A385006 The sum of the biquadratefree divisors of n.

1, 3, 4, 7, 6, 12, 8, 15, 13, 18, 12, 28, 14, 24, 24, 15, 18, 39, 20, 42, 32, 36, 24, 60, 31, 42, 40, 56, 30, 72, 32, 15, 48, 54, 48, 91, 38, 60, 56, 90, 42, 96, 44, 84, 78, 72, 48, 60, 57, 93, 72, 98, 54, 120, 72, 120, 80, 90, 60, 168, 62, 96, 104, 15, 84, 144

Offset: 1

Author

Amiram Eldar, Jun 15 2025

Keywords

• nonn
• easy
• mult

Comments

First differs from A365682 and A366992 at n = 32.

The number of these divisors is A252505(n), and the largest of them is A058035(n).

Links

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

Crossrefs

Cf. A046100, A048250, A058035, A073185, A252505, A365682, A366992.

The sum of divisors d of n such that d is: A000593 (odd), A033634 (exponentially odd), A035316 (square), A038712 (power of 2), A048250 (squarefree), A072079 (3-smooth), A073185 (cubefree), A113061 (cube), A162296 (nonsquarefree), A183097 (powerful), A186099 (5-rough), A353900 (exponentially 2^n), A385005 (cubefull), this sequence (biquadratefree).

Programs

Mathematica

f[p_, e_] := (p^Min[e+1, 4] - 1)/(p - 1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

PARI

a(n) = {my(f = factor(n), p, e); prod(i = 1, #f~, p = f[i,1]; e = f[i,2]; (p^min(e+1, 4) - 1)/(p - 1));}

Formula

Multiplicative with a(p^e) = (p^min(e+1, 4) - 1)/(p - 1).

In general, the sum of the k-free (numbers that are not divisible by a k-th power larger than 1) divisors of n is multiplicative with a(p^e) = (p^min(e+1, k) - 1)/(p - 1).

Dirichlet g.f.: zeta(s) * zeta(s-1) /zeta(4*s-4).

In general, the sum of the k-free divisors of n has Dirichlet g.f.: zeta(s)*zeta(s-1)/zeta(k*s-k).

Sum_{k=1..n} a(k) ~ (15/(2*Pi^2)) * n^2.

In general, the sum of the k-free divisors of n has an average order (Pi^2/(12*zeta(k))) * n^2.