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A345833 Numbers that are the sum of eight fourth powers in exactly one ways.

%I A345833 #6 Jul 31 2021 21:33:24
%S A345833 8,23,38,53,68,83,88,98,103,113,118,128,133,148,163,168,178,183,193,
%T A345833 198,213,228,243,248,258,328,338,353,368,403,408,418,433,468,483,488,
%U A345833 498,568,578,593,608,632,643,647,648,658,662,663,673,677,692,707,708,712
%N A345833 Numbers that are the sum of eight fourth powers in exactly one ways.
%C A345833 Differs from A003342 at term 26 because 263 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4.
%H A345833 Sean A. Irvine, <a href="/A345833/b345833.txt">Table of n, a(n) for n = 1..10000</a>
%e A345833 23 is a term because 23 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4.
%o A345833 (Python)
%o A345833 from itertools import combinations_with_replacement as cwr
%o A345833 from collections import defaultdict
%o A345833 keep = defaultdict(lambda: 0)
%o A345833 power_terms = [x**4 for x in range(1, 1000)]
%o A345833 for pos in cwr(power_terms, 8):
%o A345833     tot = sum(pos)
%o A345833     keep[tot] += 1
%o A345833     rets = sorted([k for k, v in keep.items() if v == 1])
%o A345833     for x in range(len(rets)):
%o A345833         print(rets[x])
%Y A345833 Cf. A003342, A345783, A345823, A345834, A345843, A346326.
%K A345833 nonn
%O A345833 1,1
%A A345833 _David Consiglio, Jr._, Jun 26 2021