A345775 Numbers that are the sum of seven cubes in exactly three ways.

222, 229, 248, 255, 262, 281, 283, 285, 318, 346, 370, 374, 377, 379, 381, 396, 400, 407, 412, 419, 426, 433, 437, 438, 444, 451, 463, 472, 475, 477, 489, 494, 501, 505, 507, 510, 522, 529, 533, 536, 559, 564, 566, 568, 570, 577, 578, 584, 585, 592, 594, 596

Offset

1,1

Comments

Differs from A345521 at term 28 because 470 = 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 6^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 6^3 + 6^3 = 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 = 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3.

Likely finite.

Links

Sean A. Irvine, Table of n, a(n) for n = 1..390

Example

229 is a term because 229 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 5^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 = 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, 7):
    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. A048931, A345521, A345774, A345776, A345785, A345825.

Sequence in context: A043612 A371423 A345521 * A361628 A101955 A063352

Adjacent sequences: A345772 A345773 A345774 * A345776 A345777 A345778

Keyword

nonn

Author

David Consiglio, Jr., Jun 26 2021

Status

approved