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a(-3) = a(-2) = a(-1) = 0, a(0) = 1, a(n) = a(n-1) + 2*a(n-2) + 2*a(n-3) + a(n-4), for n>0.
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%I #17 Mar 01 2018 13:09:06

%S 0,0,0,1,1,3,7,16,37,86,199,461,1068,2474,5731,13276,30754,71242,

%T 165033,382301,885605,2051515,4752360,11008901,25502256,59076293,

%U 136850967,317016966,734373742,1701185901,3940818284,9128954536,21147336648

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

%H G. C. Greubel, <a href="/A123392/b123392.txt">Table of n, a(n) for n = -3..1000</a>

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

%F a(n) = Sum_{k, 0<=k<=[n/2]} A016095(n-k,k).

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

%F a(n) = Sum_{k=0..n} C(n-k,k)*F(n-k+1). - _Paul Barry_, Mar 01 2010

%t Join[{0, 0, 0}, CoefficientList[Series[1/(1 - x - 2*x^2 - 2*x^3 - x^4), {x, 0, 50}], x]] (* or *) Join[{0, 0, 0}, Table[Sum[ Binomial[n - k, k]*Fibonacci[n - k + 1], {k, 0, n}], {n, 0, 50}]] (* _G. C. Greubel_, Oct 13 2017 *)

%t LinearRecurrence[{1,2,2,1},{0,0,0,1},40] (* _Harvey P. Dale_, Mar 01 2018 *)

%o (PARI) x='x+O('x^50); concat([0,0,0], Vec(1/(1-x-2*x^2-2*x^3-x^4))) \\ _G. C. Greubel_, Oct 13 2017

%K nonn

%O -3,6

%A _Philippe Deléham_, Oct 14 2006