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%I #24 Dec 29 2019 06:22:32
%S 0,1,2,11,134,3897,272924,46128053,18820518548,18538256897833,
%T 44083993723565422,253086226505245985535,3507775143446703083080522,
%U 117373664327956358368203332177,9481679355248745685146904663002936,1849164516374760291573731440270350925577
%N a(n) = P(n)*a(n-1) + a(n-2), with a(0) = 0, a(1) = 1 where P(n) is the n-th Pell number.
%C The ratio of the ratios of the consecutive numbers of this sequence tends to the silver ratio (A014176 = 1 + sqrt(2)). This property can be proven.
%F a(n) = A000129(n)*a(n-1) + a(n-2) for n > 1.
%F a(n)*a(n-2)/a(n-1)^2 ~ 1 + sqrt(2).
%F a(n) ~ c * (1 + sqrt(2))^(n*(n+1)/2) / 2^(3*n/2), where c = 1.2795822677496757181586660872019163393334688780614201258263927413618707127... - _Vaclav Kotesovec_, Dec 29 2019
%t Nest[Append[#1, Fibonacci[#2, 2] #1[[-1]] + #1[[-2]] ] & @@ {#, Length@ #} &, {0, 1}, 14] (* _Michael De Vlieger_, Dec 22 2019 *)
%o (PARI) seq(n)={my(u=Vec(1/(1 - 2*x - x^2) + O(x^n)), v=vector(#u+1)); v[2]=1; for(n=2, #u, v[n+1] = u[n]*v[n] + v[n-1]); v} \\ _Andrew Howroyd_, Dec 22 2019
%Y Cf. A000129 (Pell numbers), A014176 (silver ratio), A135829.
%K nonn
%O 0,3
%A _Archit Handa_, Dec 22 2019
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