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a(n) = a(n-1) + a(n-2) + 1, with a(0) = 1 and a(1) = 8.
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%I #13 Aug 25 2017 23:36:03

%S 1,8,10,19,30,50,81,132,214,347,562,910,1473,2384,3858,6243,10102,

%T 16346,26449,42796,69246,112043,181290,293334,474625,767960,1242586,

%U 2010547,3253134,5263682,8516817

%N a(n) = a(n-1) + a(n-2) + 1, with a(0) = 1 and a(1) = 8.

%H G. C. Greubel, <a href="/A022322/b022322.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 (2,0,-1).

%F From R. J. Mathar, Apr 07 2011: (Start)

%F G.f.: (1+6*x-6*x^2)/((1-x)*(1-x-x^2)).

%F a(n) = A022114(n) - 1. (End)

%F a(n) = 2*F(n+2) + 5*F(n) - 1, where F = A000045. - _G. C. Greubel_, Aug 25 2017

%t LinearRecurrence[{2,0,-1},{1,8,10},40] (* _Harvey P. Dale_, Oct 14 2012 *)

%o (PARI) x='x+O('x^50); Vec((1+6*x-6*x^2)/((1-x)*(1-x-x^2))) \\ _G. C. Greubel_, Aug 25 2017

%K nonn

%O 0,2

%A _N. J. A. Sloane_