A261029 Number of ways to write n in the form F(x,y,z) = x^3 + y^3 + z^3 - 3xyz, where 0 <= x <= y <= z and z >= x+1.

0, 1, 1, 0, 1, 1, 0, 1, 2, 1, 1, 1, 0, 1, 1, 0, 2, 1, 1, 1, 2, 0, 1, 1, 0, 1, 1, 2, 3, 1, 0, 1, 2, 0, 1, 2, 1, 1, 1, 0, 2, 1, 0, 1, 2, 1, 1, 1, 0, 2, 1, 0, 2, 1, 3, 1, 3, 0, 1, 1, 0, 1, 1, 1, 3, 2, 0, 1, 2, 0, 2, 1, 2, 1, 1, 0, 2, 2, 0, 1, 2, 3, 1, 1, 0, 1, 1

Offset

0,9

Comments

The following is a short proof of the corresponding 1915 result of R. D. Carmichael for a weaker restriction.

If n is in A074232, then a(n) >= 1, in view of the following identities: if n == 1 (mod 3), then n = F((n-1)/3, (n-1)/3, (n+2)/3); if n == 2 (mod 3), then n = F((n-2)/3, (n+1)/3, (n+1)/3); if n == 0 (mod 9), then n = F(n/9-1, n/9, n/9+1). QED

Further, if n > 1 is the cube of a positive number or the sum of two positive cubes, except for 2 and 9, then a(n) >= 2.

The sequence is unbounded.

Proof. We use the homogeneity of F(x,y,z) of degree 3. By induction, show that a(8^k) >= k+1. It is evident for k=0. Suppose that it is true for some value of k. Take k+1 triples (x_i,y_i,z_i) such that 8^k = F(x_i, y_i, z_i), i=1,...,k+1. Then for k+1 triples of even numbers (2*x_i, 2*y_i, 2*z_i) we have 8^(k+1) = F(2*x_i, 2*y_i, 2*z_i). But there is always a triple of not all even numbers x=(n-1)/3, y=(n-1)/3, z=(n+2)/3) or x=((n-2)/3, y=(n+1)/3, z=(n+1)/3), where n= 8^(k+1), for which 8^(k+1) = F(x,y,z). So a(8^(k+1)) >= k+2. QED

Theorem. For every n there exists k such that a(k)=n. For a proof, see [Shevelev] link.

Smallest such k are presented in sequence A260935.

Links

Peter J. C. Moses and Chai Wah Wu, Table of n, a(n) for n = 0..10000 (terms for n = 0..999 from Peter J. C. Moses)

R. D. Carmichael, On the representation of numbers in the form x^3+y^3+z^3-3xyz, Bull. Amer. Math. Soc. 22 (1915), 111-117.

Peter J. C. Moses, List of triples up to a(2000)

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

Formula

For positive n, a(n)=0, if and only if n == 3 or 6 (mod 9); if p is prime, other than 3, then a(p) = a(2*p) = 1.

For n >= 1, a(8^(n-1)) = n.

Mathematica

r[n_] := Reduce[0 <= x <= y <= z && z >= x+1 && n == x^3 + y^3 + z^3 - 3 x y z, {x, y, z}, Integers];
a[n_] := Which[rn = r[n]; rn === False, 0, rn[[0]] === And, 1, rn[[0]] === Or, Length[rn], True, Print["error ", rn]];
Array[a, 100, 0] (* Jean-François Alcover, Nov 06 2018 *)

Crossrefs

Cf. A072670, A074232, A260803, A260804, A260965.

Sequence in context: A340219 A260413 A053252 * A117195 A156606 A324606

Adjacent sequences: A261026 A261027 A261028 * A261030 A261031 A261032

Keyword

nonn

Author

Vladimir Shevelev, Aug 22 2015

Extensions

More terms from Peter J. C. Moses, Aug 22 2015

Status

approved