cp's OEIS Frontend

A345537 Numbers that are the sum of eight cubes in seven or more ways.

%I A345537 #6 Aug 05 2021 15:20:59
%S A345537 902,908,921,938,958,963,970,977,982,984,991,996,1003,1008,1010,1017,
%T A345537 1019,1028,1029,1033,1047,1054,1055,1058,1061,1062,1070,1073,1075,
%U A345537 1080,1087,1090,1091,1094,1096,1097,1099,1104,1106,1108,1110,1111,1113,1115,1116
%N A345537 Numbers that are the sum of eight cubes in seven or more ways.
%H A345537 Sean A. Irvine, <a href="/A345537/b345537.txt">Table of n, a(n) for n = 1..10000</a>
%e A345537 908 is a term because 908 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 6^3 + 7^3 = 1^3 + 1^3 + 2^3 + 4^3 + 4^3 + 5^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 5^3 = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 7^3 = 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 6^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 7^3 = 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3.
%o A345537 (Python)
%o A345537 from itertools import combinations_with_replacement as cwr
%o A345537 from collections import defaultdict
%o A345537 keep = defaultdict(lambda: 0)
%o A345537 power_terms = [x**3 for x in range(1, 1000)]
%o A345537 for pos in cwr(power_terms, 8):
%o A345537     tot = sum(pos)
%o A345537     keep[tot] += 1
%o A345537     rets = sorted([k for k, v in keep.items() if v >= 7])
%o A345537     for x in range(len(rets)):
%o A345537         print(rets[x])
%Y A345537 Cf. A345494, A345525, A345536, A345538, A345546, A345582, A345789.
%K A345537 nonn
%O A345537 1,1
%A A345537 _David Consiglio, Jr._, Jun 20 2021