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%I #40 Sep 08 2022 08:44:49
%S 1,4,9,21,46,99,209,436,901,1849,3774,7671,15541,31404,63329,127501,
%T 256366,514939,1033449,2072676,4154701,8324529,16673534,33386671,
%U 66837421,133778524,267724809,535721061,1071881326,2144473299,4290096449,8582053396,17167117141
%N a(n) = T(n,n) + T(n,n+1) + ... + T(n,2n), T given by A027960.
%H Vincenzo Librandi, <a href="/A027973/b027973.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-1,-2).
%F With a different offset: recurrence: a(-1)=a(0)=1 a(n+2) = a(n+1) + a(n) + 2^n; formula: a(n-2) = floor(2^n - phi^n) - (1-(-1)^n)/2. - _Benoit Cloitre_, Sep 02 2002
%F a(n) = A101220(4, 2, n+1) - A101220(4, 2, n). - _Ross La Haye_, Aug 05 2005
%F a(n) = 2*a(n-1) + Fibonacci(n+1) - Fibonacci(n-3) for n>=1; a(0)=1. - _Emeric Deutsch_, Nov 29 2006
%F O.g.f.: 4/(1-2*x) - (x+3)/(1-x-x^2). - _R. J. Mathar_, Nov 23 2007
%F a(n) = 2^(n+2) + F(n) - F(n+4) with F(n)=A000045(n). - _Johannes W. Meijer_, Aug 15 2010
%F Eigensequence of an infinite lower triangular matrix with the Lucas series (1, 3, 4, 7, ...) as the left border and the rest ones. - _Gary W. Adamson_, Jan 30 2012
%F a(n) = 2^(n+2) - Lucas(n+2). - _Vincenzo Librandi_, May 05 2017, corrected by _Greg Dresden_, Sep 13 2021
%p with(combinat): a[0]:=1: for n from 1 to 30 do a[n]:=2*a[n-1]+fibonacci(n+1)-fibonacci(n-3) od: seq(a[n],n=0..30); # _Emeric Deutsch_, Nov 29 2006
%t Table[2^n - LucasL[n], {n, 2, 50}] (* _Vincenzo Librandi_, May 05 2017 *)
%o (Magma) [2^n-Lucas(n): n in [2..40]]; // _Vincenzo Librandi_, May 05 2017
%o (PARI) vector(40, n, f=fibonacci; 2^(n+1) - f(n+2) - f(n) ) \\ _G. C. Greubel_, Sep 26 2019
%o (Sage) [2^(n+2) - lucas_number2(n+2,1,-1) for n in (0..40)] # _G. C. Greubel_, Sep 26 2019
%o (GAP) List([0..40], n-> 2^(n+2) - Lucas(1,-1,n+2)[2]); # _G. C. Greubel_, Sep 26 2019
%Y Cf. A000032, A000045, A027960.
%K nonn
%O 0,2
%A _Clark Kimberling_
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