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a(n) = a(n-1) + a(n-2) + 1, with a(0) = 1 and a(1) = 9.
2

%I #11 Aug 25 2017 23:36:14

%S 1,9,11,21,33,55,89,145,235,381,617,999,1617,2617,4235,6853,11089,

%T 17943,29033,46977,76011,122989,199001,321991,520993,842985,1363979,

%U 2206965,3570945,5777911,9348857

%N a(n) = a(n-1) + a(n-2) + 1, with a(0) = 1 and a(1) = 9.

%H G. C. Greubel, <a href="/A022323/b022323.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+7*x-7*x^2)/((1-x)*(1-x-x^2)).

%F a(n) = A022367(n) - 1. (End)

%F a(n) = 2*F(n+2) + 6*F(n) - 1, where F = A000045. - _G. C. Greubel_, Aug 25 2017

%t LinearRecurrence[{2,0,-1}, {1,9,11}, 50] (* _G. C. Greubel_, Aug 25 2017 *)

%o (PARI) x='x+O('x^50); Vec((1+7*x-7*x^2)/((1-x)*(1-x-x^2))) \\ _G. C. Greubel_, Aug 25 2017

%K nonn

%O 0,2

%A _N. J. A. Sloane_