A345519 Numbers that are the sum of six cubes in ten or more ways.

3104, 3215, 3222, 3267, 3313, 3339, 3374, 3430, 3465, 3491, 3493, 3528, 3547, 3584, 3645, 3654, 3698, 3708, 3736, 3745, 3752, 3754, 3761, 3771, 3780, 3789, 3817, 3824, 3843, 3862, 3869, 3878, 3880, 3888, 3906, 3915, 3923, 3943, 3950, 3969, 3995, 4004, 4014

Offset: 1

Author

David Consiglio, Jr., Jun 20 2021

Keywords

• nonn

Examples

			3215 is a term because 3215 = 1^3 + 1^3 + 1^3 + 4^3 + 6^3 + 13^3 = 1^3 + 1^3 + 2^3 + 2^3 + 9^3 + 12^3 = 1^3 + 1^3 + 3^3 + 8^3 + 8^3 + 11^3 = 1^3 + 2^3 + 3^3 + 5^3 + 8^3 + 12^3 = 1^3 + 3^3 + 3^3 + 3^3 + 10^3 + 11^3 = 1^3 + 3^3 + 8^3 + 8^3 + 8^3 + 9^3 = 2^3 + 3^3 + 6^3 + 6^3 + 8^3 + 11^3 = 3^3 + 3^3 + 3^3 + 8^3 + 9^3 + 10^3 = 3^3 + 5^3 + 5^3 + 5^3 + 6^3 + 12^3 = 6^3 + 6^3 + 6^3 + 6^3 + 7^3 + 10^3.
		

Links

• Sean A. Irvine, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A345187, A345477, A345506, A345518, A345567, A345772.

Programs

Python

from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**3 for x in range(1, 1000)]
for pos in cwr(power_terms, 6):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 10])
    for x in range(len(rets)):
        print(rets[x])