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%I #18 Sep 08 2022 08:45:42
%S 2,3,5,8,13,30,39,48,57,66,75,84,93,102,111,120,129,138,147,156,165,
%T 174,183,192,201,210,219,228,237,246,255,264,273,282,291,300,309,318,
%U 327,336,345,354,363,372,381,390,399,408,417,426,435
%N a(n) = Fibonacci(n+3) for n < 5 and 9*n - 15 otherwise.
%C This sequence is a possible answer to Marus du Satoy's puzzle sequence, see reference.
%D Marcus Du Sautoy, Symmetry: A Journey into the Patterns of Nature, Harper (March 11, 2008), page 96.
%H G. C. Greubel, <a href="/A158083/b158083.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1).
%F a(n) = a(n-1) + 9 for n > 5, a(n-1) + a(n-2) + 9 for n = 5, and Fibonacci(n+3) for n < 5.
%F From _R. J. Mathar_, Apr 10 2009: (Start)
%F G.f.: (2 -x +x^2 +x^3 +2*x^4 +12*x^5 -8*x^6)/(1-x)^2.
%F a(n) = 9*n - 15, n>4. (End)
%t a[n_]:= a[n]= If[n<5, Fibonacci[n+3], If[n==5, a[n-1] +a[n-2] +9, a[n-1] +9]];
%t Table[a[n], {n, 0, 50}] (* modified by _G. C. Greubel_, May 14 2021 *)
%t Join[{2,3,5,8,13},NestList[#+9&,30,50]] (* _Harvey P. Dale_, Nov 18 2012 *)
%t Table[If[n<5, Fibonacci[n+3], 9*n-15], {n,0,50}] (* _G. C. Greubel_, May 14 2021 *)
%o (Magma)
%o A158083:= func< n | n lt 5 select Fibonacci(n+3) else 3*(3*n-5) >;
%o [A158083(n): n in [0..50]]; // _G. C. Greubel_, May 14 2021
%o (Sage)
%o def A158083(n): return fibonacci(n+3) if (n<5) else 3*(3*n-5)
%o [A158083(n) for n in (0..50)] # _G. C. Greubel_, May 14 2021
%K nonn,easy
%O 0,1
%A _Roger L. Bagula_, Mar 12 2009
%E Edited by _G. C. Greubel_, May 14 2021
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