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%I #29 Oct 25 2021 08:06:57
%S 10,25,40,55,70,85,90,100,105,115,120,130,135,145,150,160,165,170,180,
%T 185,195,200,210,215,225,230,245,250,260,265,275,280,290,295,310,325,
%U 330,340,345,355,360,370,375,385,390,400,405,410,420,425,435,440,450,455,465
%N Numbers that are the sum of 10 positive 4th powers.
%H David A. Corneth, <a href="/A003344/b003344.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="http://mathworld.wolfram.com/BiquadraticNumber.html">Biquadratic Number.</a>
%e From _David A. Corneth_, Aug 03 2020: (Start)
%e 5176 is in the sequence as 5176 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 5^4 + 5^4 + 5^4 + 5^4 + 7^4.
%e 6901 is in the sequence as 6901 = 1^4 + 4^4 + 4^4 + 5^4 + 5^4 + 5^4 + 5^4 + 6^4 + 6^4 + 6^4.
%e 8502 is in the sequence as 8502 = 1^4 + 3^4 + 4^4 + 5^4 + 5^4 + 5^4 + 6^4 + 6^4 + 6^4 + 7^4. (End)
%o (Python)
%o from itertools import count, takewhile, combinations_with_replacement as mc
%o def aupto(limit):
%o pows4 = list(takewhile(lambda x: x <= limit, (i**4 for i in count(1))))
%o sum10 = set(sum(c) for c in mc(pows4, 10) if sum(c) <= limit)
%o return sorted(sum10)
%o print(aupto(465)) # _Michael S. Branicky_, Oct 25 2021
%Y Cf. A000583, A003333, A003343, A003355, A047721, A345595, A345853.
%K nonn,easy
%O 1,1
%A _N. J. A. Sloane_
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