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a(n) = 5*a(n-1) - 4*a(n-3) for n >= 4, where a(1) = 1, a(2) = 4, a(3) = 18.
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%I #7 Feb 16 2021 01:07:26

%S 1,4,18,86,414,1998,9646,46574,224878,1085806,5242734,25314158,

%T 122227566,590166894,2849577838,13758978926,66434227054,320772823918,

%U 1548828203886,7478404111214,36108929260398,174349333486446,841833050987374,4064729537895278

%N a(n) = 5*a(n-1) - 4*a(n-3) for n >= 4, where a(1) = 1, a(2) = 4, a(3) = 18.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (5,0,-4).

%F Let f(n) = floor(r*floor(s*n)) = A022804(n), where r = sqrt(2) and s = r + 2. Let a(1) = 1. Then a(n) = f(a(n-1)) for n >= 2.

%F G.f.: x*(1 - x - 2*x^2)/(1 - 5*x + 4*x^3). - _Stefano Spezia_, Feb 13 2021

%t z = 40; r = Sqrt[2]; s = 2 + Sqrt[2]; f[x_] := Floor[r*Floor[s*x]];

%t Table[f[n], {n, 1, z}] (* A022804 *)

%t a[1] = 1; a[n_] := f[a[n - 1]];

%t t = Table[a[n], {n, 1, z}] (* A341248 *)

%Y Cf. A339828, A341240, A341250.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Feb 13 2021