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

1,1

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.

Links

Sean A. Irvine, Table of n, a(n) for n = 1..344

Example

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.

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 == 6])
    for x in range(len(rets)):
        print(rets[x])

Crossrefs

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

Sequence in context: A055158 A187635 A345524 * A076528 A260067 A031903

Adjacent sequences: A345775 A345776 A345777 * A345779 A345780 A345781

Keyword

nonn

Author

David Consiglio, Jr., Jun 26 2021

Status

approved