A345519 - OEIS

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

LINKS

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

EXAMPLE

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.

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 >= 10])

for x in range(len(rets)):

print(rets[x])