A359944 Number of divisors d of n such that d-1 is a cube.

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

Offset: 1

Author

Seiichi Manyama, Jan 19 2023

Keywords

• nonn

Comments

The Cartesian equation for the Folium of Descartes is given as x^3 + y^3 = 3*k*x*y. If we set 3*k = n, then a(n)-1 is the number of integer solutions such that x,y > 0 and y >= x. Let d = m^3+1 be a divisor of n, then x = 3*k*m/(m^3+1); y = 3*k*m^2/(m^3+1) is a solution. - Thomas Scheuerle, Aug 07 2024

Crossrefs

Cf. A001093, A061704, A339606.

Programs

Mathematica

a[n_] := DivisorSum[n, 1 &, IntegerQ[Surd[#-1, 3]] &]; Array[a, 100] (* Amiram Eldar, Aug 09 2023 *)

PARI

a(n) = sumdiv(n, d, ispower(d-1, 3));

Formula

G.f.: Sum_{k>=0} x^(k^3+1)/(1 - x^(k^3+1)).

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{k>=0} 1/(k^3+1) = 1 + A339606 = 1.686503... . - Amiram Eldar, Jan 01 2024