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a(n) = a(n-1) + a(n-2) + 1, with a(0) = 1 and a(1) = 7.
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%I #18 May 19 2019 19:51:16

%S 1,7,9,17,27,45,73,119,193,313,507,821,1329,2151,3481,5633,9115,14749,

%T 23865,38615,62481,101097,163579,264677,428257,692935,1121193,1814129,

%U 2935323,4749453,7684777,12434231

%N a(n) = a(n-1) + a(n-2) + 1, with a(0) = 1 and a(1) = 7.

%H G. C. Greubel, <a href="/A022321/b022321.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+5*x-5*x^2)/((1-x)*(1-x-x^2)).

%F a(n) = 2*A000285(n) - 1. (End)

%F a(n) = 2*F(n+2) + 4*F(n) - 1, where F = A000045. - _G. C. Greubel_, Aug 25 2017

%t a[0]=1;a[1]=7;a[n_]:=a[n]=a[n-1]+a[n-2]+1; Table[a[n],{n,0,40}] (* _Harvey P. Dale_, Jan 23 2011 *)

%o (PARI) x='x+O('x^50); Vec((1+5*x-5*x^2)/((1-x)*(1-x-x^2))) \\ _G. C. Greubel_, Aug 25 2017

%K nonn

%O 0,2

%A _N. J. A. Sloane_