A346137 Numbers k such that k^3 = x^3 + y^3 + z^3, x > y > z >= 0, has at least 2 distinct solutions.

18, 36, 41, 46, 54, 58, 60, 72, 75, 76, 81, 82, 84, 87, 88, 90, 92, 96, 100, 108, 114, 116, 120, 123, 126, 132, 134, 138, 140, 142, 144, 145, 150, 152, 156, 159, 160, 162, 164, 168, 170, 171, 174, 176, 178, 180, 184, 185, 186, 189, 190, 192, 198, 200, 201, 202, 203

Offset: 1

Author

Sebastian Magee, Jul 30 2021

Keywords

• nonn

Comments

This sequence is based on a generalization of Fermat's last theorem with n=3, in which three terms are added. Fermat's Theorem states that there are no solution with only two terms, this sequence shows there are many integers for which there are multiple solutions if three terms are allowed. The sequence is also related to the Taxicab numbers.

Examples

			41 is in the sequence because 41^3 = 33^3 + 32^3 + 6^3 = 40^3 + 17^3 + 2^3.
		

Crossrefs

Subsequence of A023042.

Cf. A001235, A346071.

Programs

Mathematica

q[k_] := Count[IntegerPartitions[k^3, {3}, Range[0, k-1]^3], ?(UnsameQ @@ # &)] > 1; Select[Range[200], q] (* _Amiram Eldar, Sep 03 2021 *)

Python

from itertools import combinations
from collections import Counter
from sympy import integer_nthroot
def icuberoot(n): return integer_nthroot(n, 3)[0]
def aupto(kmax):
    cubes = [i**3 for i in range(kmax+1)]
    cands, cubesset = (sum(c) for c in combinations(cubes, 3)), set(cubes)
    c = Counter(s for s in cands if s in cubesset)
    return sorted(icuberoot(s) for s in c if c[s] >= 2)
print(aupto(203)) # Michael S. Branicky, Sep 04 2021