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