A345805 - OEIS

A345805

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

197, 239, 246, 253, 260, 267, 277, 279, 281, 293, 295, 298, 300, 302, 303, 305, 309, 312, 316, 317, 319, 324, 326, 329, 330, 335, 336, 338, 340, 343, 344, 345, 351, 352, 354, 358, 361, 362, 364, 365, 368, 370, 379, 386, 387, 388, 392, 394, 395, 396, 402, 406

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

Differs from A345551 at term 2 because 225 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3 = 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3.

Likely finite.

LINKS

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

EXAMPLE

225 is a term because 225 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 = 1^3 + 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 == 3])

for x in range(len(rets)):

print(rets[x])