A345768 - OEIS

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

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OFFSET

1,1

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.

LINKS

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

EXAMPLE

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.

PROG

(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])