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a(n) = Fibonacci(n+9) - Fibonacci(9).
5

%I #31 Aug 24 2024 16:25:14

%S 0,21,55,110,199,343,576,953,1563,2550,4147,6731,10912,17677,28623,

%T 46334,74991,121359,196384,317777,514195,832006,1346235,2178275,

%U 3524544,5702853,9227431,14930318,24157783,39088135,63245952,102334121

%N a(n) = Fibonacci(n+9) - Fibonacci(9).

%C The a(n+1) (terms doubled) are the Kn18 sums of the Fibonacci(n) triangle A104763. See A180662 for information about these knight and other chess sums.

%H Vincenzo Librandi, <a href="/A180674/b180674.txt">Table of n, a(n) for n = 0..275</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-1).

%F a(n) = F(n+9) - F(9) with F = A000045.

%F a(n) = a(n-1) + a(n-2) + 34 for n>1, a(0)=0, a(1)=21, and where 34 = F(9).

%F G.f.: x*(21 + 13*x)/((1 - x)*(1 - x - x^2)). - _Ilya Gutkovskiy_, Feb 24 2017

%F a(n) = 21*A000071(n+2) + 13*A000071(n+1). - _Bruno Berselli_, Feb 24 2017

%F From _Colin Barker_, Feb 24 2017: (Start)

%F a(n) = (-34 + (2^(-n)*((1-sqrt(5))^n*(-38+17*sqrt(5)) + (1+sqrt(5))^n*(38+17*sqrt(5)))) / sqrt(5)).

%F a(n) = 2*a(n-1) - a(n-3) for n>2. (End)

%p nmax:=31: with(combinat): for n from 0 to nmax do a(n):=fibonacci(n+9)-fibonacci(9) od: seq(a(n),n=0..nmax);

%t Fibonacci[9 +Range[0, 40]] -34 (* _G. C. Greubel_, Jul 13 2019 *)

%t LinearRecurrence[{2,0,-1},{0,21,55},40] (* _Harvey P. Dale_, Aug 24 2024 *)

%o (Magma) [Fibonacci(n+9) - Fibonacci(9): n in [0..40]]; // _Vincenzo Librandi_, Apr 24 2011

%o (PARI) concat(0, Vec(x*(21+13*x)/((1-x)*(1-x-x^2)) + O(x^40))) \\ _Colin Barker_, Feb 24 2017

%o (PARI) a(n) = fibonacci(n+9) - fibonacci(9) \\ _Charles R Greathouse IV_, Feb 24 2017

%o (Sage) [fibonacci(n+9)-34 for n in (0..40)] # _G. C. Greubel_, Jul 13 2019

%o (GAP) List([0..40], n-> Fibonacci(n+9)-34); # _G. C. Greubel_, Jul 13 2019

%Y Cf. A000045, A000071.

%Y Cf. A131524 (Kn11), A001911 (Kn12), A006327 (Kn13), A167616 (Kn14), A180671 (Kn15), A180672 (Kn16), A180673 (Kn17), A180674 (Kn18).

%K nonn,easy

%O 0,2

%A _Johannes W. Meijer_, Sep 21 2010