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A295690 a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = 2, a(1) = 2, a(2) = 1, a(3) = 1.

%I A295690 #14 Sep 08 2022 08:46:20
%S A295690 2,2,1,1,5,8,10,16,29,47,73,118,194,314,505,817,1325,2144,3466,5608,
%T A295690 9077,14687,23761,38446,62210,100658,162865,263521,426389,689912,
%U A295690 1116298,1806208,2922509,4728719,7651225,12379942,20031170,32411114,52442281,84853393
%N A295690 a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = 2, a(1) = 2, a(2) = 1, a(3) = 1.
%C A295690 a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622), so that a( ) has the growth rate of the Fibonacci numbers (A000045).
%H A295690 Clark Kimberling, <a href="/A295690/b295690.txt">Table of n, a(n) for n = 0..2000</a>
%H A295690 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,1,1).
%F A295690 a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = 2, a(1) = 2, a(2) = 1, a(3) = 1.
%F A295690 G.f.: (-2 + x^2 + 2 x^3)/(-1 + x + x^3 + x^4).
%F A295690 From _Peter Bala_, Nov 12 2019: (Start)
%F A295690 a(2*n) = (3/5)*Lucas(2*n) + (4/5)*(-1)^n.
%F A295690 a(2*n+1) = (3/5)*Lucas(2*n+1) + (7/5)*(-1)^n.
%F A295690 a(2*n) = a(2*n-1) + a(2*n-2) + 3*(-1)^n.
%F A295690 a(2*n+1) = a(2*n) + a(2*n-1) + 2*(-1)^n.
%F A295690 a(2*n+1)*F(n+3) - a(2*n+3)*F(n-1) = 3*F(n+1)^3, where F(n) = A000045(n). (End)
%t A295690 LinearRecurrence[{1, 0, 1, 1}, {2, 2, 1, 1}, 100]
%o A295690 (Magma) a:=[2,2,1,1]; [n le 4 select a[n] else Self(n-1) + Self(n-3) + Self(n-4):n in [1..40]]; // _Marius A. Burtea_, Nov 13 2019
%Y A295690 Cf. A001622, A000045, A000032.
%K A295690 nonn,easy
%O A295690 0,1
%A A295690 _Clark Kimberling_, Nov 29 2017