A345789 Numbers that are the sum of eight cubes in exactly seven ways.

902, 908, 921, 938, 958, 963, 982, 991, 996, 1003, 1008, 1010, 1017, 1019, 1028, 1029, 1033, 1047, 1055, 1058, 1061, 1062, 1070, 1087, 1091, 1094, 1096, 1097, 1104, 1108, 1111, 1113, 1115, 1116, 1118, 1120, 1122, 1123, 1127, 1134, 1141, 1143, 1145, 1152, 1153

Offset

1,1

Comments

Differs from A345537 at term 7 because 970 = 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 6^3 + 7^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 6^3 + 6^3 + 8^3 = 1^3 + 1^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 + 7^3 = 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 5^3 + 9^3 = 1^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 6^3 + 7^3 = 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3 + 7^3 = 2^3 + 2^3 + 3^3 + 3^3 + 5^3 + 6^3 + 6^3 + 7^3 = 2^3 + 2^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 + 8^3.

Likely finite.

Links

Sean A. Irvine, Table of n, a(n) for n = 1..174

Example

908 is a term because 908 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 6^3 + 7^3 = 1^3 + 1^3 + 2^3 + 4^3 + 4^3 + 5^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 5^3 = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 7^3 = 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 6^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 7^3 = 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3.

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 == 7])
    for x in range(len(rets)):
        print(rets[x])

Crossrefs

Cf. A345537, A345779, A345788, A345790, A345799, A345839.

Sequence in context: A268585 A224611 A345537 * A031528 A158406 A323143

Adjacent sequences: A345786 A345787 A345788 * A345790 A345791 A345792

Keyword

nonn

Author

David Consiglio, Jr., Jun 26 2021

Status

approved