login
A345526
Numbers that are the sum of seven cubes in eight or more ways.
8
1385, 1496, 1515, 1552, 1557, 1585, 1587, 1603, 1613, 1622, 1648, 1655, 1665, 1674, 1681, 1704, 1711, 1718, 1719, 1720, 1737, 1739, 1741, 1746, 1753, 1755, 1765, 1767, 1772, 1774, 1781, 1782, 1793, 1800, 1802, 1805, 1809, 1811, 1818, 1819, 1826, 1828, 1830
OFFSET
1,1
LINKS
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])
KEYWORD
nonn
AUTHOR
STATUS
approved