A345768 Numbers that are the sum of six cubes in exactly six ways.

1377, 1488, 1586, 1595, 1647, 1673, 1677, 1738, 1764, 1799, 1829, 1836, 1837, 1862, 1881, 1890, 1911, 1953, 1955, 2007, 2011, 2014, 2018, 2025, 2044, 2070, 2079, 2097, 2107, 2108, 2142, 2153, 2170, 2177, 2203, 2214, 2216, 2222, 2223, 2226, 2229, 2252, 2258

Offset: 1

Author

David Consiglio, Jr., Jun 26 2021

Keywords

• nonn

Comments

Differs from A345515 at term 8 because 1710 = 1^3 + 1^3 + 5^3 + 5^3 + 9^3 + 9^3 = 1^3 + 2^3 + 3^3 + 6^3 + 9^3 + 9^3 = 1^3 + 2^3 + 4^3 + 5^3 + 8^3 + 10^3 = 1^3 + 4^3 + 4^3 + 5^3 + 5^3 + 11^3 = 2^3 + 2^3 + 2^3 + 7^3 + 7^3 + 10^3 = 2^3 + 3^3 + 4^3 + 4^3 + 6^3 + 11^3 = 4^3 + 4^3 + 5^3 + 6^3 + 8^3 + 9^3.

Examples

			1488 is a term because 1488 = 1^3 + 1^3 + 1^3 + 3^3 + 8^3 + 8^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 10^3 = 1^3 + 2^3 + 3^3 + 6^3 + 6^3 + 8^3 = 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 10^3 = 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 9^3 = 3^3 + 5^3 + 5^3 + 6^3 + 6^3 + 6^3.
		

Links

• Sean A. Irvine, Table of n, a(n) for n = 1..1338

Crossrefs

Cf. A345175, A345515, A345767, A345769, A345778, A345818.

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 == 6])
    for x in range(len(rets)):
        print(rets[x])