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A022093 Fibonacci sequence beginning 0, 10.

%I A022093 #44 Jun 03 2025 14:57:11
%S A022093 0,10,10,20,30,50,80,130,210,340,550,890,1440,2330,3770,6100,9870,
%T A022093 15970,25840,41810,67650,109460,177110,286570,463680,750250,1213930,
%U A022093 1964180,3178110,5142290,8320400,13462690,21783090,35245780,57028870,92274650,149303520,241578170
%N A022093 Fibonacci sequence beginning 0, 10.
%D A022093 A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 15.
%H A022093 Vincenzo Librandi, <a href="/A022093/b022093.txt">Table of n, a(n) for n = 0..1000</a>
%H A022093 Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H A022093 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1,1).
%F A022093 a(n) = 10*F(n) = F(n+4) + F(n+2) + F(n-2) + F(n-4) for n>3, where F = A000045.
%F A022093 a(n) = round((4*phi-2)*phi^n) for n>4. - _Thomas Baruchel_, Sep 08 2004
%F A022093 G.f.: 10*x/(1 - x - x^2). - _Philippe Deléham_, Nov 20 2008
%F A022093 a(n) = F(n+5) + F(n-5) - 5*F(n) for n>0. - _Bruno Berselli_, Dec 29 2016
%F A022093 a(n) = Lucas(n+3) + Lucas(n-3), where Lucas(-n) = (-1)^n*Lucas(n) for the negative indices. - _Bruno Berselli_, Jun 13 2017
%t A022093 LinearRecurrence[{1, 1}, {0, 10}, 40] (* _Bruno Berselli_, Dec 30 2016 *)
%t A022093 Table[Fibonacci[n + 5] + Fibonacci[n - 5] - 5 Fibonacci[n], {n, 1, 40}] (* _Bruno Berselli_, Dec 30 2016 *)
%t A022093 Table[10 Fibonacci[n], {n, 0, 100}] (* _Vincenzo Librandi_, Dec 31 2016 *)
%o A022093 (Magma) [10*Fibonacci(n): n in [0..40]]; // _Vincenzo Librandi_, Dec 31 2016
%o A022093 (SageMath)
%o A022093 A022093=BinaryRecurrenceSequence(1,1,0,10)
%o A022093 [A022093(n) for n in range(51)] # _G. C. Greubel_, Jun 02 2025
%Y A022093 Cf. A000032, A000045.
%K A022093 nonn,easy
%O A022093 0,2
%A A022093 _N. J. A. Sloane_