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%I #43 Jan 09 2020 05:47:29
%S 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,
%T 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,
%U 6,0,9,3,12,3,3,0,6,9,0,3,6,12,3,3,0,3
%N 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.
%C Some results coming from the Alarcon and Duval reference.
%C For n = 0, there are infinitely many solutions because every triple (k,k,k) with k >= 0 satisfies the equation.
%C a(n) = 0 iff 3 divides n and 9 doesn't divide n (equivalent to n is in A016051).
%C When n belongs to A074232 (complement of A016051), a(n) is always a multiple of 3 because
%C 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,
%C 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).
%C For prime p <> 3, a(p) = a(2*p) = 3.
%C An inequality: (n/4)^(1/3) <= max(x, y, z) <= (n+2)/3.
%C This sequence is unbounded.
%C A261029 gives the number of triples without counting the permutations and, in link, a list of primitive triples up to n = 2000.
%D 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.
%H Vladimir Shevelev, <a href="https://arxiv.org/abs/1508.05748">Representation of positive integers by the form x^3+y^3+z^3-3xyz</a>, arXiv:1508.05748 [math.NT], 2015.
%F If n = 3*k + 1, then (k, k, k+1) is a solution for k >= 0.
%F If n = 3*k - 1, then (k, k, k-1) is a solution for k >= 1.
%F If n = 9*k, then (k-1, k, k+1) is a solution for k >= 1.
%F If n = k^3, then (k, 0, 0) is a solution for k >= 0.
%F If n = 2*k^3, then (k, k, 0) is a solution for k >= 0.
%e 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.
%e 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).
%t 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 *)
%Y Cf. A016051, A074232.
%Y Cf. A261029 (primitive triples without the permutations).
%Y Cf. A050787, A050791, A212420 (other cubic Diophantine equations).
%K nonn
%O 1,1
%A _Bernard Schott_, Nov 27 2019
%E More terms from _Giovanni Resta_, Nov 28 2019
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