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editing
approved
editing
approved
Likely finite.
Sean A. Irvine, <a href="/A345796/b345796.txt">Table of n, a(n) for n = 1..124</a>
allocated for Sean A. Irvine
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
1,1
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.
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.
(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])
allocated
nonn,new
David Consiglio, Jr., Jun 26 2021
approved
editing
allocated for Sean A. Irvine
allocated
approved