A345548 Numbers that are the sum of nine cubes in nine or more ways.
859, 861, 896, 903, 922, 929, 935, 939, 959, 966, 971, 973, 978, 985, 992, 997, 999, 1004, 1009, 1011, 1016, 1018, 1020, 1022, 1023, 1027, 1029, 1030, 1034, 1035, 1036, 1037, 1041, 1046, 1048, 1055, 1056, 1059, 1060, 1062, 1063, 1064, 1065, 1066, 1067, 1071
Offset: 1
Keywords
Examples
861 is a term because 861 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 8^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 5^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 4^3 + 4^3 + 4^3 + 4^3 + 6^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 7^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 8^3 = 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 5^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3 + 5^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 7^3 = 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3.
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, 9): 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])