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%I #8 Mar 26 2019 19:43:17
%S 244,532344,3368260700,22612933199856,152086272228543556,
%T 1022930598444844458600,6880230727163694887472044,
%U 46276430783447828140913432544,311255266556463822927417656862100,2093502876579815396773605056889844056,14080900036620070975438887843327087279356
%N Numbers that can be represented as sum of two fifth powers in two ways using Gaussian integers generated from Pell numbers.
%C Campbell and Zujev showed that if P(n) = A000129(n-1) are the Pell numbers, then a sequence of solutions to a^5 + b^5 = c^5 + d^5 can be generated using a = P(2n+3) + 1, b = P(2n+3) - 1, c = P(2n+3) + i*(P(2n+3) + P(2n+2)) and d = P(2n+3) - i*(P(2n+3) + P(2n+2)) (where i is the imaginary unit).
%H Geoffrey B. Campbell and Aleksander Zujev, <a href="http://arxiv.org/abs/1511.07424">Gaussian integer solutions for the fifth power taxicab number problem</a>, arXiv:1511.07424 [math.NT], 2015.
%e P(3) = 2 generates 244 = 3^5 + 1^5 = (2 + 3i)^5 + (2 - 3i)^5.
%t p[ n_] := With[ {s = Sqrt@2}, ((1 + s)^n - (1 - s)^n) / (2 s)] // Simplify; p0[n_] := p[n - 1]; Table[(p0[2n+3]-1)^5 + (p0[2n+3]+1)^5, {n, 0, 15}] (* after _Michael Somos_ at A000129 *)
%Y Cf. A000129.
%K nonn
%O 1,1
%A _Amiram Eldar_, Mar 18 2019
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