A345777 Numbers that are the sum of seven cubes in exactly five ways.

627, 768, 838, 845, 857, 864, 874, 881, 894, 900, 920, 937, 950, 962, 976, 981, 983, 990, 1002, 1009, 1011, 1016, 1027, 1054, 1060, 1061, 1063, 1089, 1096, 1098, 1102, 1105, 1109, 1115, 1124, 1128, 1133, 1135, 1137, 1140, 1144, 1151, 1153, 1154, 1159, 1163

Offset

1,1

Comments

Differs from A345523 at term 14 because 955 = 1^3 + 1^3 + 1^3 + 2^3 + 6^3 + 6^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 9^3 = 1^3 + 3^3 + 3^3 + 5^3 + 6^3 + 6^3 + 7^3 = 1^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 + 8^3 = 2^3 + 2^3 + 4^3 + 4^3 + 5^3 + 7^3 + 7^3 = 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3 + 8^3.

Likely finite.

Links

Sean A. Irvine, Table of n, a(n) for n = 1..351

Example

768 is a term because 768 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 8^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 6^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 7^3 = 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 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, 7):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 5])
    for x in range(len(rets)):
        print(rets[x])

Crossrefs

Cf. A345523, A345767, A345776, A345778, A345787, A345827.

Sequence in context: A098262 A031523 A345523 * A129974 A031703 A158382

Adjacent sequences: A345774 A345775 A345776 * A345778 A345779 A345780

Keyword

nonn

Author

David Consiglio, Jr., Jun 26 2021

Status

approved