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

Author

David Consiglio, Jr., Jun 26 2021

Keywords

• nonn

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.

Examples

			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.
		

Links

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

Crossrefs

Cf. A048931, A345521, A345774, A345776, A345785, A345825.

Programs

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])