A345786 Numbers that are the sum of eight cubes in exactly four ways.

256, 347, 382, 401, 408, 427, 434, 438, 445, 464, 478, 480, 490, 499, 502, 506, 511, 516, 523, 530, 532, 534, 537, 560, 565, 567, 569, 571, 578, 586, 593, 595, 600, 602, 604, 605, 611, 612, 616, 619, 621, 624, 626, 643, 645, 656, 660, 663, 664, 668, 675, 679

Offset: 1

Author

David Consiglio, Jr., Jun 26 2021

Keywords

• nonn

Comments

Differs from A345534 at term 11 because 471 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 6^3 + 6^3 = 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 = 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3 + 5^3 + 5^3.

Likely finite.

Examples

			347 is a term because 347 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3.
		

Links

• Sean A. Irvine, Table of n, a(n) for n = 1..207

Crossrefs

Cf. A345534, A345776, A345785, A345787, A345796, A345836.

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, 8):
    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])