A345779 Numbers that are the sum of seven cubes in exactly seven ways.

1072, 1170, 1235, 1261, 1268, 1305, 1392, 1396, 1411, 1440, 1441, 1448, 1450, 1459, 1489, 1502, 1504, 1513, 1538, 1540, 1547, 1559, 1564, 1565, 1566, 1567, 1576, 1592, 1593, 1594, 1600, 1602, 1606, 1620, 1621, 1625, 1626, 1628, 1629, 1639, 1658, 1664, 1667

Offset

1,1

Comments

Differs from A345525 at term 7 because 1385 = 1^3 + 1^3 + 2^3 + 2^3 + 7^3 + 8^3 + 8^3 = 1^3 + 1^3 + 2^3 + 5^3 + 5^3 + 5^3 + 10^3 = 1^3 + 2^3 + 2^3 + 3^3 + 5^3 + 6^3 + 10^3 = 1^3 + 2^3 + 6^3 + 6^3 + 6^3 + 6^3 + 8^3 = 1^3 + 4^3 + 5^3 + 5^3 + 5^3 + 6^3 + 9^3 = 2^3 + 3^3 + 3^3 + 5^3 + 7^3 + 7^3 + 8^3 = 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 6^3 + 9^3 = 3^3 + 3^3 + 3^3 + 4^3 + 6^3 + 8^3 + 8^3.

Likely finite.

Links

Sean A. Irvine, Table of n, a(n) for n = 1..345

Example

1170 is a term because 1170 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 4^3 + 9^3 = 1^3 + 1^3 + 2^3 + 5^3 + 5^3 + 5^3 + 7^3 = 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 8^3 = 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 8^3 = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 7^3 + 7^3 = 3^3 + 3^3 + 4^3 + 5^3 + 5^3 + 5^3 + 6^3 = 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 + 7^3.

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

Crossrefs

Cf. A345525, A345769, A345778, A345780, A345789, A345829.

Sequence in context: A345904 A252256 A345525 * A252257 A252260 A085337

Adjacent sequences: A345776 A345777 A345778 * A345780 A345781 A345782

Keyword

nonn

Author

David Consiglio, Jr., Jun 26 2021

Status

approved