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a(n) = 4a(n-1) - 6a(n-2) + 4a(n-3), n > 3; a(0) = 3, a(1) = 2, a(2) = a(3) = 0.
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%I #10 Sep 23 2015 15:42:47

%S 3,2,0,0,8,32,80,160,288,512,960,1920,3968,8192,16640,33280,66048,

%T 131072,261120,522240,1046528,2097152,4198400,8396800,16785408,

%U 33554432,67092480,134184960,268402688,536870912,1073807360,2147614720

%N a(n) = 4a(n-1) - 6a(n-2) + 4a(n-3), n > 3; a(0) = 3, a(1) = 2, a(2) = a(3) = 0.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (4, -6, 4).

%F Sequence is identical to its fourth differences.

%F a(n) = 2^n + 2^[(n+3)/2]*cos((n+1)Pi/4); a(n)=2^n + (1+i)^(n+1) + (1-i)^(n+1), where i=sqrt(-1). - _Emeric Deutsch_, Oct 14 2007

%F G.f.: -(3-10*x+10*x^2)/(2*x-1)/(2*x^2-2*x+1). - _R. J. Mathar_, Nov 14 2007

%p a[0]:=3: a[1]:=2: a[2]:=0: a[3]:=0; for n from 4 to 27 do a[n]:=4*a[n-1]-6*a[n-2]+4*a[n-3] end do: seq(a[n],n=0..27); # _Emeric Deutsch_, Oct 14 2007

%t a = {3, 2, 0, 0}; Do[AppendTo[a, 4*a[[ -1]] - 6*a[[ -2]] + 4*a[[ -3]]], {30}]; a (* _Stefan Steinerberger_, Oct 14 2007 *)

%t LinearRecurrence[{4, -6, 4},{3, 2, 0},32] (* _Ray Chandler_, Sep 23 2015 *)

%Y Cf. A131470, A009116, A099087.

%K nonn,easy

%O 0,1

%A _Paul Curtz_, Oct 11 2007

%E More terms from _Stefan Steinerberger_ and _Emeric Deutsch_, Oct 14 2007