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A345797 Numbers that are the sum of nine cubes in exactly five ways.

%I A345797 #6 Jul 31 2021 22:32:48
%S A345797 409,413,428,435,439,446,465,479,491,502,512,517,526,531,533,535,538,
%T A345797 540,559,561,563,566,568,570,576,579,580,587,594,600,601,603,613,615,
%U A345797 617,620,622,627,632,633,635,638,646,651,653,664,665,668,670,675,680,683
%N A345797 Numbers that are the sum of nine cubes in exactly five ways.
%C A345797 Differs from A345544 at term 8 because 472 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 6^3  = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 6^3 + 6^3  = 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3  = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3  = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3 + 5^3 + 5^3  = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3.
%C A345797 Likely finite.
%H A345797 Sean A. Irvine, <a href="/A345797/b345797.txt">Table of n, a(n) for n = 1..108</a>
%e A345797 413 is a term because 413 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 5^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3.
%o A345797 (Python)
%o A345797 from itertools import combinations_with_replacement as cwr
%o A345797 from collections import defaultdict
%o A345797 keep = defaultdict(lambda: 0)
%o A345797 power_terms = [x**3 for x in range(1, 1000)]
%o A345797 for pos in cwr(power_terms, 9):
%o A345797     tot = sum(pos)
%o A345797     keep[tot] += 1
%o A345797     rets = sorted([k for k, v in keep.items() if v == 5])
%o A345797     for x in range(len(rets)):
%o A345797         print(rets[x])
%Y A345797 Cf. A345544, A345787, A345796, A345798, A345807, A345847.
%K A345797 nonn
%O A345797 1,1
%A A345797 _David Consiglio, Jr._, Jun 26 2021