A345538 Numbers that are the sum of eight cubes in eight or more ways.

970, 977, 984, 1054, 1073, 1075, 1080, 1090, 1099, 1106, 1110, 1125, 1129, 1136, 1148, 1160, 1166, 1171, 1178, 1181, 1185, 1186, 1188, 1192, 1197, 1204, 1206, 1211, 1217, 1218, 1223, 1225, 1230, 1232, 1234, 1236, 1237, 1242, 1243, 1249, 1262, 1263, 1269, 1273

Offset

1,1

Links

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

Example

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.

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, 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])

Crossrefs

Cf. A345495, A345526, A345537, A345539, A345547, A345583, A345790.

Sequence in context: A175991 A030251 A252196 * A345790 A257975 A217161

Adjacent sequences: A345535 A345536 A345537 * A345539 A345540 A345541

Keyword

nonn

Author

David Consiglio, Jr., Jun 20 2021

Status

approved