A345806 Numbers that are the sum of ten cubes in exactly four ways.

225, 232, 251, 258, 265, 272, 284, 286, 291, 307, 310, 314, 321, 323, 328, 342, 347, 356, 363, 366, 373, 375, 377, 380, 389, 391, 398, 399, 405, 412, 419, 422, 424, 427, 434, 438, 441, 445, 450, 451, 458, 459, 461, 464, 469, 471, 476, 478, 481, 484, 488, 489

Offset

1,1

Comments

Differs from A345552 at term 9 because 288 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 5^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3.

Likely finite.

Links

Sean A. Irvine, Table of n, a(n) for n = 1..93

Example

232 is a term because 232 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 = 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3.

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, 10):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 4])
    for x in range(len(rets)):
        print(rets[x])

Crossrefs

Cf. A345552, A345796, A345805, A345807, A345856.

Sequence in context: A265432 A247885 A345552 * A044871 A202005 A359598

Adjacent sequences: A345803 A345804 A345805 * A345807 A345808 A345809

Keyword

nonn

Author

David Consiglio, Jr., Jun 26 2021

Status

approved