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a(n) = 4*a(n-1) - 2*a(n-2) + a(n-3) - 4*a(n-4) + a(n-5) for n >= 6, where a(1) = 1, a(2) = 3, a(3) = 5, a(4) = 16, a(5) = 53.
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%I #7 Feb 15 2021 16:28:08

%S 1,2,5,16,53,179,610,2081,7103,24250,82793,282671,965098,3295049,

%T 11249999,38409898,131139593,447738575,1528675114,5219223305,

%U 17819542991,60839725354,207719815433,709199811023,2421359613226,8267038830857,28225436096975,96367666726186

%N a(n) = 4*a(n-1) - 2*a(n-2) + a(n-3) - 4*a(n-4) + a(n-5) for n >= 6, where a(1) = 1, a(2) = 3, a(3) = 5, a(4) = 16, a(5) = 53.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (4,-2,1,-4,2).

%F Let f(n) = floor(r*floor(s*n)) = A184922(n), where r = sqrt(2) and s = r + 1. Let a(1) = 1. Then a(n) = f(a(n-1)) for n >= 2.

%F Also, a(n) = 4*a(n-1) - 2*a(n-2) + a(n-3) - 4*a(n-4) + a(n-5) for n >= 6, where a(1) = 1, a(2) = 3, a(3) = 5, a(4) = 16, a(5) = 53.

%F G.f.: x*(-x^4 + x^3 + x^2 + 2*x - 1)/((x - 1)*(x^2 + x + 1)*(2*x^2 - 4*x + 1)). - _Chai Wah Wu_, Feb 15 2021

%t z = 40; r = Sqrt[2]; s = 1 + Sqrt[2]; f[x_] := Floor[r*Floor[s*x]];

%t Table[f[n], {n, 1, z}] m (* A184922 *)

%t a[1] = 1; a[n_] := f[a[n - 1]]; Table[a[n], {n, 1, z}] (* A339828 *)

%Y Cf. A184922, A341239, A341240.

%K nonn

%O 1,2

%A _Clark Kimberling_, Feb 07 2021