%I #14 Dec 07 2019 12:18:25
%S 2,2,28,80,496,1952,9856,43136,204544,926720,4307968,19736576,
%T 91168768,419176448,1932378112,8894873600,40978284544,188695052288,
%U 869129519104,4002599665664,18434753560576,84900703109120
%N a(0)=2, a(1)=2, a(n) = 2*a(n-1) + 12*a(n-2).
%C If A091914(n-1)=F(n) the Fibonacci-like sequence, then a(n) is the Lucas-type sequence.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,12).
%F a(n) = ((1+sqrt(13))^n - (1-sqrt(13)^n))/(2*sqrt(13)).
%F G.f.: 2*(1-x)/(1-2*x-12*x^2).
%F E.g.f.: exp((1+sqrt(13))*x) + exp((1-sqrt(13))*x).
%F a(n) = A091914(n) + 12*A091914(n-2).
%F a(n) = 2*A125816(n). - _Alois P. Heinz_, Mar 03 2018
%p a[0]:=2:a[1]:=2:for i from 2 to 40 do a[i]:=2*a[i-1]+12*a[i-2] od: seq(a[n],n=0..40);
%t LinearRecurrence[{2,12},{2,2},30] (* _Harvey P. Dale_, May 24 2017 *)
%o (Sage) [lucas_number2(n,2,-12) for n in range(0, 22)] # _Zerinvary Lajos_, Apr 30 2009
%Y Cf. A091914, A125816.
%K easy,nonn
%O 0,1
%A _Miklos Kristof_, Mar 27 2007