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%I #14 Oct 06 2021 03:51:36
%S 8,128,648,2048,4913,5000,10368,19208,32768,52488,78608,80000,117128,
%T 165888,228488,307328,397953,405000,524288,551368,668168,839808,
%U 912673,1042568,1257728,1280000,1555848,1874048,2238728,2654208,3070625,3125000,3655808,4251528
%N Numbers k such that k^3 is the sum of two nonzero 4th powers.
%C If a^4 + b^4 = m, then (m^2 * a)^4 + (m^2 * b)^4 = m^9 = (m^3)^3 is a cube. Therefore A003336(n)^3 are terms of this sequence.
%C When k is in this sequence, k(n^4), for n > 1, is also in this sequence.
%e 8^3 = 2^9 = 4^4 + 4^4, so 8 is in the sequence.
%e 4913^3 = 17^9 = 17^8 * (1 + 2^4) = 289^4 + 578^4, so 4913 is in the sequence.
%t fourthPowerFlags = Union@Flatten@Table[a^4 + b^4 && GCD[a, b] == 1, {a, 4}, {b, a, 4}]; Take[Union@Flatten@Table[k^4 * fourthPowerFlags[[j]]^3, {k, 27}, {j, 6}], 34]
%Y Cf. A000404, A004431, A003325, A050801, A003336, A003347.
%K nonn
%O 1,1
%A _XU Pingya_, Sep 04 2017