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A345784 Numbers that are the sum of eight cubes in exactly two ways.

%I A345784 #6 Jul 31 2021 22:37:08
%S A345784 132,139,158,160,167,174,181,186,193,195,197,200,212,216,219,238,244,
%T A345784 251,258,265,272,277,288,296,298,300,301,303,307,314,315,317,321,322,
%U A345784 327,328,329,333,334,336,338,340,341,348,350,352,356,359,360,361,363,366
%N A345784 Numbers that are the sum of eight cubes in exactly two ways.
%C A345784 Differs from A345532 at term 16 because 223 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 6^3  = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3  = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3.
%C A345784 Likely finite.
%H A345784 Sean A. Irvine, <a href="/A345784/b345784.txt">Table of n, a(n) for n = 1..173</a>
%e A345784 139 is a term because 139 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 4^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3.
%o A345784 (Python)
%o A345784 from itertools import combinations_with_replacement as cwr
%o A345784 from collections import defaultdict
%o A345784 keep = defaultdict(lambda: 0)
%o A345784 power_terms = [x**3 for x in range(1, 1000)]
%o A345784 for pos in cwr(power_terms, 8):
%o A345784     tot = sum(pos)
%o A345784     keep[tot] += 1
%o A345784     rets = sorted([k for k, v in keep.items() if v == 2])
%o A345784     for x in range(len(rets)):
%o A345784         print(rets[x])
%Y A345784 Cf. A345532, A345774, A345783, A345785, A345794, A345834.
%K A345784 nonn
%O A345784 1,1
%A A345784 _David Consiglio, Jr._, Jun 26 2021