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%I #31 Mar 23 2024 12:20:04
%S 2,16,254,4048,64514,1028176,16386302,261152656,4162056194,
%T 66331746448,1057145886974,16848002445136,268510893235202,
%U 4279326289318096,68200709735854334,1086932029484351248,17322711762013765634,276076456162735898896
%N a(n) = 16a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 16.
%C Numbers n such that (n^2-4)/7 is a square. - _Colin Barker_, Mar 17 2014
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H <a href="/index/Rea#recur1">Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (16,-1).
%F a(n) = (8+sqrt(63))^n + (8-sqrt(63))^n.
%F a(n)^2 = a(2n) + 2.
%F G.f.: (2-16*x)/(1-16*x+x^2). - _Philippe Deléham_, Nov 02 2008
%F a(n) = 2 * A001081(n). - _R. J. Mathar_, Nov 30 2008
%t a[0] = 2; a[1] = 16; a[n_] := 16a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 15}] (* _Robert G. Wilson v_, Jan 30 2004 *)
%t LinearRecurrence[{16, -1}, {2, 16}, 20] (* _T. D. Noe_, Mar 17 2014 *)
%o (Sage) [lucas_number2(n,16,1) for n in range(0,20)] # _Zerinvary Lajos_, Jun 26 2008
%Y Cf. A080246.
%K easy,nonn
%O 0,1
%A Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 18 2004
%E More terms from _Robert G. Wilson v_, Jan 30 2004
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