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A345596 Numbers that are the sum of ten fourth powers in three or more ways.

%I A345596 #6 Jul 31 2021 17:25:42
%S A345596 520,535,550,600,615,680,775,790,855,1030,1144,1159,1224,1365,1380,
%T A345596 1399,1445,1540,1555,1605,1620,1635,1685,1700,1768,1795,1815,1830,
%U A345596 1860,1875,1895,1989,2070,2164,2229,2244,2439,2485,2580,2595,2645,2660,2675,2680,2695
%N A345596 Numbers that are the sum of ten fourth powers in three or more ways.
%H A345596 Sean A. Irvine, <a href="/A345596/b345596.txt">Table of n, a(n) for n = 1..10000</a>
%e A345596 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 A345596 (Python)
%o A345596 from itertools import combinations_with_replacement as cwr
%o A345596 from collections import defaultdict
%o A345596 keep = defaultdict(lambda: 0)
%o A345596 power_terms = [x**4 for x in range(1, 1000)]
%o A345596 for pos in cwr(power_terms, 10):
%o A345596     tot = sum(pos)
%o A345596     keep[tot] += 1
%o A345596     rets = sorted([k for k, v in keep.items() if v >= 3])
%o A345596     for x in range(len(rets)):
%o A345596         print(rets[x])
%Y A345596 Cf. A345551, A345587, A345595, A345597, A345635, A345855.
%K A345596 nonn
%O A345596 1,1
%A A345596 _David Consiglio, Jr._, Jun 20 2021