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A158935 a(n)= -3a(n-1)-3a(n-2)-2a(n-3), n>3. a(0)=4, a(1)=4, a(2)=-5, a(3)=4. 1


%S 4,4,-5,4,-5,13,-32,67,-131,256,-509,1021,-2048,4099,-8195,16384,

%T -32765,65533,-131072,262147,-524291,1048576,-2097149,4194301,

%U -8388608,16777219,-33554435,67108864,-134217725,268435453,-536870912,1073741827,-2147483651

%N a(n)= -3a(n-1)-3a(n-2)-2a(n-3), n>3. a(0)=4, a(1)=4, a(2)=-5, a(3)=4.

%C The third column of the array of differences described in A153130. The first two columns are in A158916 and A158987. Taking differences like in A158926 keeps the recurrence.

%C Also the inverse binomial transform of A153130 if the first two items of A153130 are omitted.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (-3, -3, -2).

%F a(n)= A154589(n) + A099838(n+2).

%F G.f.: (4+16*x+19*x^2+9*x^3)/((2*x+1)*(1+x+x^2)). - _R. J. Mathar_, Apr 08 2009

%F a(n)=(1/2)*I*sqrt(3)*[ -(1/2)+(1/2)*I*sqrt(3)]^(n-2)-(3/2)*[ -(1/2)-(1/2)*I*sqrt(3)]^(n-2)-(3/2)*[ -(1/2)+(1/2)*I*sqrt(3)]^(n-2)-2*(-2)^(n-2)-(1/2)*I*sqrt(3)*[ -(1/2)-(1/2)*I*sqrt(3)]^(n-2)+(9/2)*[C(2*n,n) mod 2], with n>=0 [From _Paolo P. Lava_, Apr 15 2009]

%t Join[{4},LinearRecurrence[{-3,-3,-2},{4,-5,4},50]] (* _Harvey P. Dale_, May 25 2011 *)

%K sign,easy

%O 0,1

%A _Paul Curtz_, Mar 31 2009

%E Partially edited and extended by _R. J. Mathar_, Apr 08 2009

%E Edited by _N. J. A. Sloane_, Apr 08 2009

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Last modified October 17 11:44 EDT 2019. Contains 328108 sequences. (Running on oeis4.)