A345549 - OEIS

A345549

Numbers that are the sum of nine cubes in ten or more ways.

966, 971, 978, 985, 992, 1004, 1011, 1018, 1022, 1048, 1055, 1056, 1062, 1063, 1074, 1076, 1078, 1081, 1083, 1085, 1088, 1092, 1093, 1095, 1097, 1098, 1100, 1102, 1104, 1107, 1109, 1111, 1112, 1114, 1117, 1118, 1119, 1121, 1123, 1124, 1126, 1128, 1130, 1133

(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

971 is a term because 971 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 5^3 + 6^3 + 6^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 7^3 = 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 6^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 6^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 6^3 = 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 5^3 + 6^3 = 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 7^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 6^3 + 6^3 = 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 7^3.

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, 9):

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])