A345517 Numbers that are the sum of six cubes in eight or more ways.

1981, 2105, 2168, 2277, 2368, 2376, 2431, 2438, 2457, 2466, 2494, 2538, 2555, 2557, 2583, 2593, 2646, 2665, 2672, 2709, 2746, 2753, 2763, 2765, 2772, 2880, 2881, 2889, 2916, 2942, 2961, 2970, 2977, 2979, 2980, 2987, 3007, 3033, 3040, 3042, 3043, 3049, 3068

Offset: 1

Author

David Consiglio, Jr., Jun 20 2021

Keywords

• nonn

Examples

			2105 is a term because 2105 = 1^3 + 1^3 + 4^3 + 4^3 + 4^3 + 11^3 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 11^3 = 1^3 + 2^3 + 6^3 + 7^3 + 7^3 + 8^3 = 1^3 + 4^3 + 4^3 + 4^3 + 8^3 + 9^3 = 1^3 + 4^3 + 5^3 + 5^3 + 5^3 + 10^3 = 2^3 + 3^3 + 4^3 + 5^3 + 8^3 + 9^3 = 3^3 + 3^3 + 3^3 + 7^3 + 7^3 + 9^3 = 5^3 + 5^3 + 5^3 + 5^3 + 7^3 + 8^3.
		

Links

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

Crossrefs

Cf. A344812, A345183, A345516, A345518, A345526, A345565, A345770.

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 >= 8])
    for x in range(len(rets)):
        print(rets[x])