A345782 Numbers that are the sum of seven cubes in exactly ten ways.

1704, 1711, 1800, 1837, 1863, 1926, 1938, 1963, 2008, 2019, 2045, 2053, 2059, 2078, 2113, 2143, 2161, 2171, 2176, 2217, 2223, 2250, 2260, 2266, 2276, 2286, 2295, 2304, 2313, 2315, 2331, 2350, 2354, 2357, 2374, 2404, 2412, 2413, 2442, 2444, 2446, 2447, 2511

Offset

1,1

Comments

Differs from A345506 at term 3 because 1774 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 12^3 = 1^3 + 1^3 + 1^3 + 2^3 + 6^3 + 6^3 + 11^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 9^3 + 10^3 = 1^3 + 1^3 + 4^3 + 5^3 + 5^3 + 9^3 + 9^3 = 1^3 + 2^3 + 3^3 + 4^3 + 6^3 + 9^3 + 9^3 = 1^3 + 2^3 + 4^3 + 4^3 + 5^3 + 8^3 + 10^3 = 1^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 + 11^3 = 2^3 + 2^3 + 2^3 + 4^3 + 7^3 + 7^3 + 10^3 = 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3 + 11^3 = 3^3 + 3^3 + 6^3 + 6^3 + 6^3 + 7^3 + 9^3 = 4^3 + 4^3 + 4^3 + 5^3 + 6^3 + 8^3 + 9^3.

Likely finite.

Links

Sean A. Irvine, Table of n, a(n) for n = 1..328

Example

1711 is a term because 1711 = 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 8^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 5^3 + 8^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 7^3 + 9^3 = 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 10^3 = 1^3 + 2^3 + 2^3 + 2^3 + 6^3 + 6^3 + 9^3 = 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 10^3 = 1^3 + 3^3 + 3^3 + 4^3 + 5^3 + 7^3 + 8^3 = 2^3 + 2^3 + 3^3 + 5^3 + 6^3 + 6^3 + 8^3 = 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 9^3 = 4^3 + 4^3 + 5^3 + 5^3 + 6^3 + 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 == 10])
    for x in range(len(rets)):
        print(rets[x])

Crossrefs

Cf. A345506, A345772, A345781, A345792, A345832.

Sequence in context: A166400 A210121 A345506 * A157287 A029559 A222553

Adjacent sequences: A345779 A345780 A345781 * A345783 A345784 A345785

Keyword

nonn

Author

David Consiglio, Jr., Jun 26 2021

Status

approved