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A341240 a(n) = 4*a(n-1) - 2*a(n-2) + a(n-3) - 4*a(n-4) + 2*a(n-5) for n >= 7, where a(1) = 1, a(2) = 2, a(3) = 4, a(4) = 12, a(5) = 38, a(6) = 127.

%I A341240 #24 Jun 14 2022 16:04:46
%S A341240 1,2,4,12,38,127,432,1472,5023,17148,58544,199879,682428,2329952,
%T A341240 7954951,27159900,92729696,316598983,1080936540,3690548192,
%U A341240 12600319687,43020182364,146880090080,501479995591,1712159802204,5845679217632,19958397266119,68142230629212
%N A341240 a(n) = 4*a(n-1) - 2*a(n-2) + a(n-3) - 4*a(n-4) + 2*a(n-5) for n >= 7, where a(1) = 1, a(2) = 2, a(3) = 4, a(4) = 12, a(5) = 38, a(6) = 127.
%H A341240 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (4,-2,1,-4,2).
%F A341240 Let f(n) = floor(r*floor(s*n)) = A184922(n), where r = 1 + sqrt(2) and s = sqrt(2). Let a(1) = 1. Then a(n) = f(a(n-1)) for n >= 2.
%F A341240 G.f.: x*(1 - 2*x - 2*x^2 - x^3 + x^5)/(1 - 4*x + 2*x^2 - x^3 + 4*x^4 - 2*x^5). - _Stefano Spezia_, Feb 11 2021
%t A341240 z = 50; r = 1 + Sqrt[2]; s = Sqrt[2]; f[x_] := Floor[r*Floor[s*x]];
%t A341240 Table[f[n], {n, 1, z}] (* A341239 *)
%t A341240 a[1] = 1; a[n_] := f[a[n - 1]];
%t A341240 Table[a[n], {n, 1, z}] (* A341240 *)
%t A341240 LinearRecurrence[{4,-2,1,-4,2},{1,2,4,12,38,127},30] (* _Harvey P. Dale_, Jun 14 2022 *)
%Y A341240 Cf. A184922, A339828, A341239.
%K A341240 nonn,easy
%O A341240 1,2
%A A341240 _Clark Kimberling_, Feb 07 2021