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
Keywords
Examples
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).
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
- Vladimir Shevelev, Representation of positive integers by the form x^3+y^3+z^3-3xyz, arXiv:1508.05748 [math.NT], 2015.
Crossrefs
Programs
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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 *)
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.
Extensions
More terms from Giovanni Resta, Nov 28 2019
Comments