A330013 a(n) is the number of solutions with nonnegative (x,y,z) to the cubic Diophantine equation x^3+y^3+z^3 - 3*x*y*z = n.

3, 3, 0, 3, 3, 0, 3, 6, 6, 3, 3, 0, 3, 3, 0, 6, 3, 6, 3, 6, 0, 3, 3, 0, 3, 3, 9, 12, 3, 0, 3, 6, 0, 3, 9, 6, 3, 3, 0, 6, 3, 0, 3, 6, 6, 3, 3, 0, 9, 3, 0, 6, 3, 12, 3, 12, 0, 3, 3, 0, 3, 3, 6, 9, 9, 0, 3, 6, 0, 9, 3, 12, 3, 3, 0, 6, 9, 0, 3, 6, 12, 3, 3, 0, 3

Offset

1,1

Comments

Some results coming from the Alarcon and Duval reference.

For n = 0, there are infinitely many solutions because every triple (k,k,k) with k >= 0 satisfies the equation.

a(n) = 0 iff 3 divides n and 9 doesn't divide n (equivalent to n is in A016051).

When n belongs to A074232 (complement of A016051), a(n) is always a multiple of 3 because

1) if (a,a,b) [resp. (a,b,b)] with a < b is a primitive solution, then these triples generate 3 solutions with the permutations (a,a,b), (a,b,a), (b,a,a), [resp. (a,b,b), (b,b,a), (b,a,b)] and,

2) if (a,b,c) with a < b < c is a primitive solution, then this triple generates 6 solutions with the permutations (a,b,c), (b,c,a), (c,a,b), (a,c,b), (c,b,a), (b,a,c).

For prime p <> 3, a(p) = a(2*p) = 3.

An inequality: (n/4)^(1/3) <= max(x, y, z) <= (n+2)/3.

This sequence is unbounded.

A261029 gives the number of triples without counting the permutations and, in link, a list of primitive triples up to n = 2000.

References

Guy Alarcon and Yves Duval, TS: Préparation au Concours Général, RMS, Collection Excellence, Paris, 2010, chapitre 9, Problème: étude d'une équation diophantienne cubique, pages 137-138 and 147-152.

Links

Table of n, a(n) for n=1..85.

Vladimir Shevelev, Representation of positive integers by the form x^3+y^3+z^3-3xyz, arXiv:1508.05748 [math.NT], 2015.

Formula

If n = 3*k + 1, then (k, k, k+1) is a solution for k >= 0.

If n = 3*k - 1, then (k, k, k-1) is a solution for k >= 1.

If n = 9*k, then (k-1, k, k+1) is a solution for k >= 1.

If n = k^3, then (k, 0, 0) is a solution for k >= 0.

If n = 2*k^3, then (k, k, 0) is a solution for k >= 0.

Example

3^3+2^3+2^3-3*2*2*3 = 7 so (3,2,2), (2,2,3) and (2,3,2) are solutions and a(7) = 3.
When n=35, (0,1,3) is a primitive solution that generates 6 solutions and (9,9,10) is another primitive solution that generates 3 solutions, so a(35)=6+3=9 (see comments).

Mathematica

a[n_] := Length@ Solve[x^3 + y^3 + z^3 - 3 x y z == n && x >= 0 && y >= 0 && z >= 0, {x, y, z}, Integers]; Array[a, 85] (* Giovanni Resta, Nov 28 2019 *)

Crossrefs

Cf. A016051, A074232.

Cf. A261029 (primitive triples without the permutations).

Cf. A050787, A050791, A212420 (other cubic Diophantine equations).

Sequence in context: A084055 A084103 A036477 * A128164 A376032 A339702

Adjacent sequences: A330010 A330011 A330012 * A330014 A330015 A330016

Keyword

nonn

Author

Bernard Schott, Nov 27 2019

Extensions

More terms from Giovanni Resta, Nov 28 2019

Status

approved