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%I #6 Jul 31 2021 21:33:31
%S 518,2678,2693,2708,2738,2758,2773,2838,2853,2868,2883,2918,2998,3078,
%T 3108,3123,3253,3302,3317,3363,3382,3428,3477,3492,3542,3622,3732,
%U 3778,3797,3893,3926,3953,3973,3988,4018,4053,4101,4118,4133,4166,4193,4243,4258
%N Numbers that are the sum of eight fourth powers in exactly three ways.
%C Differs from A345578 at term 13 because 2933 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 6^4 + 6^4 = 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 7^4 = 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4.
%H Sean A. Irvine, <a href="/A345835/b345835.txt">Table of n, a(n) for n = 1..10000</a>
%e 2678 is a term because 2678 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 3^4 + 6^4 + 6^4 = 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4.
%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, 8):
%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. A345578, A345785, A345825, A345834, A345836, A345845, A346328.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jun 26 2021
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