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Expansion of 1/(1 - x + 4*x^2).
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%I #69 Feb 28 2024 15:59:32

%S 1,1,-3,-7,5,33,13,-119,-171,305,989,-231,-4187,-3263,13485,26537,

%T -27403,-133551,-23939,510265,606021,-1435039,-3859123,1881033,

%U 17317525,9793393,-59476707,-98650279,139256549,533857665,-23168531,-2158599191,-2065925067

%N Expansion of 1/(1 - x + 4*x^2).

%C Row sums of Riordan array (1,x(1-4x)). In general, a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(k,n-k)*r^(n-k) yields the row sums of the Riordan array (1,x(1-kx)).

%C For n >= 1, a(n) equals the determinant of the n X n matrix with 2's along the superdiagonal and the subdiagonal, and 1's along the main diagonal, and 0's everywhere else. - _John M. Campbell_, Jun 04 2011

%C For n >= 1, |a(n-1)| is the unique odd positive solution x to 4^(n+1) = 15*x^2 + y^2. The value of y is |A272931(n)|. - _Jianing Song_, Jan 22 2019

%C Define the sequence u(n) = (u(n-1) + u(n-2))/u(n-3) with u(1) = 1, u(2) = -1, u(3) = 2. Then u(4*n) = 2*(a(n-1)+4*a(n-2))*a(n-1)/(a(n)+a(n-1))/a(n), u(4*n+1) = a(n+1)/a(n), u(4*n+2) = -1, u(4*n+3) = 4*(a(n)+a(n-1))/(a(n)+a(n+1)). For example, a(2) = -3, a(3) = -7 and u(8) = 5/3, u(9) = 7/3, u(10) = -1. - _Michael Somos_, Oct 24 2023

%H T. D. Noe, <a href="/A106853/b106853.txt">Table of n, a(n) for n = 0..1000</a>

%H Paul Barry, <a href="https://arxiv.org/abs/2004.04577">On a Central Transform of Integer Sequences</a>, arXiv:2004.04577 [math.CO], 2020.

%H Taras Goy and Mark Shattuck, <a href="https://doi.org/10.2478/amsil-2023-0027">Determinants of Toeplitz-Hessenberg Matrices with Generalized Leonardo Number Entries</a>, Ann. Math. Silesianae (2023). See p. 15.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1,-4).

%F G.f.: 1/(1 - x + 4*x^2).

%F a(n) = 2^n*(cos(2*n*arctan(sqrt(15)/5))+sqrt(15)*sin(2*n*arctan(sqrt(15)/5))/15).

%F a(n) = ((1 + sqrt(-15))^(n+1) - (1 - sqrt(-15))^(n+1))/(2^(n+1)*sqrt(-15)).

%F a(n) = Sum_{k=0..n} ((-1)^(n-k)*binomial(k, n-k)*4^(n-k)).

%F a(n) = a(n-1) - 4*a(n-2), a(0) = 1, a(1) = 1. - _Philippe Deléham_, Oct 21 2008

%F a(n) = Sum_{k=0..n} A109466(n,k)*4^(n-k). - _Philippe Deléham_, Oct 25 2008

%F G.f.: 1/(1 - 2*x)^2/(1 + 3*x*G(0)/2), where G(k) = 1 + 1/(1 - x/(x + (k + 1)/(2*k + 4)/G(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, Jun 01 2013

%F For n >= 1, 15*A272931(n)^2 + a(n-1)^2 = 4^(n+1). - _Jianing Song_, Jan 22 2019

%F a(n) = Product_{k=1..n} (1 + 4*cos(k*Pi/(n+1))). - _Peter Luschny_, Nov 28 2019

%F a(n) = 2^n * U(n, 1/4), where U(n, x) is the Chebyshev polynomial of the second kind. - _Federico Provvedi_, Mar 28 2022

%e G.f. = 1 + x - 3*x^2 - 7*x^3 + 5*x^4 + 33*x^5 + 13*x^6 - 119*x^7 - 171*x^8 + ... - _Michael Somos_, Oct 24 2023

%p f:= gfun:-rectoproc({a(n)=a(n-1)-4*a(n-2), a(0)=1,a(1)=1},a(n),remember):

%p map(f, [$0..100]); # _Robert Israel_, Jan 15 2018

%t Join[{a=1,b=1},Table[c=b-4*a;a=b;b=c,{n,80}]] (* _Vladimir Joseph Stephan Orlovsky_, Jan 22 2011 *)

%t CoefficientList[Series[1/(1-x*(1-4x)),{x,0,40}],x] (* or *) LinearRecurrence[ {1,-4},{1,1},40] (* _Harvey P. Dale_, May 26 2013 *)

%t a[ n_] := 2^n * ChebyshevU[n, 1/4]; (* _Michael Somos_, Oct 24 2023 *)

%o (Sage) [lucas_number1(n,1,4) for n in range(1, 36)] # _Zerinvary Lajos_, Apr 22 2009

%o (PARI) x='x+O('x^30); Vec(1/(1-x+4*x^2)) \\ _G. C. Greubel_, Jan 14 2018

%o (Magma) I:=[1,1]; [n le 2 select I[n] else Self(n-1) - 4*Self(n-2): n in [1..30]]; // _G. C. Greubel_, Jan 14 2018

%o (PARI) {a(n) = 2^n*polchebyshev(n, 2, 1/4)}; /* _Michael Somos_, Oct 24 2023 */

%Y Cf. A106852, A272931.

%K easy,sign

%O 0,3

%A _Paul Barry_, May 08 2005