A345796 Numbers that are the sum of nine cubes in exactly four ways.

224, 257, 264, 283, 320, 348, 355, 372, 374, 376, 381, 383, 390, 400, 402, 407, 411, 414, 416, 442, 450, 453, 454, 461, 474, 476, 481, 486, 488, 500, 503, 509, 510, 514, 519, 528, 529, 537, 542, 543, 544, 545, 548, 550, 552, 554, 555, 557, 564, 572, 573, 574

Offset: 1

Author

David Consiglio, Jr., Jun 26 2021

Keywords

• nonn

Comments

Differs from A345543 at term 17 because 409 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3.

Likely finite.

Examples

			257 is a term because 257 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 4^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3.
		

Links

• Sean A. Irvine, Table of n, a(n) for n = 1..124

Crossrefs

Cf. A345543, A345786, A345795, A345797, A345806, A345846.

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, 9):
    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])