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A295687 a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = 1, a(1) = 2, a(2) = 2, a(3) = 1.

%I A295687 #13 Dec 04 2021 12:36:03
%S A295687 1,2,2,1,4,8,11,16,28,47,74,118,193,314,506,817,1324,2144,3467,5608,
%T A295687 9076,14687,23762,38446,62209,100658,162866,263521,426388,689912,
%U A295687 1116299,1806208,2922508,4728719,7651226,12379942,20031169,32411114,52442282,84853393
%N A295687 a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = 1, a(1) = 2, a(2) = 2, a(3) = 1.
%C A295687 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 A295687 Clark Kimberling, <a href="/A295687/b295687.txt">Table of n, a(n) for n = 0..2000</a>
%H A295687 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,1,1).
%F A295687 a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = 1, a(1) = 2, a(2) = 2, a(3) = 1.
%F A295687 G.f.: (-1 - x + 2*x^3)/(-1 + x + x^3 + x^4).
%F A295687 From _Peter Bala_, Nov 27 2021: (Start)
%F A295687 a(2*n) = 3*a(2*n-2) - a(2*n-4) - (-1)^n, for n >= 2;
%F A295687 a(2*n+1) = 3*a(2*n-1) - a(2*n-3) + 7*(-1)^n, for n >= 2.
%F A295687 a(2*n) = Lucas(2*n-1) - Fibonacci(n-3)*Fibonacci(n-2) = A002878(n-1) - A001654(n-3);
%F A295687 a(2*n+1) = Lucas(2*n) - Fibonacci(n-4)*Fibonacci(n) = A005248(n) - A192883(n-3).
%F A295687 a(4*n-1) = Fibonacci(2*n+1)^2 - Fibonacci(2*n)^2 + Fibonacci(2*n-1)^2 - Fibonacci(2*n-2)^2 - 3, for n >= 1;
%F A295687 a(4*n+1) = Fibonacci(2*n+2)^2 - Fibonacci(2*n+1)^2 + Fibonacci(2*n)^2 - Fibonacci(2*n-1)^2 + 3, for n >= 0.
%F A295687 Conjecture: a(2*n+7) = Fibonacci(n)^3*Sum_{k >= 1} k^2 * Fibonacci(n*k)/ Fibonacci(n+2)^(k+1), n >= 1. (End)
%t A295687 LinearRecurrence[{1, 0, 1, 1}, {1, 2, 2, 1}, 100]
%Y A295687 Cf. A000045, A001622, A001654, A002878, A005248, A192883, A295690.
%K A295687 nonn,easy
%O A295687 0,2
%A A295687 _Clark Kimberling_, Nov 29 2017