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a(n) = 2*(a(n-1)+a(n-2)+a(n-3))-a(n-4) for n >= 4, with initial terms 0,0,1,0.
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%I #10 Sep 09 2018 06:44:02

%S 0,0,1,0,2,6,15,46,132,380,1101,3180,9190,26562,76763,221850,641160,

%T 1852984,5355225,15476888,44729034,129269310,373595239,1079710278,

%U 3120420620,9018182964,26063032485,75323561860,217689133998,629133273722,1818228906675,5254779066930,15186593360656,43890069394800,126844654738097

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

%H Colin Barker, <a href="/A317976/b317976.txt">Table of n, a(n) for n = 0..1000</a>

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

%F G.f.: x^2*(1 - 2*x) / (1 - 2*x - 2*x^2 - 2*x^3 + x^4). - _Colin Barker_, Sep 09 2018

%t (See A192235.)

%o (PARI) concat(vector(2), Vec(x^2*(1 - 2*x) / (1 - 2*x - 2*x^2 - 2*x^3 + x^4) + O(x^40))) \\ _Colin Barker_, Sep 09 2018

%Y Other sequences with this recurrence but different initial conditions: A192234, A192237, A317973, A317974, A317975.

%K nonn,easy

%O 0,5

%A _N. J. A. Sloane_, Sep 03 2018