A345790 Numbers that are the sum of eight cubes in exactly eight ways.

970, 977, 1054, 1073, 1075, 1090, 1099, 1106, 1110, 1125, 1129, 1148, 1160, 1166, 1178, 1181, 1186, 1188, 1206, 1211, 1217, 1218, 1225, 1230, 1232, 1234, 1236, 1237, 1242, 1249, 1263, 1276, 1281, 1286, 1292, 1298, 1305, 1312, 1314, 1321, 1323, 1324, 1334

Offset: 1

Author

David Consiglio, Jr., Jun 26 2021

Keywords

• nonn

Comments

Differs from A345538 at term 3 because 984 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 9^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 6^3 + 9^3 = 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 5^3 + 6^3 + 8^3 = 1^3 + 1^3 + 2^3 + 2^3 + 4^3 + 6^3 + 7^3 + 7^3 = 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 9^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 6^3 + 6^3 + 8^3 = 2^3 + 2^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 + 7^3 = 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 6^3 + 7^3 = 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 + 8^3.

Likely finite.

Examples

			977 is a term because 977 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 5^3 + 8^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 5^3 + 6^3 + 6^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 8^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 7^3 = 1^3 + 2^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 8^3 = 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 6^3.
		

Links

• Sean A. Irvine, Table of n, a(n) for n = 1..176

Crossrefs

Cf. A345538, A345780, A345789, A345791, A345800, A345840.

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, 8):
    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])