A345806 - OEIS

A345806

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

225, 232, 251, 258, 265, 272, 284, 286, 291, 307, 310, 314, 321, 323, 328, 342, 347, 356, 363, 366, 373, 375, 377, 380, 389, 391, 398, 399, 405, 412, 419, 422, 424, 427, 434, 438, 441, 445, 450, 451, 458, 459, 461, 464, 469, 471, 476, 478, 481, 484, 488, 489

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OFFSET

1,1

COMMENTS

Differs from A345552 at term 9 because 288 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 5^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3.

Likely finite.

LINKS

Sean A. Irvine, Table of n, a(n) for n = 1..93

EXAMPLE

232 is a term because 232 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 = 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3.

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, 10):

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])