

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.


1



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
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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 137138 and 147152.


LINKS

Table of n, a(n) for n=1..85.
Vladimir Shevelev, Representation of positive integers by the form x^3+y^3+z^33xyz, 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, k1) is a solution for k >= 1.
If n = 9*k, then (k1, 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+2^33*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 A339702 A260636
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



