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%I #20 Jan 28 2018 02:45:43
%S 2,2,2,-7,11,-16,29,-61,128,-259,515,-1024,2045,-4093,8192,-16387,
%T 32771,-65536,131069,-262141,524288,-1048579,2097155,-4194304,8388605,
%U -16777213,33554432,-67108867,134217731,-268435456,536870909,-1073741821,2147483648
%N a(n) = -3a(n-1) - 3a(n-2) - 2a(n-3), n > 3.
%C The inverse binomial transform of A153130, after dropping A153130(0).
%C The inverse binomial transform of the full A153130 is A158916.
%C Dropping two initial terms of A153130 yields A158935, dropping three yields essentially a sign-reversed version of A158916, dropping 4 essentially the sequence here.
%H Muniru A Asiru, <a href="/A158927/b158927.txt">Table of n, a(n) for n = 0..600</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (-3, -3, -2).
%F a(n) = -3a(n-1) - 3a(n-2) - 2a(n-3), with a(0)=a(1)=a(2)=2, a(3)=-7.
%F a(n) = (-1)^(n+1)*A157823(n) - A099838(n+3).
%F G.f.: (2+8*x+14*x^2+9*x^3)/((2*x+1)*(1+x+x^2)). - _R. J. Mathar_, Apr 09 2009
%F a(0)=2; a(n) = (1/2)*(-2)^n - 3*cos(2*Pi*n/3) + sqrt(3)*sin(2*Pi*n/3) for n >= 1. - _Richard Choulet_, Apr 23 2009
%p a := proc(n) option remember: if n=0 then 2 elif n=1 then 2 elif n=2 then 2 elif n=3 then -7 elif n>=4 then -3*procname(n-1) - 3*procname(n-2) - 2*procname(n-3) fi; end:
%p seq(a(n), n=0..100); # _Muniru A Asiru_, Jan 27 2018
%o (GAP) a := [2,2,2,-7];; for n in [5..10^3] do a[n] := -3*a[n-1] - 3*a[n-2] - 2*a[n-3]; od; a; # _Muniru A Asiru_, Jan 27 2018
%Y Same recurrence as A131562, A158916, A158926.
%K sign
%O 0,1
%A _Paul Curtz_, Mar 31 2009
%E Edited and extended by _R. J. Mathar_, Apr 09 2009
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