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%I #18 Jul 18 2017 12:05:04
%S 0,1,1,1,1,1,1,-1,-7,-23,-59,-139,-311,-677,-1443,-3031,-6295,-12967,
%T -26543,-54073,-109743,-222071,-448323,-903411,-1817767,-3653245,
%U -7335147,-14716663,-29508351,-59138095,-118472607
%N a(n) = A000045(n) - A113405(n).
%C Because the inverse binomial transform of A000045 and A113405 is individually the same as the original sequence up to a sign flip of each second term, the same is true for their difference here. (The inverse binomial transform is a linear transform.)
%C The sequence and its higher order differences in subsequent rows has zeros on the main diagonal:
%C 0, 1, 1, 1, 1, 1, 1, -1, -7,-23, -59, -139, -311, -677, -1443, -3031
%C 1, 0, 0, 0, 0, 0, -2, -6,-16,-36, -80, -172, -366, -766, -1588, -3264
%C -1, 0, 0, 0, 0,-2, -4,-10,-20,-44, -92, -194, -400, -822, -1676
%C 1, 0, 0, 0,-2,-2, -6,-10,-24,-48,-102, -206, -422, -854, -1732
%C -1, 0, 0,-2, 0,-4, -4,-14,-24,-54,-104, -216, -432, -878, -1764
%C 1, 0,-2, 2,-4, 0,-10,-10,-30,-50,-112, -216, -446, -886, -1790,
%C -1,-2, 4,-6, 4,-10, 0,-20,-20,-62,-104, -230, -440, -904, -1792
%H G. C. Greubel, <a href="/A140096/b140096.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (3,-1,-3,3,-1,-2).
%F a(n)= +3*a(n-1) -a(n-2) -3*a(n-3) +3*a(n-4) -a(n-5) -2*a(n-6).
%F G.f.: -x*(-1+2*x+x^2-2*x^3+x^4) / ( (2*x-1)*(1+x)*(x^2-x+1)*(x^2+x-1) ).
%F a(n+1)-2*a(n) = -A141325(n-2), n>2.
%t CoefficientList[Series[-x*(-1 + 2*x + x^2 - 2*x^3 + x^4)/((2*x - 1)*(1 + x)*(x^2 - x + 1)*(x^2 + x - 1)), {x, 0, 50}], x] (* _G. C. Greubel_, Jul 17 2017 *)
%o (PARI) x='x+O('x^50); concat([0], Vec(-x*(-1+2*x+x^2-2*x^3+x^4)/( (2*x-1)*(1+x)*(x^2-x+1)*(x^2+x-1) ))) \\ _G. C. Greubel_, Jul 17 2017
%Y Cf. A140428, A140413.
%K sign,easy,less
%O 0,9
%A _Paul Curtz_, Jun 21 2008
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