A345781 Numbers that are the sum of seven cubes in exactly nine ways.
1496, 1648, 1720, 1737, 1772, 1781, 1802, 1835, 1844, 1882, 1891, 1898, 1900, 1907, 1912, 1919, 1945, 1952, 1954, 1961, 1996, 2000, 2012, 2026, 2071, 2080, 2098, 2107, 2110, 2115, 2116, 2132, 2134, 2136, 2139, 2150, 2152, 2168, 2178, 2185, 2187, 2195, 2205
Offset: 1
Keywords
Examples
1648 is a term because 1648 = 1^3 + 1^3 + 1^3 + 2^3 + 4^3 + 7^3 + 9^3 = 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 10^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 5^3 + 10^3 = 1^3 + 1^3 + 3^3 + 4^3 + 5^3 + 7^3 + 8^3 = 1^3 + 2^3 + 2^3 + 5^3 + 6^3 + 6^3 + 8^3 = 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 9^3 = 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 6^3 + 9^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 8^3 + 8^3 = 3^3 + 3^3 + 3^3 + 5^3 + 5^3 + 7^3 + 7^3.
Links
- Sean A. Irvine, Table of n, a(n) for n = 1..338
Programs
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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 == 9]) for x in range(len(rets)): print(rets[x])
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