A345778 Numbers that are the sum of seven cubes in exactly six ways.

955, 969, 1046, 1053, 1079, 1107, 1117, 1121, 1158, 1161, 1177, 1184, 1196, 1198, 1216, 1222, 1242, 1254, 1272, 1280, 1287, 1291, 1294, 1297, 1298, 1310, 1324, 1350, 1351, 1355, 1366, 1369, 1376, 1378, 1388, 1403, 1404, 1415, 1417, 1418, 1422, 1433, 1437

Offset: 1

Author

David Consiglio, Jr., Jun 26 2021

Keywords

• nonn

Comments

Differs from A345524 at term 5 because 1072 = 1^3 + 1^3 + 1^3 + 5^3 + 6^3 + 6^3 + 8^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 10^3 = 1^3 + 1^3 + 3^3 + 4^3 + 5^3 + 5^3 + 9^3 = 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 6^3 + 9^3 = 2^3 + 4^3 + 4^3 + 5^3 + 5^3 + 7^3 + 7^3 = 3^3 + 3^3 + 3^3 + 6^3 + 6^3 + 6^3 + 7^3 = 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 6^3 + 8^3.

Likely finite.

Examples

			969 is a term because 969 = 1^3 + 1^3 + 1^3 + 3^3 + 5^3 + 6^3 + 6^3 = 1^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 7^3 = 1^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 6^3 = 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 8^3 = 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 6^3 = 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..344

Crossrefs

Cf. A345524, A345768, A345777, A345779, A345788, A345828.

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, 7):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 6])
    for x in range(len(rets)):
        print(rets[x])