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a(n) = A113655(Fibonacci(n+1)).
1

%I #14 Dec 09 2022 07:15:02

%S 3,3,2,1,5,8,15,19,36,57,89,142,233,377,612,985,1599,2586,4181,6763,

%T 10946,17711,28659,46366,75027,121395,196418,317809,514229,832040,

%U 1346271,2178307,3524580,5702889,9227465,14930350,24157817,39088169

%N a(n) = A113655(Fibonacci(n+1)).

%H G. C. Greubel, <a href="/A102905/b102905.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,0,0,0,0,0,1,-1,-1).

%F a(n) = f(Fibonacci(n+1)), where f(n) = n-2 if (n mod 3) = 0, f(n) = n+2 if (n mod 3) = 1, otherwise f(n) = n.

%F a(n) = A113655(Fibonacci(n+1)).

%F G.f.: (3-4*x^2-4*x^3+2*x^4+2*x^5+2*x^6-4*x^7-x^8+2*x^9) / ((1-x)*(1+x)*(1+x^2)*(1-x-x^2)*(1+x^4)). - _Colin Barker_, Dec 11 2012

%F a(n) = (1 + 3*(-1)^n)/4 + Fibonacci(n+1) + (3/2)*(-1)^floor(n/2) * (n mod 2) + A014017(n) + A014017(n-1) - A014017(n-2). - _G. C. Greubel_, Dec 09 2022

%t f[n_]:= If[Mod[n,3]==0, n-2, If[Mod[n,3]==1, n+2, n]]; (* f=A113655 *)

%t Table[f[Fibonacci[n+1]], {n,0,50}]

%o (Magma)

%o A113655:= func< n | 6*Floor((n+2)/3) -(n+2) >;

%o A102905:= func< n | A113655(Fibonacci(n+1)) >;

%o [A102905(n): n in [0..50]]; // _G. C. Greubel_, Dec 09 2022

%o (SageMath)

%o def A113655(n): return 6*((n+2)//3) -(n+2)

%o def A102905(n): return A113655(fibonacci(n+1))

%o [A102905(n) for n in range(51)] # _G. C. Greubel_, Dec 09 2022

%Y Cf. A000045, A014017, A113655.

%K nonn

%O 0,1

%A _Roger L. Bagula_, Mar 16 2005

%E Edited by _G. C. Greubel_, Dec 09 2022