A385134 The sum of divisors d of n such that n/d is a biquadratefree number (A046100).

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

Offset

1,2

Links

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

Formula

a(n) = Sum_{d | n} d * A307430(n/d) = n * Sum_{d | n} A307430(d) / d.

a(n) = Sum_{d^3 | n} mu(d) * A000203(n/d^3), where mu is the Moebius function (A008683).

Multiplicative with a(p) = 1 + p, a(p^2) = 1 + p + p^2, and a(p^e) = p^(e-3) * (1 + p + p^2 + p^3), for e >= 3.

In general, the sum of divisors d of n such that n/d is k-free (not divisible by a k-th power larger than 1) is multiplicative with a(p^e) = p^max(e-k+1,0) * (p^min(e+1,k)-1)/(p-1).

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

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

Sum_{i=1..n} a(i) ~ (1575 / (2*Pi^6)) * n^2.

Mathematica

f[p_, e_] := p^(e-3)*(1 + p + p^2 + p^3); f[p_, 1] := 1 + p; f[p_, 2] := 1 + p + p^2; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

Prog

(PARI) a(n) = {my(f = factor(n), p, e); prod(i = 1, #f~, p = f[i, 1]; e = f[i, 2]; p^max(e-3, 0) * (p^min(e+1, 4)-1)/(p-1)); }

Crossrefs

Cf. A000203, A008683, A046100, A307430, A385006.

The sum of divisors d of n such that n/d is: A001615 (squarefree), A002131 (odd), A069208 (powerful), A076752 (square), A129527 (power of 2), A254981 (cubefree), A244963 (nonsquarefree), A327626 (cube), this sequence (biquadratefree), A385135 (exponentially odd), A385136 (cubefull), A385137 (3-smooth), A385138 (5-rough), A385139 (exponentially 2^n).

Sequence in context: A074847 A365170 A365174 * A325317 A325316 A227131

Adjacent sequences: A385131 A385132 A385133 * A385135 A385136 A385137

Keyword

nonn,easy,mult

Author

Amiram Eldar, Jun 19 2025

Status

approved