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A345786 Numbers that are the sum of eight cubes in exactly four ways.

%I A345786 #6 Jul 31 2021 22:37:14
%S A345786 256,347,382,401,408,427,434,438,445,464,478,480,490,499,502,506,511,
%T A345786 516,523,530,532,534,537,560,565,567,569,571,578,586,593,595,600,602,
%U A345786 604,605,611,612,616,619,621,624,626,643,645,656,660,663,664,668,675,679
%N A345786 Numbers that are the sum of eight cubes in exactly four ways.
%C A345786 Differs from A345534 at term 11 because 471 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 6^3  = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 6^3 + 6^3  = 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3  = 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3  = 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3 + 5^3 + 5^3.
%C A345786 Likely finite.
%H A345786 Sean A. Irvine, <a href="/A345786/b345786.txt">Table of n, a(n) for n = 1..207</a>
%e A345786 347 is a term because 347 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3.
%o A345786 (Python)
%o A345786 from itertools import combinations_with_replacement as cwr
%o A345786 from collections import defaultdict
%o A345786 keep = defaultdict(lambda: 0)
%o A345786 power_terms = [x**3 for x in range(1, 1000)]
%o A345786 for pos in cwr(power_terms, 8):
%o A345786     tot = sum(pos)
%o A345786     keep[tot] += 1
%o A345786     rets = sorted([k for k, v in keep.items() if v == 4])
%o A345786     for x in range(len(rets)):
%o A345786         print(rets[x])
%Y A345786 Cf. A345534, A345776, A345785, A345787, A345796, A345836.
%K A345786 nonn
%O A345786 1,1
%A A345786 _David Consiglio, Jr._, Jun 26 2021