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Numbers that are the sum of 6 positive 4th powers.
39

%I #37 Feb 16 2025 08:32:27

%S 6,21,36,51,66,81,86,96,101,116,131,146,161,166,181,196,211,226,246,

%T 261,276,291,306,321,326,336,341,356,371,386,401,406,421,436,451,466,

%U 486,501,516,531,546,561,576,581,596,611,626,630,641,645,660,661,675,676,690

%N Numbers that are the sum of 6 positive 4th powers.

%H David A. Corneth, <a href="/A003340/b003340.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 13090 is in the sequence as 13090 = 4^4 + 4^4 + 5^4 + 6^4 + 8^4 + 9^4.

%e 17539 is in the sequence as 17539 = 2^4 + 3^4 + 4^4 + 5^4 + 9^4 + 10^4.

%e 23732 is in the sequence as 23732 = 3^4 + 5^4 + 5^4 + 7^4 + 10^4 + 10^4. (End)

%t Select[Range[1000], AnyTrue[PowersRepresentations[#, 6, 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 + 5 <= limit]

%o ss = set(sum(c) for c in combs_with_rep(qd, 6))

%o return sorted(s for s in ss if s <= limit)

%o print(aupto(700)) # _Michael S. Branicky_, Jun 21 2021

%Y Cf. A000583, A003329, A003339, A003341, A003351, A345559, A345813.

%K nonn,easy,changed

%O 1,1

%A _N. J. A. Sloane_