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%I #89 Feb 22 2023 21:41:48
%S 1,-1,3,-5,11,-21,43,-85,171,-341,683,-1365,2731,-5461,10923,-21845,
%T 43691,-87381,174763,-349525,699051,-1398101,2796203,-5592405,
%U 11184811,-22369621,44739243,-89478485,178956971,-357913941,715827883,-1431655765,2863311531,-5726623061
%N Expansion of 1/((1-x)*(1+2*x)).
%C a(n+1) is the reflection of a(n) through a(n-1) on the numberline. - _Floor van Lamoen_, Aug 31 2004
%C If a zero is added as the (new) a(0) in front, the sequence represents the inverse binomial transform of A001045. Partial sums are in A077898. - _R. J. Mathar_, Aug 30 2008
%C a(n) = A077953(2*n+3). - _Reinhard Zumkeller_, Oct 07 2008
%C Related to the Fibonacci sequence by an INVERT transform: if A(x) = 1+x^2*g(x) is the generating function of the a(n) prefixed with 1, 0, then 1/A(x) = 2+(x+1)/(x^2-x+1) is the generating function of 1, 0, -1, 1, -2, 3, ..., the signed Fibonacci sequence A000045 prefixed with 1. - _Gary W. Adamson_, Jan 07 2011
%C Also: Gaussian binomial coefficients [n+1,1], or q-integers, for q=-2, diagonal k=1 in the triangular (or column r=1 in the square) array A015109. - _M. F. Hasler_, Nov 04 2012
%C With a leading zero, 0, 1, -1, 3, -5, 11, -21, 43, -85, 171, -341, 683, ... we obtain the Lucas U(-1,-2) sequence. - _R. J. Mathar_, Jan 08 2013
%C Let m = a(n). Then 18*m^2 - 12*m + 1 = A000225(2n+3). - _Roderick MacPhee_, Jan 17 2013
%H Vincenzo Librandi, <a href="/A077925/b077925.txt">Table of n, a(n) for n = 0..1000</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Lucas_sequence#Specific_names">Lucas sequence</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (-1,2)
%H <a href="/index/Ga#Gaussian_binomial_coefficients">Index entries related to Gaussian binomial coefficients</a>.
%H <a href="/index/Lu#Lucas">Index entries for Lucas sequences</a>
%F G.f.: 1/(1+x-2*x^2).
%F a(n) = (1-(-2)^(n+1))/3. - _Vladeta Jovovic_, Apr 17 2003
%F a(n) = Sum_{k=0..n} (-2)^k. - _Paul Barry_, May 26 2003
%F a(n+1) - a(n) = A122803(n). - _R. J. Mathar_, Aug 30 2008
%F a(n) = Sum_{k=0..n} A112555(n,k)*(-2)^k. - _Philippe Deléham_, Sep 11 2009
%F a(n) = A082247(n+1) - 1. - _Philippe Deléham_, Oct 07 2009
%F G.f.: Q(0)/(3*x), where Q(k) = 1 - 1/(4^k - 2*x*16^k/(2*x*4^k + 1/(1 + 1/(2*4^k - 8*x*16^k/(4*x*4^k - 1/Q(k+1)))))); (continued fraction). - _Sergei N. Gladkovskii_, May 22 2013
%F G.f.: Q(0)/2 , where Q(k) = 1 + 1/(1 - x*(4*k-1 + 2*x)/( x*(4*k+1 + 2*x) + 1/Q(k+1) )); (continued fraction). - _Sergei N. Gladkovskii_, Sep 08 2013
%F E.g.f.: (2*exp(-2*x) + exp(x))/3. - _Ilya Gutkovskiy_, Nov 12 2016
%F a(n) = A086893(n+2) - A061547(n+3), n >= 0. - _Yosu Yurramendi_, Jan 16 2017
%F a(n) = (-1)^n*A001045(n+1). - _M. F. Hasler_, Feb 13 2020
%F a(n) - a(n-1) = a(n-1) - a(n+1) = (-2)^n, a(n+1) = - a(n) + 2*a(n-1) = 1 - 2*a(n). - _Michael Somos_, Feb 22 2023
%e G.f. = 1 - x + 3*x^2 - 5*x^3 + 11*x^4 - 21*x^5 + 43*x^6 - 85*x^7 + ...
%p a:=n->sum ((-2)^j, j=0..n): seq(a(n), n=0..35); # _Zerinvary Lajos_, Dec 16 2008
%t CoefficientList[Series[(1 - x)^(-1)/(1 + 2 x), {x, 0, 50}], x] (* _Vladimir Joseph Stephan Orlovsky_, Jun 20 2011 *)
%o (Sage) [gaussian_binomial(n,1,-2) for n in range(1,35)] # _Zerinvary Lajos_, May 28 2009
%o (Magma) [(1-(-2)^(n+1))/3: n in [0..40]]; // _Vincenzo Librandi_, Jun 21 2011
%o (PARI) a(n)=(1+(-2)^n*2)/3 \\ _Charles R Greathouse IV_, Jun 21 2011
%Y Cf. A001045 (unsigned version).
%Y Cf. A014983, A014985, A014986. - _Zerinvary Lajos_, Dec 16 2008
%Y Cf. A232600, A232601, A232602.
%K sign,easy
%O 0,3
%A _N. J. A. Sloane_, Nov 17 2002
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