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A295683 a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = 2, a(1) = 1, a(2) = 0, a(3) = 1.

%I A295683 #24 Jun 22 2025 14:04:16
%S A295683 2,1,0,1,4,5,6,11,20,31,48,79,130,209,336,545,884,1429,2310,3739,6052,
%T A295683 9791,15840,25631,41474,67105,108576,175681,284260,459941,744198,
%U A295683 1204139,1948340,3152479,5100816,8253295,13354114,21607409,34961520,56568929,91530452
%N A295683 a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = 2, a(1) = 1, a(2) = 0, a(3) = 1.
%C A295683 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 A295683 Clark Kimberling, <a href="/A295683/b295683.txt">Table of n, a(n) for n = 0..2000</a>
%H A295683 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,1,1).
%F A295683 G.f.: (2 - x - x^2 - x^3)/(1 - x - x^3 - x^4).
%F A295683 a(n) = (2/5) * A000045(n) + (4/5) * A000045(n-1) + (6/5) * A056594(n) + (3/5) * A056594(n-1) for n >= 1. - _Robert Israel_, Apr 03 2019
%p A295683 f:= gfun:-rectoproc(a(n) = a(n-1) + a(n-3) + a(n-4), a(0) = 2, a(1) = 1, a(2) = 0, a(3) = 1}, a(n), remember):
%p A295683 map(f, [$0..100]); # _Robert Israel_, Apr 03 2019
%t A295683 LinearRecurrence[{1, 0, 1, 1}, {2, 1, 0, 1}, 45]
%t A295683 CoefficientList[Series[(2-x-x^2-x^3)/(1-x-x^3-x^4),{x,0,40}],x] (* _Harvey P. Dale_, Jun 22 2025 *)
%o A295683 (PARI) my(x='x+O('x^45)); Vec((-2 + x + x^2 + x^3)/(-1 + x + x^3 + x^4)) \\ _Georg Fischer_, Apr 03 2019
%o A295683 (Magma) I:=[2,1,0,1]; [n le 4 select I[n] else Self(n-1) +Self(n-3) +Self(n-4): n in [1..45]]; // _G. C. Greubel_, Apr 03 2019
%o A295683 (Sage) ((2-x-x^2-x^3)/(1-x-x^3-x^4)).series(x, 45).coefficients(x, sparse=False) # _G. C. Greubel_, Apr 03 2019
%Y A295683 Cf. A001622, A000045, A056594.
%K A295683 nonn,easy
%O A295683 0,1
%A A295683 _Clark Kimberling_, Nov 29 2017
%E A295683 a(0) = 2 amended by _Georg Fischer_, Apr 03 2019