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a(n) = a(n-1) + 3*a(n-2) -2*a(n-3) - 2*a(n-4), where a(0) = -2, a(1) = 1, a(2) = 0, a(3) = 1.
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%I #12 Aug 27 2021 21:06:51

%S -2,1,0,1,3,4,11,15,34,49,99,148,279,427,770,1197,2095,3292,5643,8935,

%T 15090,24025,40139,64164,106351,170515,280962,451477,740631,1192108,

%U 1949123,3141231,5123122,8264353,13453011,21717364,35301447,57018811,92582402

%N a(n) = a(n-1) + 3*a(n-2) -2*a(n-3) - 2*a(n-4), where a(0) = -2, a(1) = 1, a(2) = 0, a(3) = 1.

%C 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 Clark Kimberling, <a href="/A295860/b295860.txt">Table of n, a(n) for n = 0..2000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1,3,-2,-2)

%F G.f.: (-2 + 3 x + 5 x^2 - 6 x^3)/(1 - x - 3 x^2 + 2 x^3 + 2 x^4).

%F a(n) = -A016116(n+1) + 4*A000045(n) - A000045(n+1). - _Robert Israel_, Jan 12 2018

%p f:= n -> - 2^floor((n+1)/2) + 4*combinat:-fibonacci(n) - combinat:-fibonacci(n+1):

%p map(f, [$0..30]); # _Robert Israel_, Jan 12 2018

%t LinearRecurrence[{1, 3, -2, -2}, {-2, 1, 0, 1}, 100]

%Y Cf. A001622, A000045, A016116.

%K easy,sign

%O 0,1

%A _Clark Kimberling_, Jan 07 2018