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A024551 a(n) = floor(a(n-1)/(sqrt(5) - 2)) for n > 0 and a(0) = 1.

%I A024551 #23 Apr 15 2019 12:38:16
%S A024551 1,4,16,67,283,1198,5074,21493,91045,385672,1633732,6920599,29316127,
%T A024551 124185106,526056550,2228411305,9439701769,39987218380,169388575288,
%U A024551 717541519531,3039554653411,12875760133174,54542595186106,231046140877597
%N A024551 a(n) = floor(a(n-1)/(sqrt(5) - 2)) for n > 0 and a(0) = 1.
%H A024551 Clark Kimberling, <a href="/A024551/b024551.txt">Table of n, a(n) for n = 0..250</a>
%H A024551 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (5,-3,-1).
%F A024551 a(n) = 5*a(n-1) - 3*a(n-2) - a(n-3). - _Clark Kimberling_, Aug 16 2012
%F A024551 G.f.: (-x^2-x+1)/[(1-x)(1-4x-x^2)].
%F A024551 a(n) = (3*Fibonacci(3*n+2) + 1)/4 = 1 + 3*Sum_{k=0..n} A001076(k). - _Ehren Metcalfe_, Apr 15 2019
%t A024551 a[0] = 1;
%t A024551 a[n_] := Floor[a[n - 1]/FractionalPart[Sqrt[5]]]
%t A024551 Table[a[n], {n, 0, 60}]
%t A024551 (* _Clark Kimberling_, Aug 16 2012 *)
%t A024551 a[0]=1;
%t A024551 a[1]=4;
%t A024551 a[2]=16;
%t A024551 a[n_]:=Floor[a[n-1]^2/a[n-2]]+3
%t A024551 Table[a[n],{n,0,60}]
%t A024551 With[{c=Sqrt[5]-2},NestList[Floor[#/c]&,1,30]] (* _Harvey P. Dale_, Jul 18 2018 *)
%o A024551 (PARI) a(n)=([0,1,0; 0,0,1; -1,-3,5]^n*[1;4;16])[1,1] \\ _Charles R Greathouse IV_, Jan 20 2017
%o A024551 (PARI) step(n)=2*n + sqrtint(5*n^2)
%o A024551 a(n)=if(n, step(a(n-1)), 1) \\ _Charles R Greathouse IV_, Jan 20 2017
%K A024551 nonn,easy
%O A024551 0,2
%A A024551 _Clark Kimberling_