A345770 Numbers that are the sum of six cubes in exactly eight ways.

1981, 2105, 2168, 2277, 2368, 2376, 2431, 2466, 2538, 2557, 2583, 2646, 2665, 2672, 2746, 2753, 2763, 2765, 2880, 2881, 2916, 2961, 2970, 2977, 2979, 2987, 3007, 3040, 3042, 3049, 3068, 3088, 3141, 3159, 3169, 3185, 3248, 3278, 3311, 3312, 3367, 3384, 3393

Offset: 1

Author

David Consiglio, Jr., Jun 26 2021

Keywords

• nonn

Comments

Differs from A345517 at term 8 because 2438 = 1^3 + 2^3 + 2^3 + 2^3 + 6^3 + 13^3 = 1^3 + 2^3 + 4^3 + 5^3 + 8^3 + 12^3 = 1^3 + 5^3 + 5^3 + 9^3 + 9^3 + 9^3 = 2^3 + 2^3 + 2^3 + 7^3 + 7^3 + 12^3 = 2^3 + 2^3 + 3^3 + 4^3 + 10^3 + 11^3 = 2^3 + 3^3 + 6^3 + 9^3 + 9^3 + 9^3 = 2^3 + 4^3 + 5^3 + 8^3 + 9^3 + 10^3 = 4^3 + 4^3 + 5^3 + 5^3 + 9^3 + 11^3 = 6^3 + 7^3 + 7^3 + 8^3 + 8^3 + 8^3.

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..1347

Crossrefs

Cf. A345184, A345517, A345769, A345771, A345780, A345820.

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])