%I #36 Feb 16 2025 08:32:27
%S 5,20,35,50,65,80,85,100,115,130,145,165,180,195,210,245,260,275,290,
%T 305,320,325,340,355,370,385,405,420,435,450,500,515,530,545,560,580,
%U 595,610,625,629,644,659,674,675,689,690,709,724,739,754,755,770,785,789,800
%N Numbers that are the sum of 5 positive 4th powers.
%H David A. Corneth, <a href="/A003339/b003339.txt">Table of n, a(n) for n = 1..10000</a> (first 1000 terms from T. D. Noe)
%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/BiquadraticNumber.html">Biquadratic Number.</a>
%e From _David A. Corneth_, Aug 04 2020: (Start)
%e 22418 is in the sequence as 22418 = 1^4 + 2^4 + 7^4 + 10^4 + 10^4.
%e 30004 is in the sequence as 30004 = 2^4 + 3^4 + 5^4 + 11^4 + 11^4.
%e 39028 is in the sequence as 39028 = 5^4 + 5^4 + 7^4 + 11^4 + 12^4. (End)
%t Select[Range[1000], AnyTrue[PowersRepresentations[#, 5, 4], First[#]>0&]&] (* _Jean-François Alcover_, Jul 18 2017 *)
%o (Python)
%o from itertools import combinations_with_replacement as combs_with_rep
%o def aupto(limit):
%o qd = [k**4 for k in range(1, int(limit**.25)+2) if k**4 + 4 <= limit]
%o ss = set(sum(c) for c in combs_with_rep(qd, 5))
%o return sorted(s for s in ss if s <= limit)
%o print(aupto(800)) # _Michael S. Branicky_, May 20 2021
%Y Supersequence of A047716.
%Y Cf. A000583, A003328, A003338, A003340, A003350, A344190, A344238.
%K nonn,easy,changed
%O 1,1
%A _N. J. A. Sloane_