Also, a(n) = (sqrt(5)*phi)^n + (sqrt(5)/phi)^n, where phi = golden ratio. -
N. J. A. Sloane, Aug 08 2014
Let S(n, m)=sum(k=0, n, binomial(n, k)*fibonacci(m*k)), then for n>0 a(n)= S(2*n, 2)/S(n, 2). -
Benoit Cloitre, Oct 22 2003
a(n)= 5*a(n-1) - 5*a(n-2).
G.f.: (2-5*x)/(1-5*x+5*x^2). (End)
a((2*m+1)*k)/a(k) = Sum_{i=0..m-1} (-1)^(i*(k+1))*a(2*(m-i)*k) + 5^(m*k).
a(m+r)*a(n+s) + a(m+s)*a(n+r) = 2*a(m+n+r+s) + 5^(n+s)*a(m-n)*a(r-s).
a(n)^2 - a(n+1)*a(n-1) = -5^n.
a(n)^2 - a(n+r)*a(n-r) = -5^(n-r+1)*
A093131(r)^2.
a(m)*a(n+1) - a(m+1)*a(n) = -5^(n+1)*
A093131(m-n).
a(m+n) + 5^(n)*a(m-n) = a(m)*a(n).