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%I #37 Jan 31 2023 05:57:05
%S 1,1,17,33,305,833,5713,19041,110449,415105,2182289,8823969,43740593,
%T 184924097,884773585,3843559137,17999936497,79496882689,367495866641,
%U 1639445989665,7519379855921,33750515690561,154060593385297
%N G.f.: 1/(1-x-16*x^2).
%C The ratio a(n+1)/a(n) converges to (1+sqrt(65))/2 as n approaches infinity. - _Felix P. Muga II_, Mar 12 2014
%H Vincenzo Librandi, <a href="/A168579/b168579.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=1, b=4.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1,16).
%F a(0)=a(1)=1, a(n) = a(n-1) + 16*a(n-2) for n>1.
%F a(n) = (1/sqrt(65))*( ((1+sqrt(65))/2)^(n+1) - ((1-sqrt(65))/2)^(n+1) ), for n >= 0 [Binet representation] - _Felix P. Muga II_, Mar 12 2014
%F E.g.f.: (1/sqrt(65))*exp(x/2)*( sinh((sqrt(65)/2)*x) + sqrt(65)*cosh((sqrt(65)/2) *x) ). - _G. C. Greubel_, Jul 26 2016
%t Join[{a=1,b=1},Table[c=1*b+16*a;a=b;b=c,{n,60}]] (* _Vladimir Joseph Stephan Orlovsky_, Feb 01 2011 *)
%t CoefficientList[Series[1/(1 - x - 16 x^2), {x, 0, 40}], x] (* _Vincenzo Librandi_, Mar 14 2014 *)
%t LinearRecurrence[{1,16},{1,1},30] (* _Harvey P. Dale_, Aug 14 2014 *)
%o (PARI) a(n)=([0,1; 16,1]^n*[1;1])[1,1] \\ _Charles R Greathouse IV_, Jul 26 2016
%K nonn,easy
%O 0,3
%A _Philippe Deléham_, Nov 30 2009
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