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%I #6 Jul 31 2021 20:00:10
%S 520,535,550,600,615,680,775,790,855,1030,1144,1159,1224,1365,1380,
%T 1399,1445,1540,1555,1605,1635,1685,1700,1768,1795,1815,1830,1860,
%U 1875,1895,1989,2070,2164,2229,2244,2439,2485,2580,2595,2645,2675,2680,2695,2710,2755
%N Numbers that are the sum of ten fourth powers in exactly three ways.
%C Differs from A345596 at term 21 because 1620 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4.
%H Sean A. Irvine, <a href="/A345855/b345855.txt">Table of n, a(n) for n = 1..10000</a>
%e 535 is a term because 535 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4.
%o (Python)
%o from itertools import combinations_with_replacement as cwr
%o from collections import defaultdict
%o keep = defaultdict(lambda: 0)
%o power_terms = [x**4 for x in range(1, 1000)]
%o for pos in cwr(power_terms, 10):
%o tot = sum(pos)
%o keep[tot] += 1
%o rets = sorted([k for k, v in keep.items() if v == 3])
%o for x in range(len(rets)):
%o print(rets[x])
%Y Cf. A345596, A345805, A345845, A345854, A345856, A346348.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jun 26 2021
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