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A341250 a(n) = 5*a(n-1) - 4*a(n-3) for n >= 4, where a(1) = 1, a(2) = 3, a(3) = 13.

%I A341250 #10 Feb 15 2021 18:52:08
%S A341250 1,3,13,61,293,1413,6821,32933,159013,767781,3707173,17899813,
%T A341250 86427941,417311013,2014955813,9729067301,46976092453,226820639013,
%U A341250 1095186925861,5288030259493,25532868741413,123283596003621,595265858980133,2874197819935013
%N A341250 a(n) = 5*a(n-1) - 4*a(n-3) for n >= 4, where a(1) = 1, a(2) = 3, a(3) = 13.
%H A341250 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (5,0,-4).
%F A341250 Let f(n) = floor(r*floor(s*n)) = A341249(n), where r = 2 + sqrt(2) and s = sqrt(2).  Let a(1) = 1. Then a(n) = f(a(n-1)) for n >= 2.
%F A341250 a(n) = (A218989(n-2) + 1)/2. - _Hugo Pfoertner_, Feb 13 2021
%F A341250 G.f.: x*(-2*x^2 - 2*x + 1)/(4*x^3 - 5*x + 1). - _Chai Wah Wu_, Feb 15 2021
%t A341250 z = 40; r = 2 + Sqrt[2]; s = Sqrt[2]; f[x_] := Floor[r*Floor[s*x]];
%t A341250 Table[f[n], {n, 1, z}] (* A341249 *)
%t A341250 a[1] = 1; a[n_] := f[a[n - 1]];
%t A341250 Table[a[n], {n, 1, z}] (* A341250 *)
%Y A341250 Cf. A218989, A339828, A341240, A341249.
%K A341250 nonn,easy
%O A341250 1,2
%A A341250 _Clark Kimberling_, Feb 13 2021