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A320534 a(n) = ((1 + sqrt(4*n^2 + 1))^n + (1 - sqrt(4*n^2 + 1))^n)/2^n.

%I A320534 #16 Feb 16 2025 08:33:56
%S A320534 2,1,9,28,577,3251,105193,857501,37831169,403541596,22550351001,
%T A320534 297238464799,20106709638337,315569447182601,25059144736026633,
%U A320534 456277507970965876,41600491470425952257,862007599260004863571,88733427132980061934777,2061632980592377284802309
%N A320534 a(n) = ((1 + sqrt(4*n^2 + 1))^n + (1 - sqrt(4*n^2 + 1))^n)/2^n.
%C A320534 a(0) = 2 assuming 0^0 = 1, or using the limit for n -> 0 (assuming n is a real variable); the same value for a(0) arises from other formulae for this sequence.
%H A320534 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/LucasPolynomial.html">Lucas Polynomial</a>
%H A320534 Wikipedia, <a href="https://en.wikipedia.org/wiki/Fibonacci_polynomials">Fibonacci polynomials</a>
%F A320534 a(n) = 2^(1 - n) * Sum_{k=0..floor(n/2)} binomial(n, 2*k)*(4*n^2 + 1)^k.
%F A320534 a(n) = 2^(1 - n) * hypergeom([(1 - n)/2, -n/2], [1/2], 4*n^2 + 1).
%F A320534 For n > 0, a(n) = n^n * L_n(1/n), where L_n(x) is the Lucas polynomial.
%F A320534 For n > 0, a(n) = 2*(-i*n)^n*cos(n*arcsin(sqrt(4*n^2+1)/(2*n))). - _Peter Luschny_, Oct 14 2018
%t A320534 Table[2^(1 - n) Hypergeometric2F1[(1 - n)/2, -n/2, 1/2, 4 n^2 + 1], {n, 0, 19}]
%t A320534 (* or *)
%t A320534 a[0] = Limit[n^n LucasL[n, 1/n], n -> 0]; (* a[0] = 2 *)
%t A320534 a[n_] := a[n] = n^n LucasL[n, 1/n];
%t A320534 Table[a[n], {n, 0, 19}]
%Y A320534 Cf. A084844, A320519.
%K A320534 nonn
%O A320534 0,1
%A A320534 _Vladimir Reshetnikov_, Oct 14 2018