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A345779 Numbers that are the sum of seven cubes in exactly seven ways.

%I A345779 #6 Jul 31 2021 22:39:27
%S A345779 1072,1170,1235,1261,1268,1305,1392,1396,1411,1440,1441,1448,1450,
%T A345779 1459,1489,1502,1504,1513,1538,1540,1547,1559,1564,1565,1566,1567,
%U A345779 1576,1592,1593,1594,1600,1602,1606,1620,1621,1625,1626,1628,1629,1639,1658,1664,1667
%N A345779 Numbers that are the sum of seven cubes in exactly seven ways.
%C A345779 Differs from A345525 at term 7 because 1385 = 1^3 + 1^3 + 2^3 + 2^3 + 7^3 + 8^3 + 8^3  = 1^3 + 1^3 + 2^3 + 5^3 + 5^3 + 5^3 + 10^3  = 1^3 + 2^3 + 2^3 + 3^3 + 5^3 + 6^3 + 10^3  = 1^3 + 2^3 + 6^3 + 6^3 + 6^3 + 6^3 + 8^3  = 1^3 + 4^3 + 5^3 + 5^3 + 5^3 + 6^3 + 9^3  = 2^3 + 3^3 + 3^3 + 5^3 + 7^3 + 7^3 + 8^3  = 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 6^3 + 9^3  = 3^3 + 3^3 + 3^3 + 4^3 + 6^3 + 8^3 + 8^3.
%C A345779 Likely finite.
%H A345779 Sean A. Irvine, <a href="/A345779/b345779.txt">Table of n, a(n) for n = 1..345</a>
%e A345779 1170 is a term because 1170 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 4^3 + 9^3 = 1^3 + 1^3 + 2^3 + 5^3 + 5^3 + 5^3 + 7^3 = 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 8^3 = 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 8^3 = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 7^3 + 7^3 = 3^3 + 3^3 + 4^3 + 5^3 + 5^3 + 5^3 + 6^3 = 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 + 7^3.
%o A345779 (Python)
%o A345779 from itertools import combinations_with_replacement as cwr
%o A345779 from collections import defaultdict
%o A345779 keep = defaultdict(lambda: 0)
%o A345779 power_terms = [x**3 for x in range(1, 1000)]
%o A345779 for pos in cwr(power_terms, 7):
%o A345779     tot = sum(pos)
%o A345779     keep[tot] += 1
%o A345779     rets = sorted([k for k, v in keep.items() if v == 7])
%o A345779     for x in range(len(rets)):
%o A345779         print(rets[x])
%Y A345779 Cf. A345525, A345769, A345778, A345780, A345789, A345829.
%K A345779 nonn
%O A345779 1,1
%A A345779 _David Consiglio, Jr._, Jun 26 2021