A345785 Numbers that are the sum of eight cubes in exactly three ways.

223, 230, 237, 249, 263, 270, 275, 282, 284, 286, 289, 291, 293, 308, 310, 312, 319, 326, 345, 349, 354, 364, 371, 373, 375, 378, 380, 385, 386, 387, 389, 397, 399, 404, 406, 410, 412, 413, 415, 420, 423, 439, 441, 443, 446, 449, 452, 453, 459, 460, 465, 473

Offset

1,1

Comments

Differs from A345533 at term 5 because 256 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 6^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 5^3.

Likely finite.

Links

Sean A. Irvine, Table of n, a(n) for n = 1..198

Example

230 is a term because 230 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 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, 8):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 3])
    for x in range(len(rets)):
        print(rets[x])

Crossrefs

Cf. A345533, A345775, A345784, A345786, A345795, A345835.

Sequence in context: A220474 A243767 A345533 * A359449 A098591 A138665

Adjacent sequences: A345782 A345783 A345784 * A345786 A345787 A345788

Keyword

nonn

Author

David Consiglio, Jr., Jun 26 2021

Status

approved