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%I #41 Sep 08 2022 08:44:39
%S 1,-3,13,-51,205,-819,3277,-13107,52429,-209715,838861,-3355443,
%T 13421773,-53687091,214748365,-858993459,3435973837,-13743895347,
%U 54975581389,-219902325555,879609302221,-3518437208883,14073748835533
%N a(n) = (1 - (-4)^n)/5.
%C q-integers for q=-4.
%C In Penrose's book, presented as partial sums of the series for 1/(1-x^2) evaluated at x=2. - _Olivier Gérard_, May 22 2009
%C Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-3, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=2, a(n-1)=(-1)^n*charpoly(A,1). - _Milan Janjic_, Jan 27 2010
%D Roger Penrose, "The Road to Reality, A complete guide to the Laws of the Universe", Jonathan Cape, London, 2004, pages 79-80. - _Olivier Gérard_, May 22 2009
%H Vincenzo Librandi, <a href="/A014985/b014985.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (-3,4).
%F a(n) = a(n-1) + q^{(n-1)} = {(q^n - 1) / (q - 1)}, with q=-4.
%F From _Paul Barry_, Jan 12 2007: (Start)
%F G.f.: x/(1+3*x-4*x^2).
%F a(n) = Sum_{k=0..floor(n/2)} C(n-k,k)*4^k*(-3)^(n-2k). (End)
%F a(n) = -3*a(n-1) +4*a(n-2). - _Vincenzo Librandi_, Oct 21 2012
%p a:=n->sum ((-4)^j, j=0..n): seq(a(n), n=0..25); # _Zerinvary Lajos_, Dec 16 2008
%t LinearRecurrence[{-3, 4}, {1, -3}, 50] (* or *) CoefficientList[ Series[ 1/((1-x)*(1+4*x)), {x,0,30}], x] (* _Vincenzo Librandi_, Oct 21 2012 *)
%o (Sage) [gaussian_binomial(n,1,-4) for n in range(1,24)] # - _Zerinvary Lajos_, May 28 2009
%o (Magma) I:=[1, -3]; [n le 2 select I[n] else -3*Self(n-1)+4*Self(n-2): n in [1..30]]; // _Vincenzo Librandi_, Oct 21 2012
%o (PARI) a(n)=(1-(-4)^n)/5 \\ _Charles R Greathouse IV_, Sep 24 2015
%Y Cf. A077925, A014983, A014986, A014987, A014989-A014994.
%K sign,easy
%O 1,2
%A _Olivier Gérard_
%E G.f. adapted to the offset by _Vincenzo Librandi_, Oct 21 2012
%E Better name from _Ralf Stephan_, Jul 14 2013
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