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A345802 Numbers that are the sum of nine cubes in exactly ten ways.

%I A345802 #6 Jul 31 2021 22:33:06
%S A345802 966,971,978,1004,1018,1022,1055,1056,1062,1063,1074,1076,1078,1085,
%T A345802 1088,1092,1093,1095,1098,1100,1104,1111,1112,1114,1117,1119,1124,
%U A345802 1130,1134,1135,1139,1140,1142,1147,1149,1153,1160,1167,1168,1170,1180,1181,1182,1183
%N A345802 Numbers that are the sum of nine cubes in exactly ten ways.
%C A345802 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 A345802 Likely finite.
%H A345802 Sean A. Irvine, <a href="/A345802/b345802.txt">Table of n, a(n) for n = 1..111</a>
%e A345802 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 A345802 (Python)
%o A345802 from itertools import combinations_with_replacement as cwr
%o A345802 from collections import defaultdict
%o A345802 keep = defaultdict(lambda: 0)
%o A345802 power_terms = [x**3 for x in range(1, 1000)]
%o A345802 for pos in cwr(power_terms, 9):
%o A345802     tot = sum(pos)
%o A345802     keep[tot] += 1
%o A345802     rets = sorted([k for k, v in keep.items() if v == 10])
%o A345802     for x in range(len(rets)):
%o A345802         print(rets[x])
%Y A345802 Cf. A345549, A345792, A345801, A345812, A345852.
%K A345802 nonn
%O A345802 1,1
%A A345802 _David Consiglio, Jr._, Jun 26 2021