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A345776 Numbers that are the sum of seven cubes in exactly four ways.

%I A345776 #6 Jul 31 2021 22:39:17
%S A345776 470,496,503,603,634,653,659,685,690,692,711,712,747,751,754,761,766,
%T A345776 773,775,777,780,783,787,792,794,812,813,829,831,836,842,843,859,867,
%U A345776 871,875,883,885,890,892,899,901,904,906,907,911,913,918,919,927,930,936
%N A345776 Numbers that are the sum of seven cubes in exactly four ways.
%C A345776 Differs from A345522 at term 5 because 627 = 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 6^3 + 7^3  = 1^3 + 1^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3  = 1^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 6^3  = 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3  = 2^3 + 2^3 + 3^3 + 3^3 + 5^3 + 6^3 + 6^3.
%C A345776 Likely finite.
%H A345776 Sean A. Irvine, <a href="/A345776/b345776.txt">Table of n, a(n) for n = 1..360</a>
%e A345776 496 is a term because 496 = 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 6^3 = 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3.
%o A345776 (Python)
%o A345776 from itertools import combinations_with_replacement as cwr
%o A345776 from collections import defaultdict
%o A345776 keep = defaultdict(lambda: 0)
%o A345776 power_terms = [x**3 for x in range(1, 1000)]
%o A345776 for pos in cwr(power_terms, 7):
%o A345776     tot = sum(pos)
%o A345776     keep[tot] += 1
%o A345776     rets = sorted([k for k, v in keep.items() if v == 4])
%o A345776     for x in range(len(rets)):
%o A345776         print(rets[x])
%Y A345776 Cf. A345522, A345766, A345775, A345777, A345786, A345826.
%K A345776 nonn
%O A345776 1,1
%A A345776 _David Consiglio, Jr._, Jun 26 2021