A345545 Numbers that are the sum of nine cubes in six or more ways.

472, 498, 505, 507, 524, 596, 598, 605, 624, 629, 631, 636, 643, 650, 655, 657, 661, 662, 669, 672, 676, 681, 687, 688, 690, 692, 694, 696, 706, 707, 713, 718, 720, 722, 725, 727, 728, 729, 731, 732, 737, 739, 742, 744, 746, 748, 749, 750, 751, 753, 755, 756

Offset

1,1

Links

Sean A. Irvine, Table of n, a(n) for n = 1..10000

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. A345503, A345536, A345544, A345546, A345554, A345590, A345798.

Sequence in context: A269749 A138132 A034284 * A345798 A235287 A260102

Adjacent sequences: A345542 A345543 A345544 * A345546 A345547 A345548

Keyword

nonn

Author

David Consiglio, Jr., Jun 20 2021

Status

approved