A345798 Numbers that are the sum of nine cubes in exactly six ways.

472, 498, 505, 507, 524, 596, 598, 605, 636, 643, 655, 661, 662, 669, 672, 676, 681, 688, 690, 692, 696, 706, 718, 722, 725, 728, 729, 731, 732, 737, 739, 742, 748, 749, 750, 751, 756, 765, 772, 782, 783, 785, 787, 788, 791, 793, 794, 800, 801, 802, 808, 810

Offset

1,1

Comments

Differs from A345545 at term 9 because 624 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 8^3 = 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 + 6^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 6^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 5^3 + 6^3 = 1^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 6^3 + 6^3 = 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3.

Likely finite.

Links

Sean A. Irvine, Table of n, a(n) for n = 1..127

Example

498 is a term because 498 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 5^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 6^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 = 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3.

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, 9):
    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. A345545, A345788, A345797, A345799, A345808, A345848.

Sequence in context: A138132 A034284 A345545 * A235287 A260102 A260135

Adjacent sequences: A345795 A345796 A345797 * A345799 A345800 A345801

Keyword

nonn

Author

David Consiglio, Jr., Jun 26 2021

Status

approved