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Numbers that are the sum of nine cubes in exactly ten ways.
6

%I #6 Jul 31 2021 22:33:06

%S 966,971,978,1004,1018,1022,1055,1056,1062,1063,1074,1076,1078,1085,

%T 1088,1092,1093,1095,1098,1100,1104,1111,1112,1114,1117,1119,1124,

%U 1130,1134,1135,1139,1140,1142,1147,1149,1153,1160,1167,1168,1170,1180,1181,1182,1183

%N Numbers that are the sum of nine cubes in exactly ten ways.

%C Differs from A345549 at term 4 because 985 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 9^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 6^3 + 9^3 = 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 5^3 + 6^3 + 8^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 4^3 + 6^3 + 7^3 + 7^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 9^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 6^3 + 6^3 + 8^3 = 1^3 + 2^3 + 2^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 + 7^3 = 1^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 6^3 + 7^3 = 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 + 8^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 5^3 + 9^3 = 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 6^3 + 7^3 = 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3 + 7^3.

%C Likely finite.

%H Sean A. Irvine, <a href="/A345802/b345802.txt">Table of n, a(n) for n = 1..111</a>

%e 971 is a term because 971 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 5^3 + 6^3 + 6^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 7^3 = 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 6^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 6^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 6^3 = 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 5^3 + 6^3 = 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 7^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 6^3 + 6^3 = 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 7^3.

%o (Python)

%o from itertools import combinations_with_replacement as cwr

%o from collections import defaultdict

%o keep = defaultdict(lambda: 0)

%o power_terms = [x**3 for x in range(1, 1000)]

%o for pos in cwr(power_terms, 9):

%o tot = sum(pos)

%o keep[tot] += 1

%o rets = sorted([k for k, v in keep.items() if v == 10])

%o for x in range(len(rets)):

%o print(rets[x])

%Y Cf. A345549, A345792, A345801, A345812, A345852.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, Jun 26 2021