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

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, 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, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 2, 1

Offset: 1

Author

Amiram Eldar, Mar 25 2025

Keywords

• nonn
• easy

Comments

Differs from A375766 and A375768 at n = 1, 31, 34, 35, 38, 39, ... .

All the terms are 3-smooth numbers (A003586).

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A003586, A005361, A013662, A046100, A082695, A382423, A382424.

Cf. A375766, A375768.

Programs

Mathematica

s[n_] := Times @@ FactorInteger[n][[;; , 2]]; biqFreeQ[n_] := Max[FactorInteger[n][[;; , 2]]] < 4; s /@ Select[Range[100], biqFreeQ]

PARI

list(kmax) = {my(e); print1(1, ", "); for(k = 2, kmax, e = factor(k)[, 2]; if(vecmax(e) < 4, print1(vecprod(e), ", "))); }

Formula

a(n) = A005361(A046100(n)).

a(n) = 2^A382423(n) * 3^A382424(n).

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... .

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).