,1,1,2,1,1,1,3,1,2,1,2,1
%N A382422 The product of exponents in the prime factorization of the biquadratefree numbers.
%C A382422 Differs from A375766 and A375768 at n = 1, 31, 34, 35, 38, 39, ... .
%C A382422 All the terms are 3-smooth numbers (A003586).
%H A382422 Amiram Eldar, <a href="/A382422/b382422.txt">Table of n, a(n) for n = 1..10000</a>
%F A382422 a(n) = A005361(A046100(n)).
%F A382422 a(n) = 2^A382423(n) * 3^A382424(n).
%F A382422 Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = zeta(4) * Product_{p prime} (1 + 1/p^2 + 1/p^3 - 3/p^4) = 1.57226906210272200398... .
%F A382422 In general, the asymptotic mean of the product of exponents in the prime factorization of the k-free numbers (numbers that are not divisible by a k-th power other than 1), for k >= 2, is zeta(k) * Product_{p prime} (1 + 1/p^2 + 1/p^3 + ... + 1/p^(k-1) - (k-1)/p^k). For k = 2 (squarefree numbers), the mean is 1 since the sequence contains only 1's. The limit when k->oo is zeta(2)*zeta(3)/zeta(6) (A082695).
%t A382422 s[n_] := Times @@ FactorInteger[n][[;; , 2]]; biqFreeQ[n_] := Max[FactorInteger[n][[;; , 2]]] < 4; s /@ Select[Range[100], biqFreeQ]
%o A382422 (PARI) list(kmax) = {my(e); print1(1, ", "); for(k = 2, kmax, e = factor(k)[, 2]; if(vecmax(e) < 4, print1(vecprod(e), ", "))); }
%Y A382422 Cf. A003586, A005361, A013662, A046100, A082695, A382423, A382424.
%Y A382422 Cf. A375766, A375768.
%K A382422 nonn,easy
%O A382422 1,4
%A A382422 _Amiram Eldar_, Mar 25 2025