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a(n+1) = 2*a(n) + 16*a(n-1), a(0)=0, a(1)=1.
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%I #25 Jan 31 2023 06:00:16

%S 0,1,2,20,72,464,2080,11584,56448,298240,1499648,7771136,39536640,

%T 203411456,1039409152,5333401600,27297349632,139929124864,

%U 716615843840,3672097685504,18810048872448,96373660712960,493708103385088,2529394778177536,12958119210516480

%N a(n+1) = 2*a(n) + 16*a(n-1), a(0)=0, a(1)=1.

%H Colin Barker, <a href="/A161007/b161007.txt">Table of n, a(n) for n = 0..1000</a>

%H J. Borowska, L. Lacinska, <a href="https://doi.org/10.17512/jamcm.2014.4.03">Recurrence form of determinant of a heptadiagonal symmetric Toeplitz matrix</a>, J. Appl. Math. Comp. Mech. 13 (2014) 19-16, remark 2 for permanent of tridiagonal Toeplitz matrices a=2, b=4.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,16).

%F a(n) = ((1+sqrt(17))^n - (1-sqrt(17))^n)/(2*sqrt(17)).

%F Limit_{n -> oo} a(n+1)/a(n) = 1 + sqrt(17).

%F G.f.: x / (1 - 2*x - 16*x^2). - _Colin Barker_, Jul 01 2015

%F a(n) = 2^(n-1)*A006131(n-1). - _R. J. Mathar_, Mar 08 2021

%F a(n) = (4*i)^n*ChebyshevU(n, -i/4). - _G. C. Greubel_, Oct 15 2022

%t LinearRecurrence[{2, 16}, {0, 1}, 50] (* _T. D. Noe_, Nov 07 2011 *)

%o (PARI) concat(0, Vec(-x/(16*x^2+2*x-1) + O(x^40))) \\ _Colin Barker_, Jul 01 2015

%o (Magma) [n le 2 select n-1 else 2*(Self(n-1) +8*Self(n-2)): n in [1..41]]; // _G. C. Greubel_, Oct 15 2022

%o (SageMath)

%o A161007=BinaryRecurrenceSequence(2,16,0,1)

%o [A161007(n) for n in range(41)] # _G. C. Greubel_, Oct 15 2022

%Y Cf. A006131, A010473 (sqrt(17)).

%K nonn,easy

%O 0,3

%A _Sture Sjöstedt_, Jun 02 2009