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A173114 a(0)=a(1)=1, a(n) = 2*a(n-1)- A010686(n), n>1.

%I A173114 #13 May 12 2025 06:45:19
%S A173114 1,1,1,-3,-7,-19,-39,-83,-167,-339,-679,-1363,-2727,-5459,-10919,
%T A173114 -21843,-43687,-87379,-174759,-349523,-699047,-1398099,-2796199,
%U A173114 -5592403,-11184807,-22369619,-44739239,-89478483,-178956967,-357913939,-715827879,-1431655763
%N A173114 a(0)=a(1)=1, a(n) = 2*a(n-1)- A010686(n), n>1.
%C A173114 The sequence in the first row and successive differences in followup rows defines the array
%C A173114   1, 1, 1, -3, -7, -19, -39, -83, -167, -339, ..
%C A173114   0, 0, -4, -4, -12, -20, -44, -84, -172, -340, ..
%C A173114   0, -4, 0, -8, -8, -24, -40, -88, -168, -344, ..
%C A173114  -4, 4, -8, 0, -16, -16, -48, -80, -176, -336, ..
%C A173114   8, -12, 8, -16, 0, -32, -32, -96, -160, -352, ..
%C A173114 The first two subdiagonals show essentially the powers of 2.
%H A173114 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,1,-2).
%F A173114 a(n) = 3 + 2*( (-1)^n-2^n )/3 = 3-A078008(n+1), n>0. - _R. J. Mathar_, Jun 30 2010
%F A173114 a(n+2)-a(n)= A154589(n+2) = -2^(n+1), n>0.
%F A173114 a(n) = 2*a(n-1)+a(n-2)-2*a(n-3), n>3.
%F A173114 G.f.: (-x-2*x^2-4*x^3+1)/( (1-x)*(1-2*x)*(1+x) ).
%F A173114 a(n) + A173078(n) = 2^n.
%F A173114 a(n) - a(n-1) = -4*A001045(n-2) = -A097074(n-1), n>1.
%K A173114 easy,sign
%O A173114 0,4
%A A173114 _Paul Curtz_, Feb 10 2010
%E A173114 Edited and extended by _R. J. Mathar_, Jun 30 2010