a(n)= (T(n, 4)(1)^n)/5, with Chebyshev's polynomials of the first kind evaluated at x=4: T(n, 4)=A001091(n)=((4+sqrt(15))^n + (4sqrt(15))^n)/2.
a(n) = 8*a(n1)  a(n2) + 2*(1)^(n+1), n>=2, a(0)=0, a(1)=1.
a(n) = 7*a(n1) + 7*a(n2)  a(n3), n>=3, a(0)=0, a(1)=1, a(2)=6.
G.f.: x*(1x)/((1+x)*(18*x+x^2)) = x*(1x)/(17*x7*x^2+x^3) (from the Stephan link, see A092184).
From Peter Bala, Mar 25 2014: (Start)
a(2*n) = 6*A001090(n)^2; a(2*n+1) = A070997(n)^2.
a(n) = u(n)^2, where {u(n)} is the Lucas sequence in the quadratic integer ring Z[sqrt(6)] defined by the recurrence u(0) = 0, u(1) = 1, u(n) = sqrt(6)*u(n1)  u(n2) for n >= 2.
Equivalently, a(n) = U(n1,sqrt(6)/2)*U(n1,sqrt(6)/2), where U(n,x) denotes the Chebyshev polynomial of the second kind.
a(n) = 1/10*( (4 + sqrt(15))^n + (4  sqrt(15))^n  2*(1)^n ).
a(n) = the bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, 4; 1, 3] and T(n,x) denotes the Chebyshev polynomial of the first kind.
See the remarks in A100047 for the general connection between Chebyshev polynomials of the first kind and 4thorder linear divisibility sequences. (End)
