A345780 - OEIS

A345780

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

1385, 1515, 1552, 1557, 1585, 1587, 1603, 1613, 1622, 1655, 1665, 1674, 1681, 1718, 1719, 1739, 1741, 1746, 1753, 1755, 1765, 1767, 1782, 1793, 1805, 1809, 1811, 1818, 1819, 1826, 1828, 1830, 1833, 1838, 1856, 1870, 1873, 1881, 1901, 1905, 1931, 1935, 1937

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

OFFSET

1,1

COMMENTS

Differs from A345526 at term 2 because 1496 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 9^3 + 9^3 = 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 4^3 + 11^3 = 1^3 + 1^3 + 4^3 + 4^3 + 5^3 + 8^3 + 9^3 = 1^3 + 2^3 + 2^3 + 4^3 + 7^3 + 7^3 + 9^3 = 1^3 + 5^3 + 5^3 + 6^3 + 7^3 + 7^3 + 7^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 11^3 = 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 7^3 + 10^3 = 2^3 + 3^3 + 6^3 + 6^3 + 7^3 + 7^3 + 7^3 = 4^3 + 4^3 + 4^3 + 4^3 + 6^3 + 8^3 + 8^3.

Likely finite.

LINKS

Sean A. Irvine, Table of n, a(n) for n = 1..343

EXAMPLE

1496 is a term because 1496 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 8^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 10^3 = 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 7^3 + 8^3 = 1^3 + 2^3 + 2^3 + 3^3 + 6^3 + 6^3 + 8^3 = 1^3 + 4^3 + 4^3 + 5^3 + 6^3 + 6^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 10^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 9^3 = 2^3 + 3^3 + 5^3 + 5^3 + 6^3 + 6^3 + 6^3 = 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 7^3 + 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, 7):

tot = sum(pos)

keep[tot] += 1

rets = sorted([k for k, v in keep.items() if v == 8])

for x in range(len(rets)):

print(rets[x])