A345538 Numbers that are the sum of eight cubes in eight or more ways.
970, 977, 984, 1054, 1073, 1075, 1080, 1090, 1099, 1106, 1110, 1125, 1129, 1136, 1148, 1160, 1166, 1171, 1178, 1181, 1185, 1186, 1188, 1192, 1197, 1204, 1206, 1211, 1217, 1218, 1223, 1225, 1230, 1232, 1234, 1236, 1237, 1242, 1243, 1249, 1262, 1263, 1269, 1273
Offset: 1
Keywords
Examples
977 is a term because 977 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 5^3 + 8^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 5^3 + 6^3 + 6^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 8^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 7^3 = 1^3 + 2^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 8^3 = 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 6^3.
Links
- Sean A. Irvine, Table of n, a(n) for n = 1..10000
Programs
-
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 >= 8]) for x in range(len(rets)): print(rets[x])
Comments