%I #32 Feb 16 2025 08:32:27
%S 8,23,38,53,68,83,88,98,103,113,118,128,133,148,163,168,178,183,193,
%T 198,213,228,243,248,258,263,278,293,308,323,328,338,343,353,358,368,
%U 373,388,403,408,418,423,433,438,453,468,483,488,498,503,518,533,548,563,568
%N Numbers that are the sum of 8 positive 4th powers.
%H David A. Corneth, <a href="/A003342/b003342.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 5396 is in the sequence as 5396 = 1^4 + 1^4 + 4^4 + 5^4 + 5^4 + 6^4 + 6^4 + 6^4.
%e 8789 is in the sequence as 8789 = 5^4 + 5^4 + 5^4 + 5^4 + 6^4 + 6^4 + 6^4 + 7^4.
%e 12469 is in the sequence as 12469 = 1^4 + 3^4 + 4^4 + 4^4 + 5^4 + 5^4 + 5^4 + 10^4. (End)
%t Select[Range[500], AnyTrue[PowersRepresentations[#, 8, 4], First[#]>0&]&] (* _Jean-François Alcover_, Jul 18 2017 *)
%o (Python)
%o from itertools import combinations_with_replacement as mc
%o from sympy import integer_nthroot
%o def iroot4(n): return integer_nthroot(n, 4)[0]
%o def aupto(lim):
%o pows4 = set(i**4 for i in range(1, iroot4(lim)+1) if i**4 <= lim)
%o return sorted(t for t in set(sum(c) for c in mc(pows4, 8)) if t <= lim)
%o print(aupto(568)) # _Michael S. Branicky_, Aug 23 2021
%Y Cf. A000583, A003331, A003341, A003343, A003353, A345577, A345833.
%K nonn,easy,changed
%O 1,1
%A _N. J. A. Sloane_