cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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.

Original entry on oeis.org

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

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Author

Vladimir Shevelev, Aug 22 2015

Keywords

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.

Crossrefs

Programs

  • 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 *)

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.

Extensions

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