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a(0)=3, a(1)=7; thereafter a(n) = a(n-1) + a(n-2) + 1.
3

%I #36 Aug 24 2022 19:40:40

%S 3,7,11,19,31,51,83,135,219,355,575,931,1507,2439,3947,6387,10335,

%T 16723,27059,43783,70843,114627,185471,300099,485571,785671,1271243,

%U 2056915,3328159,5385075,8713235,14098311,22811547,36909859,59721407,96631267,156352675

%N a(0)=3, a(1)=7; thereafter a(n) = a(n-1) + a(n-2) + 1.

%C a(n) is the minimum number of nodes required for a full binary AVL tree of height n+1 whose root node has a balance factor of 0. - _Sumukh Patel_, Jun 24 2022

%H G. C. Greubel, <a href="/A022406/b022406.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 a(n) = 4*A000045(n+2) - 1. - _Ron Knott_, Aug 25 2006

%F From _R. J. Mathar_, May 28 2008: (Start)

%F a(n) = A022403(n+1).

%F O.g.f.: (3+x-3*x^2)/((1-x)*(1-x-x^2)).

%F a(n+1) - a(n) = A022087(n+1).

%F (End)

%F a(n) = (2^(-n)*(-5*2^n + (10-6*sqrt(5))*(1-sqrt(5))^n + 2*(1+sqrt(5))^n*(5+3*sqrt(5)))) / 5. - _Colin Barker_, Mar 02 2018

%t Table[4*Fibonacci[n + 2] - 1, {n, 0, 100}] (* _Vladimir Joseph Stephan Orlovsky_, Jun 28 2011 *)

%t CoefficientList[Series[(3+x-3*x^2)/((1-x)*(1-x-x^2)), {x, 0, 50}], x] (* _G. C. Greubel_, Mar 01 2018 *)

%o (PARI) for(n=0, 40, print1(4*fibonacci(n+2) -1, ", ")) \\ _G. C. Greubel_, Mar 01 2018

%o (Magma) [4*Fibonacci(n+2) - 1: n i [0..40]]; // _G. C. Greubel_, Mar 01 2018

%Y Cf. A000045, A022087, A122195. See A022403 for a very similar sequence.

%K nonn,easy

%O 0,1

%A _N. J. A. Sloane_