a(n) = M^n(1, 1), with the matrix M= [[3, 1], [1, -1]]. Proof by Cayley-Hamilton, using S(n, -I) = (-I)^n*F(n+1), and S = A049310 and F = A000045. Motivated by A319053. - Wolfdieter Lang, Oct 08 2018
a := proc(n) option remember: if n=0 then RETURN(1) fi: if n=1 then RETURN(2) fi: 2*a(n-1) + 4*a(n-2); end: for n from 1 to 50 do printf(`%d, `, a(n)+a(n-1)) od:
f:=n-> simplify(expand((1/2)*(1+sqrt(5))^n + (1/5)*(1+sqrt(5))^n*sqrt(5) - (1/5)*sqrt(5)*(1-sqrt(5))^n + (1/2)*(1 -sqrt(5))^n )); # N. J. A. Sloane, Aug 10 2012
MATHEMATICA
a[n_]:=(MatrixPower[{{1, 5}, {1, 1}}, n].{{2}, {1}})[[2, 1]]; Table[a[n], {n, 0, 40}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2010 *)
LinearRecurrence[{2, 4}, {1, 3}, 100] (* G. C. Greubel, Feb 18 2017 *)
PROG
(PARI) { for (n=0, 200, if (n>1, a=2*a1 + 4*a2; a2=a1; a1=a, if (n, a=a1=2, a=a2=1)); if (n, write("b063782.txt", n, " ", a + a2)) ) } \\ Harry J. Smith, Aug 31 2009