A345792 - OEIS

A345792

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

1185, 1243, 1288, 1295, 1299, 1386, 1397, 1400, 1412, 1423, 1448, 1449, 1451, 1458, 1460, 1464, 1467, 1475, 1477, 1501, 1503, 1505, 1512, 1513, 1516, 1539, 1540, 1541, 1553, 1558, 1559, 1568, 1577, 1578, 1586, 1588, 1591, 1592, 1594, 1595, 1596, 1600, 1608

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

OFFSET

1,1

COMMENTS

Differs from A345540 at term 3 because 1262 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 5^3 + 5^3 + 10^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 6^3 + 10^3 = 1^3 + 1^3 + 1^3 + 4^3 + 5^3 + 5^3 + 6^3 + 9^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 7^3 + 7^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 6^3 + 6^3 + 9^3 = 1^3 + 3^3 + 3^3 + 6^3 + 6^3 + 6^3 + 6^3 + 7^3 = 1^3 + 4^3 + 4^3 + 4^3 + 5^3 + 6^3 + 6^3 + 8^3 = 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 10^3 = 2^3 + 2^3 + 4^3 + 4^3 + 6^3 + 6^3 + 7^3 + 7^3 = 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 7^3 + 7^3 + 7^3 = 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 + 9^3.

Likely finite.

LINKS

Sean A. Irvine, Table of n, a(n) for n = 1..161

EXAMPLE

1243 is a term because 1243 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 5^3 + 9^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 7^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 6^3 + 6^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 5^3 + 5^3 + 8^3 = 1^3 + 1^3 + 4^3 + 4^3 + 5^3 + 5^3 + 5^3 + 6^3 = 1^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 5^3 + 6^3 = 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 9^3 = 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 6^3 + 6^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 8^3 = 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 6^3 + 7^3.

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, 8):

tot = sum(pos)

keep[tot] += 1

rets = sorted([k for k, v in keep.items() if v == 10])

for x in range(len(rets)):

print(rets[x])